Eric Chapdelaine
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MATH1342 Calculus 2 for Science and Engineering


The Class

5 quizzes (lowest one dropped)

2 midterms

Integration

There are two methods. Don’t think of them as rules.

There are infinitely many integrals that cannot be integrated. For example $\int \frac{e^x}{x} dx$ and $\int \frac{x}{\ln x} dx$.

Substitution

Example (#26):

$\int \sin ^{10} x \cos ^3 (x) dx$

$\int \sin^{10} x \cos ^2 (x) \cos (x) dx$

$\int \sin^{10} x \left( 1-\sin^2(x) \right) \cos (x) dx$

$u = \sin x$

$du = \cos x$

$\int u^{10} (1-u^2) du$

$\int u^{10} dx - \int u^{12} du$

Example:

$\int \frac{\ln x}{x} dx$

Let $z = \ln x$

Then $dz = \frac{1}{x} dx$

$\int z dz = \frac{z^2}{2} = \frac{1}{2} \ln (x)^2 + C$

NOTE: The difference between differential and derivative

$(x^2) \prime = 2x$

$\prime$ is the same thing as $\frac{d}{dx}$

Derivative: $\frac{d}{dx} \left( x^2 \right) = \frac{d (x^2)}{dx} = 2x$

Differential: $d(x^2) = 2x dx$

Example:

$\int \frac{dx}{x-1}$

Let $u = x-1$

$du = dx$

$\int \frac{du}{u} = \ln u + C = \ln (x-1) + C$

NOTE:

$d(x \pm C) = dx$

$d(2x) = 2 dx$

Example:

$\int \frac{dx}{2x+1}$

$\frac{1}{2} \int \frac{d(2x)}{2x+1}$

Example:

$\int \frac{dx}{\sqrt{2x+1} } dx$

Let $2x+1 = u$

$du = 2 dx \implies dx = \frac{1}{2} du$

$\frac{1}{2} \int \frac{du}{\sqrt{u} } = \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = \sqrt{2x + 1} + C$

Example:

$\int \frac{\sin^{-1} x}{\sqrt{1-x^2} } dx$

Let $\sin ^{-1} x = u$

$du = \frac{dx}{\sqrt{1+ x^2} }$

$\int d du = \frac{u^2}{2} = \frac{\sin ^{-1} x ^2}{2} + C$

How to Remember Derivative of Inverse Trig:

Sin:

$(\sin^{-1})\prime = ?$

Let $\sin^{-1} (x) = z$. So we need $z’$.

$x = \sin z$

$\frac{d}{dx (x) = \frac{d}{dx} (\sin z)}$

$1 = \cos (z) \times z’ \implies z’ = \frac{1}{\cos z} \implies \frac{1}{\sqrt{1-\sin ^2 (z)} } \implies \frac{1}{\sqrt{1-x^2}}$

Tangent:

$(\tan^{-1} x)’ = z \implies x = \tan z$

$\frac{d}{dx} x = \frac{d}{dx} \tan z \implies 1 = \sec^2 z * z’ \implies z’ = \frac{1}{\sec ^2 z}$

$= \frac{1}{1+\tan^2 z} = \frac{1}{1+x^2}$

Recall that $1 + \tan ^2 (x) = \sec ^2 (x)$

Example:

$\int \frac{tan^{-1} x}{1+x^2} dx$

Example:

$\int x^2 e^{3x^3} dx$

Let $t = 3x^3$

$dt = 9x^2 dx$

$x^2 dx = \frac{1}{9} dt$

$\frac{1}{9} \int e^t dt = \frac{1}{9} e^t = \frac{1}{9} e^{3x^2} + C$

By Parts

Definition: $(u v) ‘ = v * u’ + u * v’$

$\frac{d}{dx}(u v) = v \frac{d}{dx} u + u \frac{d}{dx} v$

$\frac{d(uv)}{dx} = v \frac{du}{dx} + u \frac{dv}{dx}$

$d(uv) = v du + u dv$

$\int d (uv) = \int v du + \int u dv$

$uv = \int v du + \int u dv$

$\int u dv = uv - \int v du$

What should be $u$ and $dv$?

L Logarithmic Functions
I Inverse Trigonometric Function
A Algebraic Functions
T Trigonometric Functions
E Exponential Functions

(Not a rule, but is helpful)

Example:

$\int \ln x dx$

$u = \ln x \implies du = \frac{dx}{x}$

$dv = dx \implies v =x$

$\int \ln x dx = x \ln x - \int x \frac{dx }{x} = x \ln x - \int dx = x \ln x - x + C$

Example:

$\int x \ln x dx$

$u = \ln x \implies du = \frac{dx}{x}$

$dv = x dx \implies v = \frac{x^2}{2}$

$= \frac{x^2}{2} \ln x - \frac{1}{2} \int x^2 \frac{dx}{x} = \frac{x^2}{2} \ln x - \frac{1}{2} \ln x dx = \frac{x^2}{2} \ln x - \frac{1}{2} \frac{x^2}{2} = \frac{x^2}{2} \left( \ln x - \frac{1}{2} \right) + C$

Example:

$\int x^2 e^{5x} dx$

$u = x^2 \implies du = 2x dx$

$dv = e^{5x} dx \implies v = \frac{1}{5} e^{5x}$

$\int x^2 e^{5x} dx = \frac{1}{5} x^2 e^{5x} - \frac{2}{5} \int x e^{5x} dx$

Second round:

$u = x \implies du = dx$

$dv = e^{5x} dx \implies v = \frac{1}{5} e^{5x}$

$= \frac{1}{5} x^2 e^{5x} - \frac{2}{5}\left( \frac{1}{5}x^{5x} - \frac{1}{5}\int e^{5x} dx \right)$

$= \frac{1}{5} x^2 e^{5x} - \frac{2}{25}x e^{5x} + \frac{2}{125} e^{5x}$

$= \frac{1}{5} e^{5x} \left( x^2 - 2x + \frac{2}{25} \right) + C$

Example:

$\int x \sin x dx$

$u = x \implies du = dx$

$ dv = \sin x dx \implies v = - \cos x$

$\int x \sin x dx = -x \cos x + \int \cos x dx$

$= - x \cos x + \sin x + C$

Important Trig Substitutions

$\sqrt{q^2 + x^2} $ (tan or cotan)

$\sqrt{q^2 - x^2} $ (sin or cos)

$\sqrt{x^2 - q^2} $ (sec and csc)

Always take out the $q$.

For example: $q \sqrt{1+ \left( \frac{x}{1} \right) ^2}$

First quiz on Wednesday.

Example:

$\int \frac{\ln x}{x} dx$

$u = \ln x \implies du = \frac{dx}{x}$

$dv =\frac{dx}{x} \implies v = \ln x$

$I = \int \frac{\ln x}{x} dx = \left( \ln x \right)^2 - \int \frac{\ln x}{x} dx$

Solve for I

$2I = \left( \ln x \right) ^{2}$

$\therefore I = \frac{1}{2} \left( \ln x \right) ^2 + C$

Example:

$\int e^x \sin x dx$

$u = e^{x} \implies du = d^x dx$

$dv = \sin x dx \implies v = -\cos x$

$\int e^x \sin x dx = -ex \cos x + \int e^x \cos x dx$

Lets do this again. When you do it again, stick to your choice of u and dv (otherwise, you will be left with the $\int e^x \sin x dx = \int e^x \sin x dx$).

$u = e^x \implies du = e^x dx$

$dv = \cos x dx \implies v = \sin x dx$

$= -e^x \cos x + \left( e^x \sin x - \int e^x \sin x dx \right)$

$= e^x \sin x - e^x \cos x - \int e^x \sin x dx$

$I = e^x \sin x - e^x \cos x - I$

$\therefore I = \frac{1}{2} e^x \left( \sin x - \cos x \right) + C$

Example (on your own):

$\int e^{2x} \cos \frac{x}{3} dx$

Example:

$\int \frac{\ln x}{x^2} dx$

We can do this both by subtitiution and by parts

By Parts:

$u = \ln x \implies du = \frac{dx}{x}$

$dv = \frac{dx}{x^3} \implies v= -\frac{1}{2x^2}$

$\int \frac{\ln x}{x^2} dx = -\frac{1}{2x^2} \ln x + \frac{1}{2} \int \frac{dx}{x^3}$

By Substitution:

$- \int \frac{\ln \frac{1}{x}}{x^2} dx$

Let $u = \frac{1}{x} \implies du = -\frac{dx}{x^2}$

$= - \int u \ln u du$

You can do the rest by parts

Example:

$\int x e^{x^2} dx$ By Substitution Example:

$\int x^2 e^x dx$ By parts Example:

$u = x^2 \implies du = 2x dx$

$dv = e^x dx \implies v = e^x$

Example: $\int x^2 e^{x^2} dx$ By Parts (do on your own)

Example: $\int t^n e^t dt$

$u = t^n \implies du = ut^{u-1}$

$dv = e^t dt \implies v = e^t$

$= t^n e^t - u \int t^{u-1} e^t dt$

This is valid for all $n \ge 0$

Example:

$\int \sin ^n x dx = \int \sin^{n-1} x \sin x dx$

$u = \sin^{n-1} x \implies du = \left( n-1 \right) \sin^{n-2} x * \cos x$

$dv = \sin x dx \implies v = - \cos x$

$=-\cos x \sin^{n-1} x + \left( n-1 \right) \int \cos ^2 x \sin ^{n-2} x dx$

$=-\cos x \sin^{n-1} x + \left( n-1 \right) \left( \int \sin ^{n-2} x dx - \int \sin^u x dx \right)$

$I =- \cos x \sin^{n-1} x + \left( u-1 \right) \int \sin^{n-2} x dx - n-1) - I$

$\left( 1+ \left( n-1 \right) \right) I = n I =-\cos x \sin^{n-1} x + \left( n+1 \right) \int \sin^{n-2} x dx$

$I = -\frac{1}{n} \cos x \sin^{n-1} x + \frac{n-1}{n} \int \sin ^{n-2} x dx$ for all $n \ge 1$

Let’s now do this the traditional way

Recall $\cos^2 + \sin^2 x = 1$ and $\cos^2 x - \sin^2 x = \cos 2x$

$\int \sin^2 x dx = \frac{1}{2} \int \left( 1-\cos(2x) \right) dx$

$\int \sin^2 x dx = \frac{1}{2} x -\frac{1}{4} \sin 2x + C$

Example:

$\int \frac{\cos x \sqrt{x} }{\sqrt{x} } dx$

$u = \sqrt{x}$

$du = \frac{1}{2\sqrt{x}}$

Example:

$\int \frac{\sin \sqrt{x}}{\sqrt{x} } dx$

Substitute $z= \sqrt{x} \implies dz = \frac{1}{2\sqrt{x} } dx$

$= 2 \int \sin z dz$

$= - 2 \cos \sqrt{x} + C$

Example:

$\int \frac{x\cos\sqrt{1-2x} }{\sqrt{1-x^2} } dx$

Integration by Partial Fraction

$\frac{P(x)}{Q(x)}$, where both $P(x)$ and $Q(x)$ are polynomials. Such that $\deg [Q(x)] > \deg [P(x)]$ ($\deg$ means the degree)

There are four distinct cases:

Case 1:

$Q(x)$ is a product of distinct linear factors

$Q(x) = (x+b_1)(x+b_2) … (x+b_u)$

Then, $\frac{P(x)}{Q(x)} = \frac{A_1}{x+b_1} + \frac{A_2}{x+b_2} + … + \frac{A_u}{x+b_u}$ where $A_1,\ A_2,\ …,\ A_u$ to be found.

Example:

$\frac{x^2 + 2x - 1}{x(2x-1)(x+2)} = \frac{A}{x} + \frac{B}{2x-1} + \frac{C}{x+2}$

$x^2 + 2x - 1 = A(2x-1)(x+2) + Bx (x+2) + Cx(2x-1)$

And solve for A, B, and C.

Case 2:

$Q(x)$ is a product of linear factor, some of which are repeated.

That is, $Q(x)$ has $x+b$ repeated $r$ times i.e. $\left( x+b \right) ^r$

Example:

$\frac{x^3 - x + 1}{(x+1)x^2 (x-1)^3} = \frac{A}{x+1}+ \frac{B}{x} + \frac{C}{x^2} + \frac{D}{x-1} + \frac{E}{(x-1)^2} + \frac{F}{(x-1)^3}$

Case 3:

$Q(x)$ contains irreducible quadratic factors (none of which are repeated). That is, $Q(x)$ has factors $ax^2+bx+c$, where $b^2-4ac <0$ (that is, we have no real roots).

Example:

$\frac{x^3+1}{(x-2)(x^2+1)(x^2+x+4)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{x^2+x+4}$

Case 4:

$Q(x)$ has some repeated irreducible quadratic factors. That is $(ax^2+bx+c)^r$.

Example:

$\frac{x^3+x^2+1}{x(x-1)(x^2+x+1)(x^2+1)^3} = \frac{A}{x} + \frac{B}{x-1} + \frac{Cx+D}{x^2+x+1} + \frac{Ex+F}{x^2 + 1} + \frac{Gx+H}{(x^2+1)^2} + \frac{Ix+J}{(x^2+1)^3}$

Example:

$\frac{1}{x^6 - x^3}= \frac{1}{x^3(x^3-1)} = \frac{1}{x^3(x-1)(x^2+x+1)}$

$= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-1} + \frac{Ex+F}{x^2+x+1}$

Example:

$\frac{1}{x+x^4} = \frac{1}{x(1-x^3)} = \frac{1}{x(1-x)(1+x+x^2)}$

$=\frac{A}{x} + \frac{B}{1-x} + \frac{Cx+D}{1+x+x^2}$

Example:

$\frac{x^2+1}{(x-1)^2(2x^2+1)(3x^2+2x+1)^3}$

$= \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{2x^2+1} + \frac{Ex+F}{3x^2+2x+1} + \frac{Gx+H}{(3x^2+2x+1)^2} + \frac{Ix+J}{(3x^2+2x+1)^3}$

$\frac{x^2-1}{(x-1)(x^2+2x+2)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2x+2}$

Full Solution Example (#68 on handout): Expect this kind of question on the final

$\int \frac{10}{(x+1)(x^2+4)} dx$

$\frac{1}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4}$

$10 = A(x^2+4) + (x+1)(Bx+C) = Ax^2 + 4A + Bx^2 + Cx + Bx + C$

$0 = A+B \implies A=-B$

$0=C+B \implies C+B=0$

$10 = 4A+C \implies C-4B=10$

$A=2$

$C=-B=2$

$B=-2$

$\int \frac{2}{x+1} dx + \int \frac{-2x+2}{x^2+4} dx$ (substitution)

$= 2 \ln (x+1) - 2 \left( \int \frac{x dx}{4+x^2}- \int \frac{dx}{4+x^2} \right)$

$= 2 \ln (x+1) - 2 \left( \int \frac{x dx}{4+x^2}- \frac{1}{4} \int \frac{dx}{1+\left( \frac{x}{2} \right) ^2} \right)$

$= 2 \ln (x+1) - \ln u + \tan^{-1} \left( \frac{x}{2} \right)$

$= 2 \ln (x+1) - \ln (4+x^2) + \tan ^{-1} \left( \frac{x}{2} \right) + C$

For another example, do #71 on handout.

