175 – Midterm 2

Here is the second midterm.

Question 1: Find the volume of the solid obtained by rotating about the ${y}$-axis the region between the ${x}$-axis and the curve ${y=\sec^2x}$ for ${0\le x\le\pi/4}$.

Here is a graph of ${y=\sec^2x}$ in the specified region. To compute the volume, we use the shell method, and see that it is given by

$\displaystyle V=\int_0^{\pi/4} 2\pi x\cdot\sec^2x\,dx.$

We evaluate this integral using integration by parts.

We take: $\displaystyle u=2\pi x,\mbox{ so }du=2\pi\,dx,$ and $dv=\sec^2x\,dx,$ so $v=\tan x;$ so $\displaystyle V=\left.2\pi x\tan x\right|_0^{\pi/4}-\int_0^{\pi/4}2\pi\tan x\,dx$

$\displaystyle=\left.2\pi x\tan x+2\pi\ln(\cos x)\right|_0^{\pi/4}$

$\displaystyle=\left(2\pi\cdot\frac{\pi}4\cdot1+2\pi\ln\left(\frac1{\sqrt2}\right)\right)-(0+2\pi\ln(1))$

$\displaystyle=\frac{\pi^2}2-\pi\ln2.$

Question 2: Solve the initial value problem ${\displaystyle (t^2+3t+3)\frac{dx}{dt}=1}$, ${x(3)=0}$.

We use the technique of separation of variables, and rewrite the given differential equation as:

$\displaystyle dx=\frac{dt}{t^2+3t+3},$

so

$\displaystyle x=\int \frac{dt}{t^2+3t+3}.$

To solve this integral, we begin by completing the square in the denominator of the given fraction:

$\displaystyle t^2+3t+3=\left(t+\frac32\right)^2+\left(3-\frac94\right)=\left(t+\frac32\right)^2+\frac34.$

This indicates that to solve the integral we want to use the trigonometric substitution

$\displaystyle t+\frac32=\frac{\sqrt3}2\tan\theta,$

or

$\displaystyle \theta=\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right).$

This gives: $\displaystyle \left(t+\frac32\right)^2+\frac34=\frac34\sec^2\theta,$

$\displaystyle dt=\frac{\sqrt3}2\sec^2\theta\,d\theta,\mbox{ and}$

$\displaystyle \int \frac{dt}{t^2+3t+3}=\int\frac{(\sqrt3/2)\,d\theta}{3/4}$

$\displaystyle=\frac2{\sqrt3}\theta+C$

$\displaystyle=\frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right)+C.$

We conclude by finding the value of ${C}$, for which we use that ${x(3)=0.}$ This gives us

$\displaystyle 0=x(3)= \frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}3+\sqrt3\right)+C,$

or

$\displaystyle C=-\frac2{\sqrt3}\tan^{-1}(3\sqrt3).$

Putting this together, we have

$\displaystyle x(t)= \frac2{\sqrt3}\tan^{-1}\left(\frac2{\sqrt3}t+\sqrt3\right)-\frac2{\sqrt3}\tan^{-1}(3\sqrt3).$

Question 3: Find ${\displaystyle \int\frac{x^3+5}{x^4-x^2}\,dx}$.

To solve this integral, we use the technique of partial fractions decomposition, and begin by factoring

$\displaystyle x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1).$

This means we need to find constants ${A,B,C,}$ and ${D}$ such that

$\displaystyle \frac{x^3+5}{x^4-x^2}=\frac Ax+\frac B{x^2}+\frac C{x-1}+\frac D{x+1}.$

As seen in class, there are several ways of doing this. For example, we can begin by adding the fractions on the right hand side, to obtain

$\displaystyle \frac{Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)}{x^4-x^2}.$

