Calculus II-Final Exam

Solutions

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Section 1

Evaluate the following definite and indefinite integrals using any appropriate technique.

1) $$\int 4x\sqrt{1-x^2}\;dx$$

Solution:

Let u=4-x² and then $du=-2x\longrightarrow -2du=4xdx$.

$$\begin{eqnarray*}\displaystyle\int 4x\sqrt{1-x^2}\;dx&=&-2\displaystyle\int\sqrt... ...\;C\\ \\ [1mm] &=&-\displaystyle\frac{4}{3}(4-x^2)^{3/2}+\;C\\ \end{eqnarray*}$$

2) $$\int\displaystyle\frac{1}{1+\sin(\theta)}\;d\theta$$

Solution:

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{1}{1+\sin(\theta)}\;d\theta... ...{-1}(\theta)+\;C\\ \\ [1mm] &=&\tan(\theta)+\sec(\theta)+\;C\\ \end{eqnarray*}$$

3) $$\int\displaystyle\frac{\ln(2x)}{x}\;dx$$

Solution:

Let $u=\ln(2x)$ and so $du=\displaystyle\frac{1}{x}\;dx$. Therefore

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{\ln(2x)}{x}\;dx&=& \display... ...}{2}+\;C\\ \\ [1mm] &=&\displaystyle\frac{(\ln 2x)^2}{2}+\;C\\ \end{eqnarray*}$$

4) $$\int\displaystyle\frac{t^2}{t^2+1}\;dt$$

Solution:

After long division

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{t^2}{t^2+1}\;dt&=& \display... ...yle\frac{1}{t^2+1}\right)\;dt\\ \\ [1mm] &=&t-\arctan(t)+\;C\\ \end{eqnarray*}$$

5) $$\int e^{x}\cos(2x)\;dx$$

Solution:

$$u=e^x\;\;\;\;\;\;\;dv=\cos(2x)\;dx$$

$$du=e^x\;dx\;\;\;\;\;\;v=\displaystyle\frac{\sin(2x)}{2}$$

$$\begin{eqnarray*}\displaystyle\int e^{x}\cos(2x)\;dx&=& \displaystyle\frac{1}{2}... ...x)-\displaystyle\frac{1}{2}\displaystyle\int\ e^x\sin(2x)\;dx\\ \end{eqnarray*}$$

Using parts again

$$u=e^x\;\;\;\;\;\;dv=\sin(2x)\;dx$$

$$du=e^x\;dx\;\;\;\;\;v=-\displaystyle\frac{\cos(2x)}{2}$$

$$\begin{eqnarray*}\displaystyle\int e^{x}\cos(2x)\;dx&=& \displaystyle\frac{1}{2}... ...frac{2}{5}e^x\sin(2x)+\displaystyle\frac{1}{5}e^x\cos(2x)+\;C\\ \end{eqnarray*}$$

6) $$\int\sin^3(\theta)\cos^3(\theta)\;d\theta$$

Solution:

$$\begin{eqnarray*}\displaystyle\int\sin^3(\theta)\cos^3(\theta)\;d\theta&=& \disp... ...style\frac{\sin^4(x)}{4}-\displaystyle\frac{\sin^6(x)}{6}+\;C\\ \end{eqnarray*}$$

7) $$\int\displaystyle\frac{2}{x^2+4x+3}\;dx$$

Solution:

$$\begin{eqnarray*}\displaystyle\frac{2}{x^2+4x+3}&=&\displaystyle\frac{2}{(x+1)(x... ...\\ \\ [1mm] &=&\displaystyle\frac{x(A+B)+(3A+B)}{(x+1)(x+3)}\\ \end{eqnarray*}$$

Therefore,

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

$$3A+B=2\longrightarrow 3A-A=2\longrightarrow A=1;,B=-1$$

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{2}{x^2+4x+3}\;dx&=& \displa... ...] &=&\ln\left\vert\displaystyle\frac{x+1}{x+3}\right\vert+\;C\\ \end{eqnarray*}$$