Example:

$\int \frac{3x^2-2x+3}{x(x^2+1)}$

$\frac{3x^2-2x+3}{x(x^2+1) = \frac{A}{x}+ \frac{Bx+C}{x^2}}$

$3x^2 - 2x + 3 = Ax^2 + A + Bx^2 + Cx$

$3=A+B \implies B = 0$

$C=-2$

$A=3$

$= \int \frac{3}{x} dx + \int -\frac{2}{x^2+1} dx$

$= 3 \ln x - 2 \tan^{-1}(x) dx$

You don’t always have to jump to partial fractions, however. Here is an example:

$\int \frac{dx}{x^4-x} dx = \int \frac{dx}{x(x^3-1)}$

$= \int \frac{dx}{x(x-1)(x^2+x+1)} dx$

$\frac{1}{x(x-1)(x^2+x+1)} = \frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+x+1}$

$1 = A(x-1)(x^2-1)(x^2+x+1) + Bx(x^2+x+1) + (Cx+D)\left( x \right) (x-1)$

$1 = Ax^3 - A + Bx^3 + Bx^2 + Bx + (Cx^2 + Dx)(x-1)$

$1 = Ax^3 - A + Bx^3 + Bx^2 + Bx + Cx^3-Cx^2+Dx^2-Dx$

$0 = A+B+C \implies B+C=1$

$0 = B-C+D \implies B-C+D=0$

$0 = B-D \implies B=D$

$1 = -A \implies A=-1$

$B = \frac{1}{3}$

$D = \frac{1}{3}$

$C = \frac{2}{3}$

$= -\ln x + \frac{1}{3} \ln (x-1) + \int \frac{\frac{2}{3}x + \frac{1}{3}}{x^2+x+1} dx$

$= -\ln x + \frac{1}{3} \ln (x-1) + \frac{1}{3}\int \frac{2x + 1}{x^2+x+1} dx$

In this case, $2x+1$ cancels with substitution. However, if this isn’t the case, you would need to complete the square.

$= - \ln x + \frac{1}{3} \ln (x-1) + \frac{1}{3} \ln (x^2+x+1)$

$= \frac{1}{3} \ln (x^3 - 1) - \frac{1}{3} x^3$

$= \frac{1}{3} \ln \frac{x^3-1}{x^3}$

Example:

$\int \frac{dx}{x^4 - x}= \int \frac{dx}{x^4 \left( 1-\frac{1}{x^3} \right) }$

Let $z = 1- \frac{1}{x^3}$

$dz = \frac{3}{x^3} dx$

$= \frac{1}{3} \int \frac{dz}{z}$

$= \frac{1}{3} \ln \left( \frac{x^3-1}{x^3} \right)$

NOTE:

$\int \frac{dx}{ax^n \pm bx} = \int \frac{dx}{x^n \left( a+\frac{b}{x^{n-1}} \right)}$ for $n \ge 2$

Example:

$\int \frac{x+1}{(x-2)(x^2+3)} dx$

$\frac{x+1}{(x-2)(x^2+3)} = \frac{A}{x-1}+\frac{Bx+C}{x^2+3}$

$x+1 = Ax^2 + 3A + Bx^2 - Bx + Cx - C$

$0 = A+B \implies A=-B \implies A=\frac{1}{2},\ B=-\frac{1}{2}$

$1 = -B + C$

$1 = 3A- C \implies C = 1-A = \frac{1}{2}$

$I = \frac{1}{2} \int \frac{dx}{x-1} + \int \frac{-\frac{1}{2}x+\frac{1}{2}}{x^2+3} dx$

$I = \frac{1}{2} \ln (x-1) - \frac{1}{2} \int \frac{x-1}{x^2+3} dx$

$I = \frac{1}{2} \ln (x-1) - \frac{1}{2} \left( \int \frac{x}{3+x^2}dx - \int \frac{dx}{3+x^2} \right)$

$I = \frac{1}{2} \ln (x-1) - \frac{1}{2} \left( \int \frac{x}{3+x^2}dx - \frac{1}{3}\int \frac{dx}{1+\left( \frac{x}{\sqrt{3} } \right) ^2} \right)$

Now do substitution with both integrals ($u = x^2+3$ and $z = \frac{x}{\sqrt{3} }$ respectively)

$\frac{1}{2} \ln (x-1) - \frac{1}{4} \ln (3+x^2) + \frac{1}{2\sqrt{3} } \tan ^{-1} \left( \frac{x}{\sqrt{3} } \right) +C$

Example

$\int \frac{e^{\frac{1}{x}}}{x^3} dx$

$z = \frac{1}{x} \implies dz = -\frac{1}{x^2} dx$

$\int z e^z (-dz)$

$u = e^{\frac{1}{x}} \implies du = e^{\frac{1}{x}} \left( -\frac{1}{x^2} dx \right)$

$dv = \frac{dx}{x^3} \implies v = -\frac{1}{2} x^{-2}$

$-\frac{1}{2} x^{-2} e^{\frac{1}{x}} + \int \frac{1}{2}x^{-2} e^{\frac{1}{x}} \left( \frac{1}{x^2} dx\right)$

DNF

Improper Integrals

$\int_{1}^{3} \frac{dx}{x} = \ln x \mid _{-1}^{3}$

We cannot do this because you cannot do a log of a negative number.

This makes sense because we cannot divide by zero (in the integral).

$\int_{0}^{\infty} e^x dx$

Type 1: Infinite Integrals

Consider infinite region under $g=\frac{1}{x^2}$ above x-axis and to the right from the line $x$.

$\int_{1}^{\infty} \frac{1}{x^2}dx$

To evaluate this integral, we use the following:

$\int_{1}^{\infty} \frac{Ax}{x^2} = \lim_{t \to \infty} \int_{1}^{t} \frac{dx}{x^2}$

$= \lim_{t \to \infty} \left( -\frac{1}{x} \right) \mid _{1}^{t}$

$=\lim_{t \to \infty} - \frac{1}{t} + \frac{1}{1} = 1$

This converges to 1.

General:

  1. If $\int_{a}^{b} f(x) dx$ exists for every number $b\ge a$, then $\int_{a}^{\infty} f(x) dx = \lim_{b \to \infty} \int_{a}^{b} f(x) dx$ goes to $\infty$ provided that the limit exists as a finite number.
  2. If $\int_{a}^{b} f(x) dx$ exists for every number $a\le b$, then $\int_{-\infty}^{b} f(x) dx = \lim_{a \to -\infty} \int_{a}^{b} f(x) dx$ provided that the limit exists as a finite number.

The improper integral $\int_{a}^{\infty} f(x) dx$ and $\int_{-\infty}^{b} f(x) dx$ are called convergent if the corresponding limits exists as finite number and divergent if the limit does not exist.

$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{a} f(x) dx + \int_{a}^{\infty} f(x) dx$

(Usually $a=0$) which brings two cases (a) and (b). If any of them are divergent, then $\int_{-\infty}^{\infty} f(x) dx$ is divergent.

Example:

$\int_{0}^{\infty} \tan^{-1} x dx$

Zero is okay, but $\infty$ is not.

Example:

Determine the convergence and divergence

$\int_{1}^{\infty} \frac{dx}{x}$ this is divergent (but any power greater than 1 is convergent)

$\lim_{t \to \infty} \int_{1}^{t} \frac{dx}{x}$

$= \lim_{t \to \infty} \ln t - \ln 1 = \infty$ diverges

Example:

$\int_{-\infty}^{0} xe^x dx = \lim_{a \to -\infty} \int_{a}^{0} xe^x dx$

$\int xe^x dx$

$u = x \implies du = dx$

$dv = e^x dx \implies v=e^x$

$\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x$

$= \lim_{a \to -\infty} \left( xe^x - e^x \right) \mid _{a}^{0}$

$= \lim_{a \to \infty} \left( -1 - \left( ae^a - e^a \right) \right)$

$= \lim_{a \to \infty} [e^a - ae^a - 1]$

L’Hopital’s Rule:

$= \lim_{a \to \infty} [e^a - \frac{a}{e^{-a}} - 1]$

$= \lim_{a \to \infty} [e^a - \frac{1}{e^{-a} \times -1} - 1]$

$= \lim_{a \to \infty} [0 - 0 - 1] = -1$ converges

Example:

$\int_{-\infty}^{\infty} \frac{1}{1+x^2}dx$

$\int_{-\infty}^{0} \frac{1}{1+x^2} dx + \int_{0}^{\infty} \frac{1}{1+x^2} dx$

$= \lim_{t \to -\infty} \tan^{-1} x \mid {t}^{0} + \lim{z \to \infty} \tan^{-1} x \mid_{0}^{z}$

$= \frac{\pi}{2} + \frac{\pi}{2} = \pi$

NOTE:

Odd functions have the property $f(-x) = -f(x)$

$\sin x$ is odd

$\cos x$ is even

$\tan x$ is off

Example:

For what values of $b$ is the integral $\int_{1}^{\infty} \frac{dx}{x^p}$?

Recall that $\int_{1}^{\infty} \frac{dx}{x^2} = 1$

And that $\int_{1}^{\infty} \frac{dx}{x} = \infty$ so it diverges.

So now we have to investigate $p>1$ and $p<1$. Lets consider the case when $p>1$.

$\int_{1}^{\infty} \frac{dx}{x^p} = \lim_{b \to \infty} \int_{1}^{b} \frac{dx}{x^p} = \lim_{b \to \infty} \frac{x^{-p+1}}{-p+1} \mid _{1}^{b}$

$= \frac{1}{1-p} \lim_{b \to \infty} x^{1-p} \mid _{1}^{b}$

$= \frac{1}{1-p} \lim_{b \to \infty} \left( \frac{1}{b^{p-1}}-1 \right)$

Now consider when $p>1$. In that case, $p-1>0$ and $b^{p-1}$ goes to $\infty$ if $b$ goes to $\infty$ and therefore converges to $\frac{1}{p-1}$.

Now, what happens when $p<1$? In that case, $\frac{1}{b^{p-1}}$ goes to $\infty$ as $b$ goes to $\infty$ and the integral diverges.

To Summarize:

$\int_{1}^{\infty} \frac{dx}{x^p}$ is convergent if $p>1$ and divergent if $p\le 1$.

Type 2: Discontinuous Integrands

Let $f(x)$ be continuous positive on $[a,\ b]$ (i.e has disc).

At $x=b$ let $S$ be the unbounded region. The shaded area between $a$ and $t$ is $A(t) = \int_{a}^{t} f(x) dx$

If $A(t)$ approaches some definite number $A$ as $t$ goes to $b^{-}$ (approaches from the negative), then the area of the region $S$ is $A$ and write:

$\int_{a}^{b} f(x) dx = \lim_{t \to b^{-}} \int_{a}^{t} f(x) dx$

Case a:

$f(x)$ is continuous on $[a,\ b]$ and discontinuous at $a=b$, then $\int_{a}^{b} f(x) dx = \lim_{t \to b^{-}} \int_{a}^{t} f(x) dx$

Case b:

$f(x)$ is continuous on $[a,\ b]$ and discontinuous at $x=a$.

$\int_{a}^{b} f(x) dx = \lim_{t \to a^{+}} \int_{t}^{b} f(x) dx$

Case c:

$f(x)$ is continuous on $[a,\ b]$ and discontinuous at $x=c$, where $a<c<b$.

$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$

$= \lim_{t \to c^{-}} \int_{a}^{t} f(x) dx + \lim_{t \to c^+} \int_{t}^{b} f(x) dx$

Example:

Find $\int_{2}^{5} \frac{dx}{x-2} = \lim_{t \to 2^+} \int_{t}^{5} \frac{dx}{\sqrt{x-2}}$

$\lim_{t \to 2^+} \int_{t}^{5} \frac{d(x-2)}{\sqrt{x-2}} = \lim_{t \to 2^+} \frac{(x-2)^{\frac{1}{2}}}{\frac{1}{2}} \mid _{t}^{5}$

$= 2 \lim_{t \to 2^+} \left( 3^{\frac{1}{2}} - (t-2)^{\frac{1}{2}} \right) = 2\sqrt{3}$ converges!


$\int_{0}^{\frac{\pi}{2}} \sec x dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x} dx$

$= \lim_{t \to \frac{\pi}{2}} \int_{0}^{t} \frac{dx}{\cos x}$

$= \lim_{t \to \frac{\pi}{2}} \ln \left( \frac{1}{\cos x}+ \tan x \right) \mid _{0}^t$

$= \lim_{t \to \left( \frac{\pi}{2} \right) ^-} \left( \ln \left( \frac{1}{\cos t} + \frac{\sin t}{\cos t} \right) - \ln 1 \right)$

$= \infty$ therefore it diverges

$\int \frac{dx}{\cos x} = \int \frac{\cos x dx}{\cos ^2 x} = \int \frac{\cos x dx}{1 - \sin x}$

Let $z = \sin x \implies dz = \cos x dx$

$= \int \frac{dz}{1-z^2} = \int \frac{dz}{(1-z)(1+z)} = \frac{1}{2} \int \left( \frac{1}{1-z}+ \frac{1}{1+z} \right) dz$

$= \frac{1}{2} \left( \ln (1+z) - \ln (1-z) \right) = \frac{1}{2} \ln \left( \frac{1+z}{1-z} \right)$

$\frac{1}{2} \ln \left( \frac{1+\sin x}{1-\sin x} \right)$

$= \frac{1}{2} \ln \left( \frac{\left( 1+\sin x \right) ^2}{\cos ^2 x} \right)$

$= \ln \left( \sec x + \tan x \right)$

Example:

$\int_{0}^{3} \frac{dx}{x-1} = \int_{0}^{1} \frac{dx}{x-1} + \int_{1}^{3} \frac{dx}{x-1}$

Doing the first part:

$\int_{0}^{1} \frac{dx}{x-1} = \lim_{t \to 1^-} \int_{0}^{t} \frac{dx}{x-1}$

$= \lim_{t \to 1^-} \ln (x-1) \mid {0}^{t} = \lim{t \to 1^-} \left( \ln \mid t-1 \mid - \ln \mid -1 \mid \right) = - \infty$ Diverges!