This means that

$\displaystyle x^3+5=Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)$

for all values of ${x}$. When ${x=0,}$ this gives

$\displaystyle 5=-B,\mbox{ or }B=-5.$

When ${x=1}$, we have

$\displaystyle 6=2C,\mbox{ or }C=3.$

When ${x=-1}$, we have

$\displaystyle 4=-2D,\mbox{ or }D=-2.$

Finally, when ${x=2}$ we have

$\displaystyle 13=6A-5\cdot3+3\cdot12-2\cdot4,$

or ${A=0.}$

Putting this together, we have

$\displaystyle \frac{x^3+5}{x^4-x^2}=-\frac 5{x^2}+\frac 3{x-1}-\frac 2{x+1},$

and

$\displaystyle \int\frac{x^3+5}{x^4-x^2}\,dx=\frac5x+3\ln|x-1|-2\ln|x+1|+C.$

[Actually, although this was not addressed in lecture, we can use different values of ${C}$ in each of the intervals ${(-\infty,-1),}$ ${(-1,0),}$ ${(0,1),}$ and ${(1,\infty)}$.]

Question 4: We have that ${\displaystyle\int_0^1\frac4{1+t^2}\,dt=\pi}$. Suppose we do not know the value of ${\pi}$. We can use the trapezoidal rule to approximate it. Find a value of ${n}$ such that the theoretical error for the trapezoidal rule using ${n}$ subdivisions is strictly smaller than ${1/10}$. Find the approximation given by the trapezoidal rule using this ${n}$.

The error ${E_T}$ using the trapezoidal rule with ${n}$ subintervals is bounded above by

$\displaystyle \frac{M_2(b-a)^3}{12n^2},$

where ${a=0,}$ ${b=1,}$ and ${M_2=\max_{0\le t\le 1}|f''(t)|}$ where ${\displaystyle f(t)=\frac4{1+t^2}.}$

To compute a bound on ${M_2,}$ we first find ${f''(t):}$ $\displaystyle f'(t)=\frac{-8t}{(1+t^2)^2},$

$\displaystyle f''(t)=\frac{32t^2-8(1+t^2)}{(1+t^2)^3}$

$\displaystyle =\frac{24t^2-8}{(1+t^2)^3}.$

An easy way of bounding this expression is to note that ${24t^2-8}$ is increasing for ${0\le t\le 1,}$ with value ${-8}$ at ${t=0,}$ and value ${16}$ for ${t=1,}$ so the maximum of its absolute value is ${16.}$ Similarly, ${\displaystyle\frac1{(1+t^2)^3}}$ is decreasing, so it maximum occurs at ${t=0}$, where it is 1.

Hence ${M_2}$ is bounded above by ${\displaystyle \frac{16}1=16,}$ and

$\displaystyle E_T\le \frac{16}{12n^2}=\frac4{3n^2}.$

If we want ${E_t<1/10,}$ it suffices that

$\displaystyle \frac4{3n^2}<\frac1{10},$

or

$\displaystyle n^2>40/3=13.33\dots,$

so ${n\ge4.}$

The approximation ${T_4}$ is given by

$\displaystyle \frac {1/4}2\left(4+\frac8{1+\frac1{16}}+\frac8{1+\frac4{16}}+\frac8{1+\frac9{16}}+2\right)\approx3.131$

which is indeed an accurate approximation to ${\pi}$ within ${1/10.}$

[Here is a graph of ${f''(t)}$ for ${0\le t\le 1.}$ A more careful computation of ${M_2}$ would have revealed that ${M_2=8,}$ which gives ${n\ge3.}$ (Note that ${\max_{0\le t\le 1} f''(t)=2,}$ but ${M_2}$ looks at the maximum of ${|f''(t)|,}$ not just ${f''(t)}$.)

The approximation ${T_3}$ gives the value ${3.123\dots}$. A less careful computation, replacing ${24t^2-8}$ with ${24t^2+8}$ (so we do not have to worry about both positive and negative numbers) gives ${n\ge6}$, with ${T_6\approx3.137.}$ Actually, ${T_2=3.1}$ works as well but the theoretical error does not predict this.]

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