8) $$\int\displaystyle\frac{x+3}{x^2+2x+2}\;dx$$

Solution:

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{x+3}{x^2+2x+2}\;dx&=& \disp... ...displaystyle\frac{1}{2}\ln\mid x^2+2x+2\mid+2\arctan(x+1)+\;C\\ \end{eqnarray*}$$

Section 2

9) Graph the following functions and then rotate the formed region about
the line x=2. Then determine the volume of the solid created.

$$y=-x^2+4\;\;\;\;\;\;y=0\;\;\;\;\;\;x=0$$

Solution:

Using the shell method our radius for each shell is r=(2-x) and the height of each shell is the function value -x²+4. This gives the following integral

$$\begin{eqnarray*}2\pi\displaystyle\int^{2}_{0}(2-x)(-x^2+4)\;dx&=& 2\pi\displays... ...rac{20}{3}\right]\\ \\ [1mm] &=&\displaystyle\frac{40\pi}{3}\\ \end{eqnarray*}$$

Section 3

10) Consider the following integral

$$\displaystyle\int^{2}_{0}5\sqrt{4-x^2}\;dx$$

First, evaluate this integral using a geometric argument. Then solve the integral using trig substitution and verify that both answers agree.

Solution:

We know that $y=\sqrt{4-x^2}$ is the equation of a semicircle with radius 2, and our limits of integration assure us that we are determining the area of 1/4 of this circle. Therefore,

$$\displaystyle\int^{2}_{0}5\sqrt{4-x^2}\;dx= 5\displaystyle\int^{2}_{0}\sqrt{4-x^2}\;dx= 5\displaystyle\frac{\pi(2)^2}{4}=5\pi$$

Using trig substitution let $u=a\sin(\theta)\longrightarrow x=2\sin(\theta)$.

$$x=2\sin(\theta)\longrightarrow dx=2\cos(\theta)\;d\theta\longrightarrow 2\cos(\theta)=\sqrt{4-x^2}$$

$$\begin{eqnarray*}5\displaystyle\int\sqrt{4-x^2}\;dx&=& 5\displaystyle\int 4\cos^... ...&10\bigg[\displaystyle\frac{\pi}{2}\bigg]\\ \\ [1mm] &=&5\pi\\ \end{eqnarray*}$$

Section 4

Determine the convergence or divergence of the following series, being sure to tell which test you are using. When possible, determine the sum of the series. If you can only estimate the sum, decide the accuracy of your approximation.

11) $$\sum^{\infty}_{n=0}\displaystyle\frac{5^n}{n^5}$$

Solution:

It is easy to show that any exponential function will eventually grow faster than any polynomila function. So $$\lim_{n \to \infty}\displaystyle\frac{5^n}{n^5}\ne 0$$.

This can also be shown through L'Hopital's rule.

$$\begin{eqnarray*}\displaystyle\lim_{n \to \infty}\displaystyle\frac{5^n}{n^5}&=&... ..._{n \to \infty}\displaystyle\frac{5^n(\ln 5)^n}{120}&=&\infty\\ \end{eqnarray*}$$

The series diverges by the n^th term test.

12) $$\sum^{\infty}_{n=1}\left(\displaystyle\frac{1}{n^2}-\displaystyle\frac{1}{2^n}\right)$$hint: $$\sum^{\infty}_{n=1}\displaystyle\frac{1}{n^2}=\displaystyle\frac{\pi^2}{6}$$

Solution:

$$\displaystyle\sum^{\infty}_{n=1}\left(\displaystyle\frac{1}{n... ...style\sum^{\infty}_{n=1}\left(\displaystyle\frac{1}{2}\right)^n$$

The geometric series converges to $$\frac{1}{1-1/2}-1=1$$ because its index starts at n=1. Therefore, knowing that 1/n² is a convergent p-series

$$\displaystyle\sum^{\infty}_{n=1}\left(\displaystyle\frac{1}{n... ...isplaystyle\frac{1}{2^n}\right)= \displaystyle\frac{\pi^2}{6}-1$$

13) $$\sum^{\infty}_{n=1}\displaystyle\frac{\sin(n^n)}{n^2}$$

Solution:

$$\left\vert\displaystyle\frac{\sin(n^n)}{n^2}\right\vert\le\displaystyle\frac{1}{n^2}$$

which forms a convergent p-series. Therefore, the series

$$\displaystyle\sum^{\infty}_{n=1}\left\vert\displaystyle\frac{\sin(n^n)}{n^2}\right\vert$$

converges by direct comparison.