Don’t need to check the other one, because if one diverges, the entire integral diverges.

So $\int_{0}^{3} \frac{dx}{x-1}$ diverges!

Be careful! Just solving the integral doesn’t work. You have to handle improper integral.

Example:

$\int_{0}^{1} \ln x dx = \lim_{z \to 0^+} \int_{z}^{1} \ln x dx$ (can only do this integral by parts)

Recall that $\int \ln x dx = \int 2 \ln x - x$

$= \lim_{z \to 0^+} \left( x \ln x - x \right) \mid {z}^1 = \lim{z \to 0^+} \left( \left( 0-1 \right) - \left( z \ln z - z \right) \right)$

$= \lim_{z \to 0^+} \left( 0-z \ln z - 1 \right)$ This is $o \times \infty$ so we need to use L’Hopital’s rule

$z \ln z = \frac{\ln z}{\left( \frac{1}{z} \right) } =L’= \frac{\frac{1}{z}}{\frac{-1}{z^2}} = -z$

$= -1$

The geometric representation of the result asymptotically goes to -1.

Example:

$\int_{1}^{\infty} \frac{dy}{\left( 4y-3 \right)^{\frac{1}{2}}}$

Let $z = 4y-3 \implies dz = 4 dy$

$\frac{1}{4} \int_{1}^{\infty} \frac{dz}{\sqrt{z}} = \frac{1}{2} \sqrt{z} = \frac{1}{2} \sqrt{4y-3}$

$= \lim_{t \to \infty} \frac{1}{2} \sqrt{4y-3} \mid {1}^{t} = \frac{1}{2} \lim{t \to \infty} \left( \sqrt{4t+3} -1 \right) = \infty$ Diverges!

Example:

$\int_{0}^{1} \frac{dt}{\sqrt{1-t ^{2}}}= \lim_{z \to 1^-} \int_{0}^{z} \frac{dt}{\sqrt{1-t^2}}$

$= \lim_{z \to 1^-} \sin^{-1} t \mid {0}^z = \lim{z \to 1^-} \left( \sin^{-1} z - 0 \right)$

$\lim_{z \to 1^{-1}} \sin ^{-1} z = \frac{\pi}{2}$ Converges

$t= \sin \alpha \implies dt \cos \alpha d \alpha$

$\int \frac{\cos \alpha d \alpha}{\sqrt{1- \sin^2 \alpha}} = \int \frac{\cos \alpha d\alpha}{\cos \alpha} = \sin ^{-1} t$

What if we say $t = \cos \alpha$?

$\int \frac{-\sin \alpha d\alpha}{\sqrt{1-\cos^2 \alpha}} = - \int \frac{\sin ^2 \alpha d \alpha}{\sin \alpha} = - \int d \alpha = -\alpha = -\cos ^{-1} t + C$

$\sin ^{-1} t + \cos ^{-1} t = \frac{\pi}{2}$

$\sin ^{-1} t = \frac{\pi}{2} - \cos ^{-1} t$

Next quiz tomorrow:

Numerical Integration

Definition:

What is a Riemann Sum? It is the definition of the definite integral:

$\int_{a}^{b} f(x) dx = \lim_{h \to \infty} \sum_{i=1}^{h} f(x_i) \Delta x$

$\Delta x = \frac{b-a}{h}$

$x_i = \frac{\text{length } i }{\text{divisions}}$

This can be any number (i.e. positive, negative, and zero). It is not the area.

Trapezoidal Rule

n - number of rectangles

$\int_{a}^{b} f(x) dx = ?$ Numerical Approximation of this Definite Integral

$T_n = \frac{R_n + L_n}{2}$ This is the trapezoidal rule.

Definition:

$T_n = \frac{\Delta x}{2}\left( f(x_1) + 2f(x_2) + 2f(x_3) + … + 2f(x_{n}) + f(x_{n+1}) \right)$

This means that if $n=4$ then you will have 5 terms.

For k number of rectangles:

$L_k = \Delta x \left( f(x_1) + f(x_2) + … + f(x_{k-2}) + f(x_{k-1}) \right)$

$R_k = \Delta x \left( f(x_2) + f(x_3) + … + f(x_{k-2}) + f(x_{k-1}) + f(x_k) \right)$

Example:

$\int_{0}^{2} e^x dx = e^x \mid _{0}^2 = e^2 - e^0 = e^2 - 1 \approx 6.389056$

For $n=4$:

$\Delta x = \frac{b-a}{n} = \frac{2-0}{4} = \frac{1}{2}$

$T_4 = \frac{\Delta x}{2} \left( f(x_1) + 2f(x_2)+2f(x_3) + 2f(x_4) + f(x_5) \right)$

$= \frac{1}{4} \left( f(0) + 2f(\frac{1}{2}) + 2f(1) + 2f\left( \frac{3}{2} \right) +f(2) \right)$

$= \frac{1}{4} \left( 1+2e^{0.5} + 2e + 2e^{1.5} + e^2 \right)$

$\approx 6.52161$

Simpsons Rule

Definition:

$\int_{a}^{b} f(x) dx \approx \frac{\Delta x}{3} \left( f(x_1) + 4f(x_2)+2f(x_3)+4f(x_4)+2f(x_5)+…+2f(x_{n-1}) + 4f(x_{n}) + f(x_{n+1})\right)$

  • First and last terms have a coefficient of 1
  • Second to last is always 4
  • The rest of the middle terms alternate between 4 and 2.
  • n has to be even

For $n=2$:

$\int_{a}^{b} f(x) dx \approx \frac{\Delta x}{3}\left( f(x_1)+4f(x_2)+f(x_3) \right)$

For $n=4$:

$\int_{a}^{b} f(x) dx \approx \frac{\Delta x}{3} \left( f(x_1)+4f(x_2)+2f(x_3)+4f(x_4)+f(x_5) \right)$

For $n=6$

$\int_{a}^{b} f(x) dx \approx \frac{\Delta x}{3} \left( f(x_1) + 4f(x_2) + 2f(x_3) + 4f(x_4) + 2f(x_5) + 4f(x_6) + f(x_7) \right)$

Example:

$\int_{0}^{2} e^x dx = e^x \mid _{0}^2 = e^2 - e^0 = e^2 - 1 \approx 6.389056$

For $n=4$:

$S_4 = \frac{\Delta x}{3}\left( f(x_1) + 4f(x_2) + 2f(x_3) + 4f(x_4) + f(x_5) \right)$

$\frac{1}{6} \left( f(0) + 4f(0.5) + 2f(1) + 4f(1.5) + f(2) \right)$

$= \frac{1}{6} \left( 1+4e^{0.5} + 2e + 4e^{1.5}+e^2 \right)$

$\approx 6.39121$ (much better than the trapezoidal approximation)

Any polynomial of a third degree gives the exact integral.

$\int_{1}^{3} \left( x^3 + x-1 \right) dx = \left( \frac{x^4}{4}+\frac{x^{2}}{2}-x \right) \mid _{1}^3 = 22$

$\Delta x = \frac{1}{2}$

$S_4 = \frac{\Delta x}{3} \left( f(x_1) + 4f(x_2) + 2f(x_3) + 4f(x_4) + f(x_5) \right)$

$S_4 = \frac{1}{6} \left( f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + f(3) \right) = 22$

Example:

$\int_{0}^{1} \frac{dy}{1+y^2} = \frac{\pi}{4}$

Try with $n=6$.

Rectilinear Motion

Vectors: have both magnitude and direction

Scalers: just have magnitude

Distance is always $= \int_{a}^{b} \mid \vec{v}(t) \mid dt = \int_{a}^{b} \sqrt{f_1^2 (t) + f _2^2(t) + f _3^2 (t)} dx$

Direction is a scaler and displacement is a vector.

Consider a spring that does not deform. If you pull it down and let it oscillate, eventually it will reach the starting point and stop. At that time, the displacement is 0, but the distance is not.

$S = \int_{a}^{b} v(t) dt$

We need to solve for $v(t) = 0$. Say $t=c$ is the solution.

On the time interval, there is some instance $t=c$, when the object has $v=0$, i.e. returns (change its direction). Therefore, we write:

$\text{Total Distance} = \mid \int_{a}^{c} v(t) dt \mid + \mid \int_{c}^{b} v(t) dt \mid$

One of these is going to be negative. So we need to pick a point on $a \le t<c$ ($p_1$) and $c<t\le b$ ($p_2$).

If $v(p_1) > 0$, then it moves in positive

If $v(p_2) < 0$, then it moves in negative.

(Or you can just take the absolute value of both)

Example:

$v(t) = \cos t e^{\sin t}$ from $t=0$ to $t=\pi$

Find displacement:

$\text{Displacement} = \int_{0}^{\pi} \cos t e^{\sin t} dt = e^{\sin t} \mid _{0}^{\pi}$

$\int e^{u}du = e^{u}$

$= e^{\sin \pi} - e^{\sin 0} = e^{-1} - 1 = \frac{1}{e} - 1$

Displacement is always less than or equal to distance.

Distance: $= \mid \int_{0}^{\frac{\pi}{2}} \cos t e^{\sin t} dt \mid + \mid \int_{\frac{\pi}{2}}^{\pi} \cos t e^{\sin t} dt \mid$

$= \mid e^{\sin t} \mid _{0}^{\frac{\pi}{2}} \mid + \mid e^{\sin t} \mid _{\frac{\pi}{2}}^{\pi} \mid$

$= \mid e^{\sin \frac{\pi}{2}} - e^{ \sin 0} \mid + \mid e^{\sin \pi} - e ^{\sin \frac{\pi}{2}} \mid$

$= e- 1- 1 + e = 2e - 2 = 2(e-1)$

Example:

$v(t) = \ln \frac{t}{2}$ on $[1,\ 4]$

$v(t) = \ln \frac{t}{2} = 0 \implies \ln t - \ln 2 = 0$

$\ln t = \ln 2 \implies t = 2$

$\text{Total Distance} = \mid \int_{0}^{2} \left( \ln \frac{t}{2} \right) dt \mid + \mid \int_{2}^{4} \ln \left( \frac{t}{2} \right) dt \mid$

$= \mid \left( t \ln t - t \ln 2\right) _{0}^{2} \mid + \mid \left( t \ln t - t - t \ln 2 \right) _{2}^{4} \mid$

$= \mid (-2 - 2) - (-1 - \ln 2) \mid + \mid (-4) - (-2) \mid$

$= \mid -2 + 1 + \ln 2 \mid + \mid -2 - 4 \ln 2 \mid$

$= \mid \ln 2 - 1 \mid + \mid 4 \ln 2 - 2 \mid = 1 - \ln 2 + 4 \ln 2 - 2$

$= \ln 8 - 1 \approx 1.08$

$\text{Displacement} = \int_{1}^{4} v(t) dt$

Parametrization

Distance is Space

The notion of vector-valued function.

Recall that $\int x dt = xt + C$ and $\int \left( ax^{2}+bx+c da \right) = \frac{a^2}{2}x^2 + (bx+c)a + C$

Two vectors are the same if their magnitude and direction are the same.

Assume we have a function $\vec{p}(t) = (x(t),\ y(t))$ or $\vec{p}(t) = (x(t),\ y(t),\ z(t))$

Then functions $x(t)$ and $y(t)$, and $z(t)$ are components of vector $\vec{p}(t)$. A function $\vec{p}(t)$ continuous and is differentiable if each each component is continuous and differentiable. In that case, $\vec{p}^{\prime}(t) = (x^{\prime}(t),\ y^{\prime}(t),\ z^{\prime}(t))$.

Suppose the position or position vector of the object is given at time $t$.

As: $\vec{p}(t) = (x(t),\ y(t))$

Then, the velocity, which is always tangent to the curve alone which the object is moving, will be:

$\vec{v}(t) = \vec{p}^{\prime}(t) = (x^{\prime}(t),\ y^{\prime}(t))$

If $\vec{v}(t)$ of the object is continuous on the interval $[a,\ b]$, then the displacement of the object between the times $t=a$ and $t=b$ is given by:

$\vec{p}(b) - \vec{p}(a) = \int_{a}^{b} \vec{v}(t) dt$

$\frac{d\vec{p}(t)}{dt} = \vec{v}(t) \implies \int d \vec{p}(t)$

$\vec{p}(b) - \vec{p}(a) = \vec{p}(t) \mid {a}^{b} = \int{a}^{b} \vec{v}(t) dt$

Example:

A particle velocity is given by $\vec{v}(t) = \left( e^{t},\ \sqrt{2},\ -e^{-t} \right) \frac{m}{s}$. What is the displacement of the particle between $t=0$ and $t=1$ second?

$\vec{p}(1) - \vec{p}(0) = \text{Displacement} = \int_{0}^{1} \vec{v}(t)dt$

$\int_{0}^{1} \left( e^{t},\ \sqrt{2},\ -e^{-t} \right) dt = \left( e^{t},\ \sqrt{2}t,\ e^{-t} \right) \mid _{0}^{1}$

$= (e,\ \sqrt{2},\ e^{-1}) - (1,\ 0,\ 1) = (e-1,\ \sqrt{2},\ e^{-1}-1)m$

The magnitude of the displacement.

Definition of Absolute Value:

$\mid A \mid = \sqrt{A^2}$

Therefore, the magnitude of the displacement is $\mid \text{Displacement vector} \mid = \mid (e-1,\ \sqrt{2},\ e^{-1}-1) \mid = \sqrt{(e-1)^{2}+(\sqrt{2})^{2} + \left( e^{-1}-1 \right)^2 }$

Distance:

If $\vec{v}(t)$ is continuous on $[a,\ b]$, then the distance traveled by the object by the object between $t=a$ and $t=b$ is given by:

$\text{Distance} = \int_{a}^{b} \mid \vec{v}(t) \mid dt$

Distance from previous example:

$\int_{0}^{1} \mid \vec{v}(t) dt = \int_{1}^{1} \mid \left( e^{t},\ \sqrt{2},\ -e^{-t} \right) \mid dt$

$= \int_{0}^{1} \sqrt{\left( e^{t} \right) ^{2} + (\sqrt{2})^2 + \left( -e^{-t} \right) ^2 } dt$

$= \int_{0}^{1} \sqrt{e^{2t} + 2 + e^{-2t}} dt$

$= \int_{0}^{1} \sqrt{\left( e^{t}+e^{-t} \right) ^{2}} dt$

$= \int_{0}^{1} \left( e^{t}+e^{-t} \right) dt$

$= \left( e^{t} - e^{-t} \right) \mid _{0}^{1} = \left( e-e^{-1} \right) - (1-1) = e-e^{-1} m$

Note: since $e+e^{-1}-3 > 0$, the distance is always greater than the magnitude of the displacement.

Displacement: $\sqrt{\left( e-e^{-1} \right) ^{2} - 2\left( e+e^{-1}-3 \right) }$

Distance: $e-e^{-1}$

Example:

$\vec{v}(t) = (3 \sin(2\pi t),\ 3 \cos(2\pi t),\ 4t)$ on $[0,\ 2]$

Find the displacement.

$\text{Displacement} = \int_{0}^{2} \vec{v}(t) dt = \int_{0}^{2} (3 \sin (2\pi t),\ 3 \cos (2\pi t),\ 4t) dt$

$= \left( \frac{3}{2\pi} (-\cos (2\pi t)),\ \frac{3}{2\pi} \sin (2\pi t),\ 2t^2 \right) \mid _{0}^{2}$

$= \left( -\frac{3}{2\pi},\ 0,\ 8 \right) - \left( -\frac{3}{2\pi},\ 0,\ 0 \right)$

$= (0,\ 0,\ 8)$

Magnitude of displacement:

$\sqrt{0^{2}+0^{2}+8^2 } = 8$

Find the distance:

$\text{Distance} = \int_{0}^{2} \mid \vec{v}(t) \mid dt = \int_{0}^{2} \mid \left( 3\sin(2\pi t),\ 3 \cos (2\pi t),\ 4t \right) dt$

$= \int_{0}^{2} \sqrt{9\sin ^2(2\pi t) + 9 \cos ^2 (2\pi t) + 16 t^2} dt$

$= \int_{0}^{2} \sqrt{9+16t^2} dt$

$= 3 \int_{0}^{2} \sqrt{1+\left( \frac{4t}{3} \right) ^{2}} dt$

Let $z=\frac{4t}{3}$

$dz = \frac{4}{3}dt \implies dt = \frac{3}{4}dz$

$= \frac{3}{4} \int \sqrt{1+z^{2}} dz$

$= \frac{3}{4} (z \frac{\sqrt{z^{2}+1} }{2} + \frac{1}{2} \ln (z+\sqrt{z^2+1}))$

$= \frac{3}{8} \left( z \sqrt{z^{2}+1} + \ln(z+\sqrt{z^2+1} \right)$

$= \frac{9}{8} \left( \frac{4t}{3}\sqrt{\frac{16t^2}{9}+1} + \ln \left( \frac{4t}{3}+ \sqrt{\frac{16t^2}{9}+1} \right) \right) \mid _{0}^{2} \approx 10.46$

Example:

$\vec{p}(t) = \left( \sin (3t),\ \cos (3t),\ 2t ^{\frac{3}{2}} \right)$ on $[0,\ \pi]$

Displacement: $\vec{p}(\pi) - \vec{p}(0)$

Distance: $\int_{0}^{\pi} \mid \vec{v}(t) \mid dt$

$\vec{v}(t) = \vec{p}^{\prime}(t) = \left( 3 \cos (3t),\ -3 \sin 3t,\ 3t ^{\frac{1}{2}} \right)$

$= \int_{0}^{\pi} \mid \left( 3 \cos (3t),\ -3 \sin(3t),\ 3 t ^{(\frac{1}{2})} \right) dt$

$= \int_{0}^{\pi} \sqrt{9\cos ^{2}(3t) + 9 \sin ^{2} (3t) + 9t} dt$

$= \int_{0}^{\pi} \sqrt{9+9t} dt$

$= 3 \int_{0}^{\pi} \sqrt{1+t}dt$

$= 3 \frac{(1+t)^{\frac{3}{2}}}{\left( \frac{3}{2} \right) } \mid _{0}^{\pi} = 2 (1+t) ^{\left( \frac{3}{2} \right) } \mid _{0}^{\pi}$

$= 2 \left( \left( 1+\pi \right) ^{\left( \frac{3}{2} \right) }- 1 \right)$

Example:

$\vec{v}(t) = (3+4t,\ 5-7t,\ 11t)$ on the interval $[-3,\ 7]$

Displacement: $\int_{-3}^{7} \vec{v}(t) dt$

Distance: $\int_{-3}^{7} \mid (3+4t,\ 5-7t,\ 11t) \mid dt$

$= \int_{-3}^{7} \sqrt{(3+4t)^{2}+(5-7t)^2+ 2t ^{2}}dt$

Example:

$\vec{p}(t) = (-t,\ t,\ 3)$ on interval $[0,\ \frac{\pi}{4}]$


Next Monday is the last quiz before the midterm.


Arc Length

Inscribed polygon in the graph above

We define the length L of the curve C given by equation $y=f(x)$ on $a\le x\le b$ as the limit of the inscribed polygon (if the limit exists).

$L = \lim_{n \to \infty} \sum_{i=1}^{n} \mid P_{i-1},\ P_i \mid$ (The difference of the two points)

$\Delta x$ is always the same. If we let $\Delta y_i = y_i - y_{i-1}$ then $\mid P_{i-1},\ P_i \mid = \sqrt{\Delta x^{2} + \Delta y_i ^{2}} = \sqrt{\left( x_i - x_{i-1} \right) ^{2} + \left( y_i - y_{i-1} \right) ^{2}}$.

Mean Value Theorem:

$m_2 = m_1$

$m_2 = \frac{f(x_i) - f(x_{i-1})}{x_i-x_{i-1}} = m_1 = f ^{\prime}(x_{i}^*)$

Using the mean value theorem on interval $[x_{i-1},\ x_i]$, there is always an number $x_i^\star$ between $x_{i-1}$ and $x_i$ such that $f(x_i) - f(x_{i-1}) = f^{\prime}\left( x_{i}^* \right)(x_i - x_{i-1})$.

$\Delta y_i = f^{\prime}(x_i^*)$, therefore we have:

$\mid P_{i-i},\ P_i \mid = \sqrt{(\Delta x)^2 + (\Delta x)^2 f ^{\prime}(x_i^\star)^2} = \sqrt{1 +f^{\prime}(x_i^*)^2} \Delta x$

Then $L = \lim_{n \to \infty} \sum_{i=1}^{n} \mid P_{i-1},\ P_i \mid$

$= \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{1+f^{\prime}(x_i^*)^2} \Delta x = \int_{a}^{b} \sqrt{1+\left( f^{\prime}(x) \right) ^{2}} dx$

Riemann Sum:

$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x$ Say that $\sin ^2 x = 1 + \left( y^{\prime} \right) ^{2}$

$y^{\prime} = \sin ^2 x - 1$

$y = \int \left( \sin ^2 - 1 \right) dx$

Example:

Find the arc length of the curve $y = x^2 - \frac{\ln (x)}{8}$ for $1\le x\le 2$

$L = \int_{1}^{3} \sqrt{1+ (y^{\prime})^2} dx$

$y^{\prime} = 2x - \frac{1}{8x} \implies \left( y^{\prime} \right)^{2} = \left( 2x - \frac{1}{8x} \right)^{2} = 4x^{2}-\frac{1}{2}+\frac{1}{64x^2}$

$\left( 2x + \frac{1}{8x} \right) ^{2} = 1 + 4x^2 - \frac{1}{2} + \frac{1}{64x^2} = 4x^2 + \frac{1}{2} + \frac{1}{64x^2}$

$= \int_{1}^{3} \sqrt{\left( 2x+\frac{1}{8x} \right) ^{2}} dx = \int_{1}^{3} \left( 2x+\frac{1}{8x} \right) dx$

$= \left( x^{2} + \frac{1}{8}\ln x \right) \mid _{1}^{3} = 8 + \frac{\ln 3}{8} \approx 8.1373$

Example:

Find L if $x(y) = \frac{e^y + e^{-y}}{2} = \cosh (y)$ for $0\le y\le \ln 3$ $L = \int_{0}^{\ln 3} \sqrt{1+ \left( x’^2 \right) } dy$

$x^{\prime} = \frac{e^2-e^{-y}}{2} \implies (x’)^2 = \left( \frac{e^y-e^{-y}}{2} \right) ^{2}$

$1+(x’)^2 = 1 + \frac{e^{2y}-2+e^{-2y}}{4}$

$= \frac{e^{2y}+2+e^{-2y}}{4} = \left( \frac{e^{y}+e^{-y}}{2} \right) ^{2}$

$L = \int_{0}^{\ln 3} \sqrt{\left( \frac{e^y+e^{-y}}{2} \right) ^{2}} dy = \frac{4}{3}$

Example:

Length of arc defined by $f(x)= \frac{x^3}{12}+\frac{1}{x}$ on the interval $[1,\ 2]$. This equals $\frac{13}{12}$

Example:

Length of arc defined by $g(x)=\ln (\sin x)$ on the interval $[\frac{\pi}{4},\ \frac{\pi}{2}]$.

$L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 + \left( \left( \ln (\sin x) \right) ‘ \right) ^{2}}$

$\left( \ln \sin x \right)’ = \frac{\cos x}{\sin x} = \cot x$

$(\ln \sin x)’ ^{2} = \cot ^2 x$

$1+\cot ^2 x = \csc ^2 x$

$L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc x dx$

$= \int \frac{dx}{\sin x} = \int \frac{\sin x}{\sin ^2 x} dx = \int \frac{\sin x}{1- \cos x} dx$

$z = \cos x$

$dz = - \sin x dx$

$= - \int \frac{dz}{1-z^2} = - \int \frac{dx}{(1-z)(1+z)}$

$= \ln (\csc x - \cot x)$

Summations:

$\sum_{i=1}^{n} C = Cn$

$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

$\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}$

Example: $\int_{1}^{3} x^{3} dx$

$\Delta x = \frac{b-a}{n}$

Left-hand side:

$x_i = 1+\frac{2i}{n}$

Volumes of Revolution

Postulates:

Do the following integral by using $\int_{1}^{3} … dy$ : $y=x^2+1$

Disk Method

Consider a pile of CD disks.

$V= \lim_{n \to \infty} \sum_{i=1}^{n} A_i(x) dx = \int_{a}^{b} A(x) dx$

Definition:

Along the x axis: $V^\text{Disk} = \int_{a}^{b} A(x) dx$

Along the y axis: $V^\text{Disk} = \int_{a}^{b} A(y) dy$

Example:

Find the volume of the shape obtained by rotating $y=\sqrt{x}$ across the x axis from from $x=0$ to $x=1$.

$V^{\text{Disk}} = \int_{0}^{1} A(x) dx$

$A(x) = \pi R^{2}$

$A(x) = \pi \left( \sqrt{x} \right) ^{2} = \pi x$

$V = \int_{0}^{1} \pi x dx = \pi \frac{x^2}{2} \mid _{0}^{1} = \frac{\pi}{2}$

Example

Find $v$, rotating $y=x^2$, $y=8$, $x=0$ above the y-axis.

$A(y) = \pi R^{2} = \pi x^2$

$x = ^{3} \sqrt{y} = y^{\frac{1}{3}}$

$V = \int_{0}^{8} A(y) dy = \pi \int_{0}^{8} y^{\frac{2}{3}}dy$

$= \pi \frac{y^{\frac{5}{3}}}{\frac{5}{3}} \mid _{0}^{8}$

$= 96 \frac{\pi}{5}$

Washer

Example:

$y=x,\ y=x^2$ about x-axis.

$y_1 = y_2 \implies x=x^2 \implies x_0 = 0,\ x=1$

$V^{\text{Disk}} = V^{\text{Washer}} = \int_{0}^{1} A(x) dx$

$R_o = y_1 = x$

$R_i = y_2 = x^2$

$A(x) = \pi R _o^2 - \pi R _i^2 = \pi \left( R _o^2 - r _i^2 \right)$

$= \pi \left( x^2 - x^4 \right)$

$V = \int_{0}^{1} \left( x^2 - x^4 \right) dx = \pi \left( \frac{x^3}{3} - \frac{x^5}{5} \right) \mid _{0}^{1}$

$= \pi \left( \left( \frac{1}{3} - \frac{1}{5} \right) -0 \right) = \frac{2\pi}{15}$

Example:

$y=x,\ y=x^2$ rotated around $y=2$

$V^{\text{Washer}} = \int_{0}^{1} A(x) dx$

$A(x) \pi \left( R _o^2 - r _i^2 \right)$

NOTE:

In positive direction, it will always be axis minus the variable. i.e. $2-x$.

In the negative direction, it will be variable minus the axis.

$R_o = 2- y_1 = 2- x^2$

$R_i = 2-y_2 = 2 - x$

$V^{\text{Disk}} = \pi \int_{0}^{1} \left( 2-x^2 \right)^2 - \left( 2-x \right)^2 dx = \frac{8\pi}{15}$

Example (same one, but rotated differently):

$y=x,\ y=x^2$ rotated around $x=-1$

$V^{\text{Disk}} = \int_{0}^{1} A(y) dy$

$R_o = x_1 - (-1) = \sqrt{y} +1$

$R_i = x_2 - (-1)= y +1$

$\pi \int_{0}^{1} \left( \sqrt{y} +1 \right) ^{2} - \left( y+1 \right) ^{2} dy = \frac{\pi}{2}$

Example (same, rotated differently):

$y=x,\ y=x^2$ rotating around $x=3$

$V^{\text{Disk}} = \int_{0}^{1} A(y) dy$

$R_o = \text{axis} - \text{var} = 3 - x_1 = 3-y$

$R_i = \text{axis} - \text{var} = 3 - x_2 = 3-\sqrt{y}$

Shell Method

Different methods of volumes of revolution.