Theorem 8.16 says that if $$\sum^{\infty}_{n=1}\mid a_n\mid$$converges, then $$\sum^{\infty}_{n=1}a_n$$ also converges.

Therefore, the original series

$$\displaystyle\sum^{\infty}_{n=1}\displaystyle\frac{\sin(n^n)}{n^2}$$

converges by Theorem 8.16.

14) $$\sum^{\infty}_{n=10}(-1)^n\displaystyle\frac{3^n}{n!}$$

Solution:

$$\displaystyle\lim_{n \to \infty}\displaystyle\frac{3^n}{n!}=0\;\;\mbox{ and }\;\;a_{n+1}\le a_n$$

The series converges by the Alternating Series Test. Using the Ratio Test,

$$\begin{eqnarray*}\displaystyle\lim_{n \to \infty}\left\vert\displaystyle\frac{3^... ...m_{n \to \infty}\displaystyle\frac{1}{n+1}\\ \\ [1mm] &=&0<1\\ \end{eqnarray*}$$

The positive series converges by the Ratio Test and so the original alternating series $$\sum^{\infty}_{n=10}(-1)^n\displaystyle\frac{3^n}{n!}$$ converges absolutely.



15) $$\sum^{\infty}_{n=1}\displaystyle\frac{10n+3}{n2^n}$$

Solution:

Using the Limit Comparison test and the convergent Geometric Series

$$\displaystyle\sum^{\infty}_{n=0}\displaystyle\frac{1}{2^n}$$

$$\begin{eqnarray*}\displaystyle\lim_{n \to \infty}\left\vert\displaystyle\frac{10... ...rt\displaystyle\frac{10n+3}{n}\right\vert\\ \\ [1mm] &=&10>0\\ \end{eqnarray*}$$

Therefore, the series converges by the Limit Comparison Test.



16) $$\sum^{\infty}_{n=1}\displaystyle\frac{1}{\sqrt{n^3+2}}$$

Solution:

$$\displaystyle\frac{1}{\sqrt{n^3+2}}\le\displaystyle\frac{1}{\sqrt{n^3}}=\displaystyle\frac{1}{n^{3/2}}$$

and

$$\displaystyle\sum^{\infty}_{n=1}\displaystyle\frac{1}{n^{3/2}}$$

is a convergent p-series. Therefore the oringinal series

$$\displaystyle\sum^{\infty}_{n=1}\displaystyle\frac{1}{\sqrt{n^3+2}}$$

converges by Direct Comparison.



Section 5

Solve the following problems concerning Taylor Polynomials and Power Series.

17) Determine the first 4 nonzero terms of the Taylor Polynomial centered
at $c=\pi$ for

$$f(x)=\sin(x)$$

Solution:

Recall that a Taylor Polynomial centered at c is of the form

$$P_k(x)=\displaystyle\sum^{k}_{n=0}\displaystyle\frac{f^{(n)}(c)}{n!}(x-c)^n$$

$$\begin{eqnarray*}f(x)=\sin(x)&\hspace{1in}&f(\pi)=0\\ f'(x)=\cos(x)&\hspace{1in... ...\pi)=0\\ F^{(vii)}(x)=-\cos(x)&\hspace{1in}&f^{(vii)}(\pi)=1\\ \end{eqnarray*}$$

Next we get

$$\begin{eqnarray*}P_k(x)&=&\displaystyle\sum^{k}_{n=0}\displaystyle\frac{f^{(n)}(... ...ystyle\frac{(x-\pi)^5}{5!}+ \displaystyle\frac{(x-\pi)^7}{7!}\\ \end{eqnarray*}$$