$V_i = A(x_i) h = A(x_i) f(x)$

$V_i - 2\pi x_i \Delta x f(x_i)$

$V = \lim_{n \to \infty} \sum_{i=1}^{n} 2\pi x_i \Delta x f(x_i)$

$\int_{a}^{b} 2\pi x f(x) dx = \int_{a}^{b} 2\pi R h dx$

Example:

$V^{\text{Shell}} = \int_{a}^{b} 2\pi R h dx$

$V^{\text{Shell}} = \int_{a}^{b} 2\pi f(x) dx$

Or, if we are going around the y axis:

$V^{\text{Shell}} = \int_{a}^{b} 2 \pi R h dy$

Example:

Find the volume obtained by rotating about y-axis, the region bounded by $y=2x^2 - x^3$ and $y=0$ (the x-axis). Note, you cannot get the inverse of that function

The red is the elemental shell.

$y =0 = 2x^2 - x^3$

$x^2 (2-x) = 0$

$\therefore x = 0,\ x=2$

$V^{\text{Shell}} = \int_{0}^{2} 2\pi R h = dx$

$h = y = 2x^2 - x^3$

$V^{\text{Shell}} = \int_{0}^{2} 2\pi x\left( 2x^2 - x^3 \right) dx = \frac{16\pi}{5}$

Can we do this using the disk method?

You can use disk method with this, however, it is overly complicated.

Example:

$y=x$ and $y=x^2$ about y-axis

$V^{\text{Shell}} = \int_{0}^{1} 2 \pi R h dx$

$h(x) = y^{\text{top}} - y^{\text{bottom}}$

$V^{\text{Shell}} = \int_{0}^{1} 2 \pi x (x-x^2) dx$

$= 2\pi \int_{0}^{1} (x^2 - x^3) dx = 2\pi \left( \frac{x^3}{3}-\frac{x^4}{4} \right) \mid _{0}^{1}$

Now let’s use disk method:

$V^{\text{Disk}} = \int_{0}^{1} A(y) dy$

$A(y) = \pi \left( R _{o}^2 - R _{i}^2 \right)$

$R_o = \sqrt{y}$

$R_i = y$

$= \int_{0}^{1} \pi \left( y-y^2 \right) dy$

$= \frac{\pi}{6}$

Example:

$y=x$ and $y=x^2$ about the x-axis

Disk method:

$V^{\text{Disk}} = \int_{0}^{1} A(x) dx$

$A(x) = \pi \left( R _o^{2} - R _i^2 \right)$

$V^{\text{Disk}} = \int_{0}^{1} \pi \left( x^2 - x^4 \right)$

$= \pi \left( \frac{x^3}{3}- \frac{x^5}{5} \right) \mid _{0}^{1} = \frac{2\pi}{15}$

Shell method:

$y=x \implies x^{\text{Left}} = y$

$y=x^2 \implies x^{\text{Right}} = \sqrt{y}$

$V^{\text{Shell}} = \int_{0}^{1} 2 \pi R h dy$

$R = y$

$h(y) = x^{\text{Right}} - x^{\text{Left}} = \sqrt{y} -x$

$V^{\text{Shell}} = \int_{0}^{1} 2 \pi y\left( \sqrt{y} -y \right) dy$

$= 2\pi \int_{0}^{1} \left( y^{\frac{3}{2}} - y^{2} \right) dy$

$= 2\pi \left( \frac{y^{5}2}{\frac{5}{2}} - \frac{y^3}{3} \right) \mid _{0}^{1}$

$= \frac{2\pi}{15}$

Example:

$y=x$ and $y=x^2$ about $y=2$

$V^{\text{Shell}} = \int_{0}^{1} 2 \pi R h dy$

$R = \left( 2-y \right)$

$h(y) = \sqrt{y} - y$

$= 2\pi \int_0^1 \left( 2-y \right) \left( \sqrt{y} -y \right) dy$

$= 2\pi \int_{0}^{1} \left( 2\sqrt{y} -2y-y^{\frac{3}{2}}+y^{2} \right) dy$

$= \frac{8\pi}{15}$

Example:

$y = \sin x$ about y-axis on $[0,\ \pi]$

$V^{\text{Shell}} = \int_{0}^{\pi} 2\pi R h dx$

$R = x$

$h= \sin x$

$= 2\pi \int_{0}^{\pi} x \sin x dx = 2\pi^{2}$

Disk method

$V^{\text{Disk}} = V_1 + V_2$

$V_1 = \int_{0}^{1} A(y) dy$

$A(y) = \pi \left( R _o^2 - R _i^2 \right)$

$R_o = \frac{\pi}{2}$

$R_i = \sin ^{-1} y$

$A(y) = \pi \left( \left( \frac{\pi}{2} \right) ^{2} - \left( \sin ^{-1} y \right) ^{2} \right)$

$V = \pi \int_{0}^{1} \left( \left( \frac{\pi}{2} \right) ^{2} - \left( \sin ^{-1}y \right) ^2 \right) dy + V_2$

$V_2 = \int_{0}^{1} A(y) dy$

$A(y) = \pi \left( R _o^2 - R _i ^2 \right)$

$= \pi \int_{0}^{1} \left( \left( \pi - \sin ^{-1}y \right) ^{2} - \left( \frac{\pi}{2} \right) ^{2} \right)dy$

$V^{\text{Disk}} = \pi \int_{0}^{1} - \left( \sin ^{-1}y \right) ^{2} + \pi^2 - 2 \pi \sin ^{-1} y + \left( \sin ^{-1} \right) ^{2} dy$

$= \pi \int_{0}^{1} \pi^{2} - 2\pi \sin ^{-1}y dy$

$= \pi \left( \pi ^2 y - 2 \pi y \sin ^{-1} y - 2\pi \sqrt{1-y^2} \right) \mid _{0}^1$

Mass and Density

Area and density. If we have a 2D region on an xy-plane, we will assume that the density is constant in each cross-sectional line segment perpendicular to x- or y-axis. For each fixed x, it is $\delta_{\text{area}}\left( x \right)$.

If we know the function $h(x)$, which is the height of the cross-sectional line segment in the region. Then the infinitesimal area of the cross-section (or infinitesimal thin rectangle) is $dA = h(x) dx$ and infinitesimal mass is $\delta = \frac{m}{v}$.

$\delta _{\text{Area}} = \frac{m}{A}$

$d m = \delta _{\text{Area}} (x) dA$

Total mass $M = \int_{a}^{b} dm = \int _{a}^{b} \delta _{\text{Area}}(x) dA = \int _{a}^{b} \delta _{\text{Area}}(x) h(x) dx$.

Example:

A thin plate occupies region in xy-plane below the line $y=2-x$.

Let the area-density be $\delta _{\text{Area}}(y) = e^{-y} \frac{\text{kg}}{\text{m}^{2}}$

Find m.

$m = \int_{0}^{2} \delta _{\text{Area}} (y) dA = \int _{0}^{2} e^{-y}h(y) dy$

$h(y) = x^R - x^L = \left( 2-y \right) - 0 = 2 - y$

$m = \int_{0}^{2} e^{-y}(2-y) dy$

$m = -2 (e^{-2}-1) - \int_{0}^{2} y e^{-y} dy = e^{-2}+1 \text{kg}$

Example:

$y=2x-6$

$\delta _{\text{Area}}(x) = x^2 + 1$

$m = \int_{0}^{3} \delta _{\text{Area}}h(x) dx$

$h(x) = y^T - y^B$

$m= \int_{0}^{3} \left( x^2 + 1 \right) \left( 0 - \left( 2x-6 \right) \right) dx = \frac{45}{2}$

Example:

$\delta _{\text{Area}} (y) = y^{\frac{3}{2}} + 2$

$y= \sqrt{x} \implies x = y^2$

$x=4$

$m = \int_{0}^{2} \delta _{\text{Area}}(y) h(y) dy$

$h(y) = x^R - x^L = 4 - y^2$

$m = \int_{0}^{2} \left( y^{\frac{3}{2}} + 2\right) \left( 4-y^2 \right) dy$

$m \approx 14.689 \text{ kg}$

Example:

$y = \sqrt{x}$

$y = x^2$

$\delta _{\text{Area}} (y) = y^{2} + 1$

$m = \int_{0}^{1} \left( y^{2}+1 \right) \left( \sqrt{y} -y^2 \right) dy$

Moments and Center of Mass

“Give me a lever long enough and a fulcrum on which to place it, and I shall move the world.” - Archimedes

Archimedes experimentally discovered that rod will balance it.

$m_1d_1 = m_2d_2$

Now suppose that rod lied along the x-axis.

$d_1 = \bar{x} - x_1$

$d_2 = x_2 - \bar{x}$

$m_1d_1 = m_2 d_2 \implies m_1 \left( \bar{x} - x_1 \right) = m_2 \left( x_2 - \bar{x} \right)$

$= m_1 \bar{x} - m_1 x_1 = m_2 x_2 - m_2 \bar{x}$

$= \bar{x} \left( m_1 + m_2 \right) = m_1 x_1 + m_2 x_2$

$= \bar{x} = \frac{m_1x_1 + m_2 x_2}{m_1 + m_2}$

Or with the third point:

$m_1 (\bar{x} - x_1) = m_2 \left( x_2 - \bar{x} \right) + m_3 \left( x_3 - \bar{x} \right)$

$m_1 \bar{x} - m_1 x_1 = m_2 x_2 - m_2 \bar{x} + m_3 x_3 - m_3 \bar{x}$

$\bar{x} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$

Where $m_1 x_1$ and $m_2x_2$ are called the moments of the masses $m_1$ and $m_2$ with respect to the origin. The equation suggests that the center of mass $\bar{x}$ is obtained by adding the moments and dividing by the total mass $m = m_1 + m_2 + …$.

To generalize for a system of $n$ particles with masses $m_1,\ m_2,\ …,\ m_n$ located at the points $x_1,\ x_2,\ …,\ x_n$ on the x-axis it can similarly be shown that the center of mass of the system is given by

$\bar{x} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} = \frac{M_x}{M}$ where

$M = \sum_{i=1}^{n} m_i$ is the total mass of the system and

$M_x = \sum_{i=1}^{n} m_i x_i$ is the sum of the moments of the system.

Suppose we have a 1D object, laid along the x-axis between $x=a$ and $x=b$ with a continuous density $\delta _l (x)$. Then the x-component of the moment is

$M_x = \int_{a}^{b} x dm$

$M_x = \lim_{n \to \infty} \sum_{i=1}^{n} x_i m_i = \lim_{n \to \infty} \sum_{i=1}^{n} x_i \Delta m_i = \int_{a}^{b} x dm$

$dm = \delta_{L}(x) dx$

$dm = \delta _{\text{Area}} (x) dA = \delta _{\text{Area}} (x) h(x) dx$

$= \int_{a}^{b} x \delta_L (x) dx$

$M = \int_{a}^{b} dm$

$= \int_{a}^{b} \delta _{l}(x) dx$

$\bar{x} = \frac{M _x}{M} = \frac{\int _{a}^{b} x \delta _L (x) dx }{\int _{a}^{b} \delta _L (x) dx }$

We can only work with 2D because 3D requires multi-variable calculus. However, if the shape is symmetrical, we can in Calculus 2.

Consider a flat plate:

Consider plate with density $\delta _{\text{Area}}(x)$. We wish to locate center of mass of this plate. If we assume we have a function $f(x)$ which gives height of the cross section of the object, then we know that infinitesimally wide vertical slab has a mass given by:

$d m = \delta _{\text{Area}} (x) dA = \delta _{\text{Area}}(x) h(x) dx = \delta _{\text{Area}}(x) f(x) dx$

In addition, the contribution of this slab of mass to the $x$-component of the moment will be $x dm = x \delta _{\text{Area}}(x) f(x) dx$ and the x-component of the moment is:

$M_x = \int_{x=a}^{x=b} x dm = \int_{a}^{b} x \delta _{\text{Area}}(x) f(x) dx$

The total mass of the object is $M = \int_{a}^{b} d m = \int_{a}^{b} \delta _{\text{Area}}(x) f(x) dx$

If we have a plate on the xy plane, lying between lines $x=a$ and $x=b$ under graph $y=f(x)$ and above graph $y=g(x)$

$f(x) \ge g(x)$ on $[a,\ b]$

Then the total mass is

$M = \int _{a}^{b} \delta _{\text{Area}}(x) h(x) dx$

$dA = h(x) dx$

$M = \int_{a}^{b} \delta _{\text{Area}}(x) \left( f(x) - g(x) \right) dx$

The x-component is:

$M_x = \int_{a}^{b} x \delta _{\text{Area}}(x) \left( f(x) - g(x) \right) dx$

$\bar{x} = \frac{\int _{a}^{b} x\delta _{\text{Area}}(x) \left( f(x) - g(x) \right) dx}{\int _{a}^{b} \delta _{\text{Area}}(x) \left( f(x) - g(x) \right) dx }$

As far as y-component of the moment, we have:

$M_y = \int_{a}^{b} \frac{f(x) + g(x)}{2} \delta_{\text{Area}}(x) \left( f(x) - g(x) \right) dx$

$M = \int_{a}^{b} \delta _{\text{Area}}(x) \left( f(x) - g(x) \right) dx$

Example:

The region is bounded by $y=x^2$, above $y=1$, and $x=2$.

Find the center of mass of the plate if $\delta _{\text{Area}}(x) = x$.

$M = \int \delta _{\text{Area}}(x) dA$

$dA = h(x) dx$

$dA = \left( y^T - y^B\right) dx = \left( x^{2}-1 \right) dx$

$M = \int_{0}^{2} x \left( x^2 - 1 \right) dx = \frac{9}{4}$

$M_x = \int_{1}^{2} x \delta _{\text{Area}}(x) \left( x^{2}-1 \right) dx$

$M_x = \int_{1}^{2} x x \left( x^2 - 1 \right) dx = \frac{58}{15}$

$M_y = \int_{1}^{2} \frac{x^2 + 1}{2} \delta _{\text{Area}}(x) \left( x^2 - 1 \right) dx$

$M_y = \int_{1}^{2} \frac{x^4-1}{2}x dx = \frac{9}{2}$

Thus, $\bar{x} = \frac{M_x}{M} = \frac{\frac{58}{15}}{\frac{9}{4}} \approx 1.7185$

$\bar{y} = \frac{M_y}{M} = \frac{\frac{9}{2}}{\frac{9}{4}} = 2$

For the exam, you only need to setup the integral, you don’t need to solve it.

Example:

$f(x) = x$, $g(x) = x^{2}$, $\delta _{\text{Area}}(x) = 2x + 1$

$M = \int _{0}^{1} \delta _{\text{Area}}(x) h(x) dx = \int _{0}^{1} (2x+1) \left( f(x) - g(x) \right) dx$

$M = \int_{0}^{1} (2x+1)(x^2-x) dx = \frac{1}{3}$

$M_x = \int _{0}^{1} x \delta _{\text{Area}}(x) h(x) dx = \int _{0}^{1} x (2x+1) \left( x^{2} - x \right) dx = \frac{11}{60}$

$M_y = \int_{0}^{1} \frac{f(x) + g(x)}{2} \delta _{\text{Area}}(x) \left( f(x) - g(x) \right) dx$

$M_y = \int_{0}^{1} \frac{x+x^2}{2}(2x+1) \left( x-x^2 \right) dx = \frac{3}{20}$

$\overline{x} = \frac{M_x}{M} = \frac{\int_{0}^{1} x(2x+1)(x-x^2) dx}{\int_{0}^{1} (2x+1)(x-x^2) dx}$ (This is all that’s required in the exam).

$\overline{y} = \frac{M_y}{M} = \frac{\int_{0}^{1} \frac{x+x^2}{2} (2x+1)\left( x-x^2 \right) dx }{\int_{0}^{1} (2x+1)\left( x-x^2 \right) dx }$

Example:

Find the center of mass for $f(x) = 3x+5$, $g(x) = x-3$, $x=-4$ (redundant), $\delta _{\text{Area}}(x) = x^{2}+3$.

$M_x = \int_{-4}^{3} x(x^2+3)((3x+5)-(x+3)) dx$

$M_x = \int_{-4}^{3} x\left( x^{2}+3 \right) \left( 2x+8 \right) dx$

$M = \int_{-4}^{3} \left( x^{2}+3 \right) \left( 2x+8 \right) dx$

$M_y = \int_{-4}^{4} \frac{4x+2}{2} \left( x^{2}+3 \right) \left( 2x+8 \right) dx$

$\overline{x} = \frac{M_x}{M} = \frac{\int_{-4}^{3} x \left( x^2+3 \right) (2x+8) dx}{\int_{-4}^{3} \left( x^2+3 \right) \left( 2x+8 \right) dx}$

$\overline{y} = \frac{\int_{-4}^{3} \left( 2x+1 \right) \left( x^2 + 3 \right) \left( 2x+8 \right) dx }{\int_{-4}^{3} \left( x^2+3 \right) \left( 2x+8 \right) dx}$

Work and Energy

We will focus on work done by emptying fluid.

Definition:

Work: (In Newton Meters or Joules)

$W = FS$

Or

$W = F \Delta x$

If object acted by Force $F$ is displaced from $x_0$ to $x_1$, then work done if $W = \int_{x_0}^{x_1} F dx$

Consider a spring force:

Hooke’s Law:

$F_{\text{Spring}} = -kx$ (Case of stretching)

$F _\text{Compression} = kx$ (Case of compression)

Example:

What is the work done by emptying fluid from this container:

You have water filling up this container up to 10 feet.

$\delta _{\text{Water}} = 62.4 \frac{\text{lb}}{\text{ft}^{2}}$

Find the work ($W$) done to empty this container.

$dV = A(t) dz = 8 (20) dz = 160 dz$

$dm = \delta dV$

$dF = g dm = g \delta _{\text{Water}} dv$

$dF = g \delta _{\text{Water}} 160 dz$

$W = dF \times \text{Distance} = \left( 30-z \right) dF$

$W = \int_{0}^{10} \left( 30-z \right) dF = \int_{0}^{10} \left( 30-z \right) g \delta _{\text{Water}} 160 dz$

$W = \int_{0}^{10} \left( 30-z \right) g (62.4) (160) dz$ (All you need to get to on the exam)

Example:

$\delta _{\text{Water}} = 62.4 \frac{\text{lb}}{\text{ft}^2}$

Consider the cone below:

$dV = A(y) dy$

$A(y) = \pi R(y)^2$

Consider a cross-section of the cone:

$\frac{R}{y} = \frac{7}{35}$ (because they are similar triangles)

$\therefore R = \frac{7y}{35}$

$A(y) = \pi \left( \frac{7y}{35} \right)^{2}$

Hence $dV = \pi \left( \frac{y}{5} \right) ^{2} dy$

$dF = g dm \implies dF = g \delta _{\text{Water}} dV$

$dF = g (62.4) \pi \left( \frac{y}{5} \right)^{2} dy$

$w = \int_{0}^{20} \left(35-y \right) dF$

$w = \int_{0}^{20} \left( 35-y \right) g (62.4) \pi \left( \frac{y}{5} \right)^{2} dy$


Exam 1:


Exam 1 Review

Rotate $y=x^2 +3$, $y=1$, $x=0$, $x=3$ about $x=-2$ using the shell method.

Approximating Polynomials

Linearization:

Linearization (or first order approximation) represents the approximation of your function by a linear function.

Consider this first-order approximation on a second-order function:

Now we can focus the approximation on a certain point $x=a$:

Linear approximation: $L(x) = f(a) + f^{\prime}(a)(x-a)$

Example

Approximate $y=\ln x$ near $x=e$

$L(x) = f(e) + f’(e)(x-e) = 1 + \frac{1}{e}(x-e) = \frac{x}{e}$

What about near $x=3$?

$L(3) = \frac{3}{e} \approx 1.103$

$\ln 3 \approx 1.098$

$\ln 2.9 \approx 1.064710$

$L(2.9) = \frac{2.9}{e} \approx 1.0668$

Second-order approximations:

$L(x) = f(a) + f’(a)(x-a) + \frac{f’‘(a)}{2!}(x-a)^2$

Third-order approximations:

$L(x) = f(a) + f’(a)(x-a) + \frac{f’‘(a)}{2!}(x-a)^2 + \frac{f’‘(a)}{3!}(x-a)^3$

A general polynomial function, in power of $x$ is defined as follows:

$p(x) = c_0 + c_1 x + c_2 x^2 + … + c_d x^d = \sum_{k=0}^{d} c_k x^k$

Where $d$ is the degree of polynomial.

When $x$ is close to zero, a polynomial $p(x)$ can be approximated “well” by leaving off (or dropping) the terms of the form $x^k$, where $k$ is large.

This polynomial in the above formula was centered at a or about a where $a=0$. Sometimes we want to have polynomial centered at any other number. That is, we want our original polynomial to be written in powers of $(x-a)$. Then we have out definition:

A polynomial centered at $a$ is a polynomial written in terms of powers $(x-a)$ in the following form:

$q(x) = b_0 + b_1 (x-a) + b_2 (x-a) + … + b_d (x-a)^d = \sum_{k=0}^{d} b_k (x-a)^k$

Example:

$p(x) = 4 - x + 5x^2 + 9x^5$

When $x$ is close to 0, the 0 degree term, 4, is the most important, $-2x$ is the second most important, $5x^2$ is the third important, and $9x^5$ is the least important if $x$ is close to zero.

Let’s assume the notation that $p^n(x)$ means the sum of all of the first $n$ terms of the polynomial. Hence, on the previous example:

$P^0(x) = 4$

$P^1(x) = 4-2x$

$p^2(x)=4-2x+5x^2$

$p^3(x)=4-2x+5x^2$

$p^4(x)=4-2x+5x^2$

$p^5(x)=4-2x+5x^2+9x^5$

Example:

Again, $p(x) = 4-2x+5x^2+9x^5$.

If $x$ is close to zero, then what power of $x$ best approximates $\frac{1}{5}\left( p(x) - r + 2x \right)$?

Solution:

We find that $\frac{1}{5}\left( p(x) -4+2x \right) =x^2+\frac{9}{5}x^5$

2nd power best approximates it because $\frac{9}{5}x^5$ is very small.

Example:

$p(x) = 1-x+2x^2-3x^3+4x^4$

If $x$ is close to zero, what power of $x$ best approximates $\frac{1}{3} \left( p(x) - 1 + x - 2x^2 \right)$?

Follows that $\frac{1}{3}\left( p(x)-1+x-2x^2 \right) =-x^3 + \frac{4}{3}x^4$

Therefore, the 3rd power of x best approximates this.

Example:

Let $p(x) = 4-2x+5x^2+9x^5$ and we want to approximate $p(x)$ by another polynomial $p(x)$, which is centered at $a=1$, namely we want to find $p(x) = b_0 + b_1(x-1) + b_2(x-1)^2 + b_3(x-1)^2+b_4(x-1)^4 + b_5(x-1)^5$. That is, we want to find $b_0,\ b_1,\ b_2,\ b_3,\ b_4,\ b_5$.

To find constants $b_i$, we use the approach:

Solution:

$p(1) = b_0 \implies q(1) = 4 - 2 + 5 + 9 = 16$

So $b_0 = 16 = \frac{p^0(1)}{0!}$

$p’(x) = b_1 + 2b_2 (x-1) + 3b_3 (x-1)^2 + 4b_4(x-1)^3 + 5b_5 (x-1)^4$

$p’(1) = b_1 \implies p’(x) = -2 + 10x+45x^4 \implies p’(1) = -2 + 10 +45 = 53$

So $b_1 = 53$ and $b_1 = \frac{p’(1)}{1!}$

$p’‘(x) = 2b_2 + 2(3) b_3 (x-1) + 3(4) (x-1)^2 + 4 (5) b_5 (x-1)^3$

$p’‘(1) = 2b_2$

So $b_2 = \frac{p’‘(1)}{2} = \frac{p’‘(1)}{2!} = 190$

$p’‘(x) = 10 + 180x^3$

$p’‘(1) = 10 + 180 + 190$

$\therefore 2b_2 = 190 \implies b_2 = 95$

$p’’‘(x) = 2(3)b_3 + 2(3)(4)(x-1) + 3(4)(5)(x-1)^2$

$p’’‘(1) = 2(3)b_3$

$b_3 = \frac{p’’‘(1)}{3!}$

$p’’‘(x) = 540x^2$

$p’’‘(1) = 540$

$6b_3 = 540 \implies b_3 = \frac{540}{6} = 90$

$p^{(4)}(x) = p’’’‘(x) = 2(3)(4) + 2(3)(4)(5)(x-1)$

$p^{(4)}(1) = 2(3)(4)b_4$

$b_4 = \frac{p^{(4)}(1)}{4!}$

$p^4(x) = 1080x$

$p^{(4)}(x) = 1080$

$4! b_4 = 1080 \implies b_4 = \frac{1080}{4!} = 45$

$p^{(5)}(x) = 5! b_5$

$p^{(5)}(1) = 5! b_5$

$b_5 = \frac{p^{(5)}(1)}{5!}$

$p^{(5)}(x) = 1080$

$b_5 = \frac{1080}{5!} = 9$

$p(x) = 16 + 53(x-1)+95(x-1)^2 + 95(x-1)^3 + 45(x-1)^4 + 9(x-1)^5$

$p(x) = 4-2x+5x^2+9x^5$

Note that these are the exact same at $x=1$ and $x=-1$.

To Generalize:

Suppose that $p(x)$ is a polynomial of degree $d$, and that $a$ is a fixed number. Then there is a unique polynomial of degree $d$, $q(x) = \sum_{k=0}^{d} b_k(x-a)^k$ centered at $a$, such that there is an equality of functions $p(x) = q(x)$ and the coefficients of this polynomial are given by $b_k = \frac{p^{(k)}(a)}{k!}$ ($k$ derivative).

Example:

$p(x) = (x-1)^3$, find $q(x)$ centered at $a=0$.

$q(x) = b_0 + b_1 x + b_2x^2 + b_3x^3$

$b_k = \frac{p^{(k)}(a)}{k!}$

$b_0 = \frac{p^{(0)}(0)}{0!} = -1$

$b_1 = \frac{p^{(1)}(0)}{1!} = \frac{3}{1!} = 3$

$b_2 = \frac{p^{(2)}(0)}{2!} = \frac{-6}{2!} = -3$

$b_3 = \frac{p^{(3)}(0)}{3!} = \frac{6}{3!} = 1$

$q(x) = -1 + 3x - 3x^2 + x^3$

Example:

$y=x^2$ centered around $a=9$

$q(x) = c_0 + c_1(x-9) + c_2(x-9)^2$

DNF

Example:

$q(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + … + a_n x^n$

$q^{n}(x)$ is the partial form of the $n$-th order approximation

i.e $q^{2}(x) = a_0 + a_1 x + a_2 x^2$

$q^{0}(x) = a_0$

General form is $q(x) = a_0 + a_1(x-a) + a_2 (x-a)^2 + … a_n (x-a)^n$

$b_k = \frac{p^{(k)}(a)}{k!}$

Remember that $p^{(k)}$ means the $k$-th derivative of $p$.

Error

Error is always absolute value (always positive)

Error in approximating $q(x)$ by it’s $n$-th order partial sum $q^{n}(x)$. The difference between them is called the remainder and defined $R^{n}(x)$ so that $R^{n}(x) = q(x) - q^{n}(x)$

The error $E^{n}(x)$ is $E^{n}(x) = \mid R^{n}(x) \mid = \mid q(x) - q^{n}(x) \mid$

Example:

$q(x) = 7+3(x+2) - 5(x+2)^4$

What is the max error that we have in approximating $q(x)$ by $q^1(x)$ by first-order partial sum?