18) Determine the radius of convergence of the following power series. Then
find the Interval of Convergence for the power series

$$\displaystyle\sum^{\infty}_{n=1}\displaystyle\frac{n(x-2)^n}{3^n}$$

Solution:

Using the Ratio Test

$$\begin{eqnarray*}\displaystyle\lim_{n \to \infty}\left\vert\displaystyle\frac{(n... ... \\ [1mm] &=&\left\vert\displaystyle\frac{x-2}{3}\right\vert\\ \end{eqnarray*}$$

Therefore,

$$\mid x-2\mid<3\;\;\longrightarrow\;\;-3<x-2<3\;\;\longrightarrow\;\;-1<x<5$$

The Radius of convergence is R=3. Next we must check the endpoints.

When x=-1

$$\displaystyle\sum^{\infty}_{n=1}\displaystyle\frac{n(-3)^n}{3... ...\displaystyle\sum^{\infty}_{n=1}(-1)^nn\;\mbox{ which diverges}$$

When x=5

$$\displaystyle\sum^{\infty}_{n=1}\displaystyle\frac{n(3)^n}{3^n}=\displaystyle\sum^{\infty}_{n=1} \;\;\mbox{which also diverges}$$

Therefore the Interval of Convergence is (-1,5)



















19) Determine the value n such that the n^thdegree Maclaurin Polynomial
approximation of e^1/2 has an error of less than .01.

Remember that $R_n=\displaystyle\frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.


hint: Let $e^{1/2}\approx 1.6$

Solution:

Clearly we will be approximating the function e^x with the Maclaurin Series and the n+1^st derivative of e^x is e^x. Knowing that this is a Maclaurin Series c=0 and we are approximating e^1/2and so x=1/2. Plugging into our error formula we get

$$\displaystyle\frac{e^z}{(n+1)!}(1/2)^{n+1}<.01$$

where z lies in between zero and 1/2.

We want to make this fraction as large as possible before we choose the value n that will work, therefore let z=1/2.

$$\displaystyle\frac{e^{1/2}}{(n+1)!}(1/2)^{n+1}\approx \displa... ...1.6}{(n+1)!}(1/2)^{n+1}= \displaystyle\frac{1.6}{(n+1)!2^{n+1}}$$

Letting n=3 we get

$$\displaystyle\frac{1.6}{4!2^4}=\displaystyle\frac{1.6}{(24)(16)}= \displaystyle\frac{1.6}{384}<.01$$

So, choosing n=3 and evaluating S₃ of the Taylor Polynomial will approxiamte e^1/2 with .01 of the true value.

Section 6

20) Find the fluid force excerted on a semi-circular tank wall, with radius 2,
when the tank is completely full of water. Remember that for water
$\omega=62.4$.

Solution:

The formula for Fluid Force is

$$\omega\displaystyle\int^{b}_{a}h(y)L(y)\;dy$$

where h(y) is depth as a function of y and L(y) is the length of each segment as a function of y. Let

$$h(y)=-y\;\;\;\;\;\;L(y)=2\sqrt{4-y^2}$$

We then arrive at the following integral

$$\begin{eqnarray*}62.4\displaystyle\int^{0}_{-2}-y(4-y^2)^{1/2}\;dy&=& 124.8\disp... ...mm] &=&\displaystyle\frac{8}{3}124.8\\ \\ [1mm] &\approx&330\\ \end{eqnarray*}$$

Section 7
21) Determine the convergence or divergence of the following improper
integral

$$\displaystyle\int^{\infty}_{2}\displaystyle\frac{1}{x(\ln x)^2}\;dx$$

Solution:

$$\begin{eqnarray*}\displaystyle\int^{\infty}_{2}\displaystyle\frac{1}{x(\ln x)^2}... ...1}{\ln(b)}\bigg]\\ \\ [1mm] &=&\displaystyle\frac{1}{\ln(2)}\\ \end{eqnarray*}$$

Because the limit exists the improper integral converges to $$\frac{1}{\ln(2)}$$