$q^{1}(x) = 7 + 3(x+2)$

$E^{1}(x) = \mid q(x) - q^{1}(x) \mid = \mid - 5 (x+2)^4 \mid$

$E^{1}(x) = 5 \mid x+2 \mid ^{4}$

If we suppose that $x$ is within 0.1 of -2, then $\mid x+2 \mid \le 0.1$. Hence $E^{1}(x) = 5 \mid x+2 \mid ^{4} \le 5.0 (0.1)^{4} = 0.0005$

Example:

$q(x) = x^2 + 5x + 6$

Rewrite $q(x)$ centered at $a=-2$

$q(x) = b_0 + b_1 (x+2) + b_2 (x+2)^2$

$b_n = \frac{q^{(n)}(a)}{n!} = \frac{q^{(n)}(-2)}{n!}$

$b_0 = \frac{q^{(0)}(-2)}{0!} = \frac{(-2)^2+5(-2)+6}{0!} = \frac{4-10+6}{0!} = 0$

$b_1 = \frac{q^{(1)}(-2)}{1!} = \frac{2x+5}{1!} = \frac{1}{1!} = 1$

$b_2 = \frac{q^{(2)}(-2)}{2!} = \frac{2}{2!} = 1$

$\therefore q(x) = (x+2) + (x+2)^2 = x^2 + 5x +6$

Example:

$f(x) = x^3 - 3x^2 + 3x - 1$ centered at $x=1$

$a_0 = \frac{f(1)}{0!} = \frac{0}{0!} = 0$

$a_1 = \frac{f^{(1)}(1)}{1!} = \frac{0}{1!} = 1$

$a_2 = \frac{f^{(2)}(1)}{2!} = \frac{1}{2!} = 1$

$f(x) = (x-1)^3$

Example:

$q(x) = 1-x+2x^2 - 3x^3 + 4x^4$ centered at $a=1$

$q(x) = \sum_{k=0}^{\infty} b_k (x-1)^k = b_0 + b_1 (x-1) + b_2 (x-1)^2 + b_3 (x-1)^3 + b_4 (x-1)^4$

Do this work. You should get:

$q(x) = 3+10(x-1) + 17(x-1)^2 + 13 (x-1)^3 + 4(x-1)^4$

Example:

Approximate $q(x) = x^2 + 5x + 6$ at $x=-2.1$ by using the re-centered 1st order partial sum.

Recall from above that the re-centered polynomial is $q(x) = (x+2) + (x+2)^2$

$q^{1}(x) = x+2$ first-order partial sum

So $q^{1}(x) = -2.1 + 2 = -0.1$

$q(-2.1) = -0.09$

$E^{1}(x) = \mid q(x) - q^{1}(x) \mid = \mid x^{2} \mid$

$E^{1}(-2.1) = \mid -0.09 - (-0.1) \mid = 0.01$

Definition:

Suppose $f(x)$ is continuous and $n$-times differentiable function at $a$. Then the polynomial $T _{f} ^{n} \left( x ; a \right) = \sum _{k=0}^{n} \frac{f^{(k)}(a)}{k!}\left( x-a \right)^{k} = \frac{f(a)}{0!} + \frac{f’(a)}{1!}(x-a) + \frac{f’‘(a)}{2!}(x-a)^2 = \frac{f’’‘(a)}{3!} (x-a)^3 + … + \frac{f^{(n)}(a)}{n!} (x-a)^n$

This is called the $n$-th order Taylor polynomial of $f(x)$ centered at $a$. When $a=0$, we simply use $T _{f} ^{n}(x)$ in place of $T _{f}^{n} (x,\ 0)$ and refer to it:

$T _{f} ^{n}(x) = \frac{f(0)}{0!} + \frac{f’(0)}{1!}x + \frac{f’‘(0)}{2!}x^2 + \frac{f’’‘(0)}{3!}x^3 + … + \frac{f^{(n)}(0)}{n!}x^4$

This is referred to as a Maclaurin polynomial of $f$.

For $n$ big enough, we expect that $T _{f}^{n}(x;\ a)$ approximates $f(x)$ well when $x$ is near $a$.

Note that first-order Taylor polynomial is $T _{f}^{1} (x,\ a) = f(a) + f’(a)(x-a)$ which is linearization of $f(x)$ at $a$. That is $L _{f}(x,\ a)$.

Example:

Let $f(x) = e^{x}$. Find $T _{f} ^{n}(x)$, the $n$-th order Taylor polynomial centered at $a=0$. Use first and third order polynomial to approximate $e^{0.1}$ and compare this approximation with the calculator value.

$T _{f} ^{n} (x) = \sum _{k=0}^{n} \frac{f^{(k)}(x)}{k!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!} + … + \frac{x^{n}}{n!}$

$e^x \approx T _{f}^{n}(x) = \frac{f(0)}{0!} + \frac{f’(0)}{1!} + \frac{f’‘(0)}{2!} x^2 + \frac{f’’‘(0)}{3!} + … + \frac{f^{(n)}(0)}{n!} x^{n}$

$f(x) = e^{x}$

$f^{(n)}(x) = e^{x}$

$\frac{f(0)}{0!} = \frac{1}{0!} = 1$

$\frac{f’(0)}{1!} = \frac{1}{1!} = 1$

$\frac{f’‘(0)}{2!} = \frac{1}{2!} = \frac{1}{2}$

$e^{x} = T _{f}^{n}(x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^3}{3!} + … + \frac{x^n}{n!}$

$T _{f}^{1} (x) = 1 + x$

$T _{f} ^{1} (0.1) = 1 + 0.1 = 1.1$

$T _{f}^{3}(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$

$T _{f}^{3} (0.1) = 1 + 0.1 + \frac{0.1^{2}}{2!} + \frac{0.1^{3}}{3!} \approx 1.1051666$

$e^{0.1} = 1.105170918$

For practice, do $e^{x}$ at $a=-1$.

Example:

Recall that $T _{f} ^{\infty} (x;\ a) = T _{f} (x;\ a)$

$= c _{o} c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 + … + c_n (x-a)^n$

$= \sum_{n=0}^{\infty} c_n (x-a)^n$ where $C_n = \frac{f^{(n)}(a)}{n!}$

Let’s find T.S. for $f(x) = \sin x$ centered at $R = pi/2$

$T _f (x;\ \frac{\pi}{2}) = c_0 + c_1 (x-\frac{\pi}{2}) + c_2 (x-\frac{\pi}{2})+ c_3 (x-\frac{\pi}{2}^3) + …$

$f(x) = \sin x$

$f’(x) = \cos x$

$f’‘(x) = -\sin x$

$f’’‘(x) = -\cos x$

$f^{(4)}(x) = \sin x$

Therefore the derivative at $\frac{\pi}{2}$ are 1, 0, -1, 0, 1, 0, -1, 0,… and so T.S. centered at $\frac{\pi}{2}$ is:

$F _f (x;\ \frac{\pi}{2}) = 1 - \frac{\left( x-\frac{\pi}{2} \right) ^{2}}{2!} + \frac{\left( x-\frac{\pi}{2} \right) ^{4}}{4!} - \frac{\left( x-\frac{\pi}{2} \right) ^{6}}{6!} + \frac{\left( x - \frac{\pi}{2} \right) ^{8}}{8!} - \frac{\left( x-\frac{\pi}{2} \right) ^{10}}{10!}$

$= \sum_{n=0}^{\infty} (-1)^{n+2} \frac{\left( x-\frac{\pi}{2} \right)^{2n} }{(2n)!}$ You get the choice of either the summation or just the expansion

Use the first two non-zero term (this phrase will be on the exam) to estimate $\sin (1.5)$ and compare with calculator.

$T _f ^{2}\left( 1.5;\ \frac{\pi}{2} \right) = 1 - \frac{\left( x-\frac{\pi}{2} \right) ^{2}}{2!}$

$\sin (1.5) \approx T _f ^{2}\left( 1.5;\ \frac{\pi}{2} \right) = 1 - \frac{\left( 1.5 - \frac{\pi}{2} \right) ^{2}}{2!} \approx 0.99749394$

Calculator value is

$\sin (1.5) = 0.99749498$

Recall this limit: $\lim_{x \to 0} \frac{\sin x}{x} = L’ = \lim_{x \to 0} \frac{\cos x}{1} = L’ = \frac{1}{1} = 1$

$\sin x \approx x$ for $x$ near 0.

$\lim_{x \to \frac{\pi}{2}} \frac{\sin \left( x- \frac{\pi}{2} \right) }{x- \frac{\pi}{2}} = 1$

$\sin x \approx x a- \frac{x^{3}}{3!} + \frac{x^5}{5!}$

The nth order Taylor Series at $a$ of $f(x)$ should approximate $f(x)$ very well when $x$ is is close to $a$ and when $n$ is big.

Example:

$f(x) = e^x$ centered at $a=0$

$T _{f} (x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$

$T _{f} ^1 (x) = 1+x$

$T _{f}^2 (x) = 1 + x + \frac{x^2}{2!}$

$T _{f}^3 (x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$

$T _{f} ^1 (2) = 1 + 2 = 3$

$T _{f} ^2 (2) = 1 + 2 + \frac{2^2}{2!}= 5$

$T _{f} ^3 (2) = 1 + 2 + 2 + \frac{8}{6} = 5 + \frac{4}{3} = 6.333…$

$T _{f} ^4 (2) = 7$

$f(2) = e^2 = 7.289$

Example:

$f(x) = \sqrt[3]{x}$ at $a=1$

$T _{f}^3 (x;\ 1) = c_0 + c_1 (x-1) + c_2(x-1)^2+c_3(x-1)^3$

$f’(x) = \frac{1}{3} x^{-\frac{2}{3}}$

$f’‘(x) = - \frac{2}{9} x^{-\frac{5}{3}}$

$f’’‘(x) = \frac{10}{27} x^{-\frac{8}{3}}$

$c_0 = \frac{f(1)}{0!} = \frac{\sqrt[3]{1}}{0!} = 1$

$c_1 = \frac{f’(1)}{1!} = \frac{1}{3} \frac{1}{1!} = \frac{1}{3}$

$c_2 = \frac{f’‘(1)}{2!} = -\frac{1}{9}$

$c_3 = \frac{f’’‘(1)}{3!} = \frac{5}{81}$

$T _{f}^{3} (x;\ 1) = 1 + \frac{1}{3}(x-1) - \frac{1}{9}(x-1)^2 + \frac{5}{81}(x-1)^3$

$T _{f}^{3} (1.2,\ 3) = 1 + \frac{0.2}{3} = 1.0666…$

$\sqrt[3]{1.2} = 1.06265$

Example:

$f(x) = \ln x$ at $a=1$

$c_0 = \frac{f(0)}{0!}$

$T _{f}^2 (x;\ 1) = c_0 + c_1(x-1) + c_2(x-1)$

$f(x) = \ln x$

$f’(x) = \frac{1}{x}$

$f’‘(x) = -\frac{1}{x^2}$

$c_0 = \frac{f(1)}{0!} = \frac{\ln 1}{0!} = 0$

$c_1 = \frac{f’(1)}{1!} = 1$

$c_2 = \frac{f’‘(1)}{2!} = -\frac{1}{2}$

$T _{f}^2 (x;\ 1) = (x-1) - \frac{(x-2)}{2}$

$T _{f}^2 (0.9;\ 1) = -0.1 - 0.05 = -0.15$

$T _{f}^2 (0.9) = (0.9 - 1) - \frac{(0.9 - 1)^2}{2} = -0.105$

Find the error:

$E = \mid \ln (0.9) - T _{f}^{2}(0.9;\ 1) \mid$

$E = \mid -0.10536 - (-0.105) \mid = \mid -0.00036 \mid = 0.00036$

Definitions

Definition: Error in Approximation By Polynomials

We refer to $n$-th order Taylor remember for $f(x)$ at $a$ as $R _{f}^{n} (x;\ a) = f(x) - T _{f}^{n}(x;\ a)$ then the $n$-th order Taylor error for $f(x)$ ordered at $a$ is:

$E _{f}^n (x;\ a) = \mid R _{f}^n (x;\ a) \mid = \mid f(x) - T _{f}^n (x;\ a) \mid$

Here we establish the upper bound on the error in approximating a function by its Taylor Series. The so called Taylor-Lagerange Theorem. Without proof, the formula that gives $n$-th order error in Lagrange Theorem is:

Definition: Lagrange Error Theorem

$E _{f}^n (x;\ a) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

Where $c$ is the number between $x$ and $a$ that result in the largest error.

It looks similar to the $n+1$ th order term in the Taylor Series with small exception that there is $C$ (unknown) instead of center of expansion $a$. Our goal is always to place an upper bound of $f^{(a+1)}(c)$ which does not have a $C$ in it, so this mysterious $C$ should be determined. and we will see that it will not constitute any problem to find this $C$. It will always be a number between $a$ and $x$.

Example:

$f(x) = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$

We found that

$T _{f}^1 (0.1) = 1.1$

$T _{f}^3 (0.1) = 1.105166$

And the calculator value was $e^{0.1} = 1.1051709$

Question is: Can we use Lagrange formula for error to obtain upper bound on the error if we approximate $e^{0.1}$ by the 1st and 3rd order Taylor Series.

$E _{f}^n (0.1) = \mid \frac{f^{(n+1)}(c)}{(n+1)!} (0.1) ^{n+1} \mid = \frac{(0.1)^{n+1}}{(n+1)} \mid f^{n+1}(c) \mid$

$f(x) = e^x$

$f^{(n)}(x) = e^x \implies f^{(n+1)}(x) = e^x$ so $f^{(n+1)}(c) = e^c = \frac{(0.1)^{n+1}}{(n+1)!} e^{0.1}$

$[a;\ x]$

$[0,\ 0.1]$

$0\le c \le 0.1$

Recall:

$E _{f}^n (x;\ a) = \mid \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \mid$

$e^{x} = 1 + x + \frac{x^2}{2!}$

$T _{f}^1 (x) = 1 + x \implies T _{f}^1 (0.1) = 1 + 0.1 = 1.1$

$T _{f}^3 (x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \implies T _{f}^3 (0.1) = 1.1051666$

Calculator value of $e^{0.1} = 1.10517091808$

Approximate $e^{0.1}$ by setting an upper bound error using Lagrange formula.

The error in approximation ($n$ order) is:

$E _{f} ^n (0.1) = \mid \frac{f^{(n+1)}(c)}{(n+1)!} (0.1)^{n+1} \mid = \frac{(0.1)^{n+1}}{(n+1)!} \mid f^{(n+1)}(c) \mid = \frac{\left( 0.1 \right) ^{n+1}}{(n+1)!}e^{c}$

$f(x) = e^x$

$f^{(n+1)}(x) = e^x \implies f^{(n+1)} (c) = e^{c}$

$c$ is between $[a;\ x]$, in our case $a=0$, $x=0.1$, therefore $e$ $c$ between $[0;\ 0.1]$. Since $e^{c} \le e^{0.1}$, then:

$E _{f}^{n}(0.1) \le \frac{(0.1)^{n+1}e^{0.1}}{(n+1)!}$

What should be the $n$ so that we would have 0.0001 error?

$E _{f}^1 (0.1) = \mid e^{0.1} - T _{f}^1 (0.1) \mid = 1.105… - 1.1 \mid = 0.00517$

$E _{f}^3 (0.1) = \mid e^{0.1} - T _{f}^{3} (0.1) \mid = 0.00000425808$ This is the actual error.

Now we want to get the upper bound error if we use 3 terms.

$E _{f}^3 (0.1) = \frac{(0.1)^{3+1}e^{0.1}}{(3+1)!} = \frac{0.0001 e^{0.1}}{24} = 0.0000046 \approx 0.5 x 10^{-6}$

On the quiz, you will be asked something like “how many terms do you need to get an error of 0.001 (or $10^{-3}$?”

To do this, you just need to check via trial-and-error

Example:

Find $n$ for an error of $\le 0.0001 = 10^{-4}$

$E _{f}^{1}(0.1) = \frac{(0.1)^{1+1}}{(1+1)!}e^{0.1} \approx 0.0055$

$E _{f}^{2}(0.1) = \frac{(0.1)^{2+1)}}{(2+1)!}e^{0.1} \approx 0.00018$

$E _{f}^{3}(0.1) = \frac{(0.1)^(3+1)}{(3+1)!} e^{0.1} \approx 0.0000046$


Question 1

Question 2

Question 2b

Question 2c

Question 2d


Example:

$f(x) = \ln x$

$f’(x) = \frac{1}{x}$

$f’‘(x) = -\frac{1}{x^2}$

$f’’‘(x) = \frac{2}{x^3}$

$f^{(4) }(x) = -\frac{1(2)(3)}{x^4}$

$f^{(5)}(x) = \frac{1(2)(3)(4)}{x^5}$

$\therefore f^{(n)}(x) = \frac{(-1)^{n+1}(n-1)!}{x^n} \forall n\ge 1$

$f^{(n+1)}(x) = \frac{(-1)^{n}(n)!}{x^{n+1}}$

$T _{f}^3 (x;\ 1) = c_0 + c_1 (x-1) + c_2 (x-1)^2 + c_3(x-1)^3$

$c_0 = \frac{f(1)}{0!} = \frac{\ln 1}{0!} = 0$

$c_1 = \frac{f’(1)}{1!} = \frac{1}{1!} = 1$

$c_2 = \frac{f’‘(1)}{2!} = \frac{-1}{2!} = -\frac{1}{2}$

$c_3 = \frac{f’’‘(1)}{3!} = \frac{2}{3!} = \frac{1}{3}$

$T _{f}^3 (x;\ 1) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}$

$E _{f}^3 (1.1;\ 1) = \mid \ln (1.1) - T _{f}^3(1.1;\ 1) \mid$

$E _{f}^3(1.1;\ 1) = \mid 0.0953101798 - \left(0.1 - \frac{(0.1)^2}{2} + \frac{(0.1)^3}{3} \right) \mid$

$= \mid -0.00002315353 \mid = 0.000023153$

$E _{f}^3 (x;\ 1) = \mid \frac{f^{(n+1)}(c)}{(n+1)!}(x-1)^{x+1} \mid = \mid \frac{(-1)^{n+2}}{(n+1)!} \frac{n!}{C^{n+1}} (x-1)^{n+1} \mid = \frac{1}{n+1} \frac{1}{C^{n+1}} \mid x-1 \mid ^{n+1}$

$= \frac{1}{(n+1)C^{n+1}} (0.1)^{n+1}$

$\mid x - 1 \mid \le 0.1$

$0.9 \le x \le 1.1$

$0.9 \le C \le 1.1$

Because we are getting the upper bound, we want the smallest value of $C$ (which is 0.9).

$C=0.9$

$E _{f}^n (x;\ 1) \le \frac{1}{(n+1)(0.9)^{n+1}} = \frac{1}{n+1}\left( \frac{1}{9} \right) ^{n+1}$

$E _{f}^3 (x;\ 1) = \frac{1}{4}\left( \frac{1}{9} \right) ^{4} \approx 0.000038103$

We already know that if $n=3$, the error is 0.000038103 and it is less than 0.0001. We should go on and check what if $n=2$ will give us again an error less than 0.0001.

Let $n=2$.

$E _{f}^2 (1.1;\ 1) = \frac{1}{2+1}\left( \frac{1}{9} \right) ^{2+1} = \frac{1}{3} \left( \frac{1}{9} \right) ^3 = \left( \frac{1}{3} \right)^{7} = 0.00045724$

No! Since 0.00045724 > 0.0001, then $n=3$ is the order that we must retain.

Example:

$f(x) = e^{3x}$

Part a:

$f’(x) = 3e^{3x}$

$f’‘(x) = 3^{2} e^{3x}$

$f’’‘(x) = 3^{3}e^{3x}$

$f^{(n)}(x) = 3^{n} e^{3x}$

Therefore, $f^{(n+1)}(x)=3^{n+1}e^{3x}$

$T _{f}^{3}(x;\ 0) = c_0 + c_1x + c_2 x^{2} + c_3 x^{3}$

$c_0 = \frac{f(0)}{0!} = \frac{e^{3(0)}}{0!} = \frac{1}{0!} = 1$

$c_1 = \frac{f’(0)}{1!} = \frac{3e^{0}}{1!} = \frac{3}{1!} = 3$

$c_2 = \frac{f’‘(0)}{2!} = \frac{9e^{0}}{2!} = \frac{9}{2}$

$c_3 = \frac{f’’‘(0)}{2!} = \frac{27e^{0}}{3!} = \frac{9}{2}$

$c_4 = \frac{81}{4!} = \frac{81}{24} = \frac{27}{8}$

Part b:

$T _f ^{3} (x) = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3$

$E _{f}^{3} (0.1) = \mid f(0.1) - T _{f}^{3}(0.1) \mid$

$= \mid e^{0.3} - \left( 1 + 0.3 + \frac{9}{2}(0.1)^{2} + \frac{9}{2}(0.1)^3 \right) \mid = 0.0003588$

Part c: Find upper bound for the error $E _{f}^{2}(x;\ 0)$ in approximating $f(x) = e^{3x}$ by $T _{f}^{3}(x;\ 0)$ if $\mid x \mid \le 0.1$

$E _{f}^{n} = \mid \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \mid$

$= \frac{3^{n+1}e^{3c}}{(n+1)!} \mid x \mid ^{n+1}$

$E _{f}^{3}(x;\ 0) = \frac{3^{4}e^{3c}}{4!} \mid x \mid ^{4} \le \frac{3^{4}e^{3c}}{4!} (0.1)^4$

$c = [0,\ 0.1]$

$e^{0} < e^{3(0.1)}$

$E _{f}^{3}(x;\ 0) = \frac{3^{4}e^{3c}}{4!} \mid x \mid ^{4} \le \frac{3^{4}e^{3c}}{4!} (0.1)^4 \le \frac{3^{4}e^{0.3}}{4!}(0.1)^4$

$E _{f}^{3} (x;\ 0) = \frac{3^{4}e^{3(c)}}{4!} \mid x \mid ^4 \le 0.0004555$

If $\mid x \mid \le 0.1$ find the smallest $n$ so that the error is less than 0.0001.

Since for $n=3$, $E _{f}^{2} = 0.0004555$ we try $n=4$

$E _{f}^{4}(0.1) = \frac{3^{5}e^{0.3}}{5!}(0.1)^5 = 0.00002732$

Example:

**Part a: For $f(x) = \ln (1+x)$, find $f^{(n+1)}(x)$ **

$f’(x) = \frac{1}{1+x}$

$f’‘(x) = \frac{-1}{(1+x)^2}$

$f’’‘(x) = \frac{2}{(1+x)^3}$

$f^{(n+1)}(x) = (-1)^{n} \frac{n!}{(1+x)^{n+1}}$

Part b: Use the answer from part A to find $T _{f}^{3}(x;\ 0)$. That is, Taylor Series at $a=0$. Also find $E _{f}^{3}(x;\ 0) = \mid f(x) - T _{f}^{3}(x;\ 0) \mid$ at $x=0.1$.

$T _{f}^{3}(x) = c_0 + c_1 x + c_2 x^{2} + c_3 x^{3}$

$c_n = \frac{f^{n}(0)}{n!}$

$c_0 = \frac{f(0)}{0!} = 0$

$c_1 = \frac{f’(0)}{1!} = 1$

$c_2 = \frac{f’‘(0)}{2!} = -\frac{1}{2}$

$c_3 = \frac{f’’‘(0)}{3!} = \frac{1}{3}$

$\therefore T _{f}^{3} (x;\ 0) = x - \frac{x^2}{2} + \frac{x^3}{3}$

$E _{f}^{3}(0.1) = \mid f(0.1) - T _{f}^{3}(0.1) \mid$

$= \mid \ln 1.1 - (0.1 - \frac{0.1^{2}}{2} + \frac{0.1^{3}}{3}) \mid = 0.00002315$ Part c: use $f^{(n+1)}(x)$ obtained in Part A to find $E _{f}^{3}(x;\ 0)$ if $\mid x \mid \le 0.1$

$f^{(n+1)}(c) = \frac{(-1)^{n}n!}{(1+c)^{n+1}}$

So $E _{f}^{n}(x) = \mid \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \mid$

$= \frac{n!}{(1+c)^{n+1}(n+1)!} \mid x \mid ^{n+1} = \frac{1}{(1+c)^{n+1}(n+1)} \mid x \mid ^{n+1}$

$E _{f}^{n}(0.1) = \frac{1}{(1+c)^{n+1}(n+1)}(0.1)^{n+1}$

$E _{f}^{3}(0.1) = \frac{1}{(1+c)^{3+1}(3+1)}(0.1)^{3+1}$

$-0.1 \le c \le 0.1$

Since we want the largest number, we choose $c=-0.1$.

$E _{f}^{3}(0.1) = \frac{1}{(1-0.1)^{3+1}(3+1)}(0.1)^{3+1} = 0.0000381039$

Part d: If $\mid x \mid \le 0.1$ find the smallest $n$ such that $E _{f}^{n}(0.1,\ 0)$ is less than $0.0001$.

Since the error for $n=3$ is 0.0000381 and it is smaller than the 0.0001, the answer is at most $n=3$. We now begin by checking $n$’s less than 3.

$E _{f}^{2}(0.1) = \frac{1}{3} \frac{1}{0.9^3}(0.1)^{3} = \frac{1}{3} \left( \frac{1}{9} \right) ^{3} = 0.00045725$

Since $E _{f}^{2}(0.1)$ is greater than 0.0001, then $n=3$ is the smallest.

Sequences and Series

We can expand our function into a Taylor Series. The main question is: for what values of $x$ this series is equal to the original function.

$e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$

The answer based on the issue of convergence of the series. If it diverges, then there is no equality, otherwise the issue never narrows down to the issue of the interval of convergence.

Example:

$\frac{1}{1-x} = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4 x^4$

$\frac{1}{1-x} = 1+x+x^2+x^3+x^4 + …(-1,\ 1)$

$\frac{1}{1-x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n}$

This means that this series is only exactly equal to $\frac{1}{1+x}$ for values between -1 and 1.

For example:

$x=0.2$

$\frac{1}{1-0.2} = 1.25$

$1 + 0.2 + 0.04 + … \approx 1.24$

What about $x=1.1$?

$\frac{1}{1-1.1} = -10$

$1 + 1.1 + … \approx 2.2$ not even close

What about $x=-0.2$?

$\frac{1}{1-(-0.2)} = \frac{1}{1.2} = 0.833$

$1-0.2 + 0.04 - 0.008 + … \approx 0.832$

Sequences

$\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} = \frac{\pi^{2}}{6}$

$\sum_{n=1}^{\infty} \ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + … (-1,\ 1]$

A harmonic series is defined by $\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + …$ and it is always divergent.

Alternating harmonic series: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + …$

Definition: Sequence and Series

Sequence:

List of numbers given in some defined order:

$a_1,\ a_2,\ a_3,\ a_4$

Where $a_1$ is the first of the sequence, $a_2$ is the second term, and $an$ is the $n$ th term of the sequence. Notion is either {$a _n$} or $a_n$.

A sequence $a_n$ has limit $L$ and we write $\lim_{n \to \infty} a_n = L$, meaning if $n \to \infty$ and $a_n \to L$ then it converges. Otherwise, it diverges.

Series:

If we add the terms of the sequence $a_1 + a_2 + a_3 + … + a_n$, then we get infinite series (or series).

$\sum_{n=1}^{\infty} a _n$

Example sequences:

$a_n = \frac{n}{n+1} = \frac{1}{2},\ \frac{2}{3},\ \frac{3}{4},\ \frac{4}{5}, …, \frac{n}{n+1},\ \frac{n+1}{n+2}$

$a_n = \frac{(-1)^{n}(n+1)}{3^{n}} = -\frac{2}{3},\ \frac{3}{9},\ -\frac{4}{27},\ \frac{5}{81}$

$1,\ 1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ …$

$\frac{3}{5},\ -\frac{4}{25},\ \frac{5}{125},\ -\frac{6}{625},\ …$

$a_n = \frac{(-1)^{n}(n+2)}{5^{n}}$

$a_n = \frac{n^2}{n+1}$

$\lim_{n \to \infty} \frac{n^2}{n+1} = \infty$ so it diverges

Example series:

Consider the sequence: $1,\ \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4} = a_n = \frac{1}{n}$

The series: $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$

$1+2+3+4+5+…+n$, if we start to add, we get cumulative sum: $1,\ 3,\ 6,\ 10,\ 15,\ 21 = \frac{n(n+1)}{2}$

$\sum_{n=1}^{\infty} i = \frac{n(n+1)}{2}$