Calculus II-Midterm 2

Solutions

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Section 1

Evaluate the following indefinite integrals using any appropriate technique. Don't forget to look for ``u-substitutions.''

1) $$\int 4x\sqrt{3-x^2}\;dx$$

Solution:

Let u=3-x². Then $du=-2x\;dx\;\longrightarrow\;-2du=4x\;dx$

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

2) $$\int\ln(x)\;dx$$

Solution:

$$u=\ln(x)\;\;\;\;\;dv=dx$$

$$du=\displaystyle\frac{1}{x}\;dx\;\;\;\;v=x$$

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

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

Solution:

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

4) $$\int\displaystyle\frac{x^3}{\sqrt{x^2-9}}\;dx$$

Solution:

$$u=x\;\;\;\;a=3$$

Let $x=3\sec(\theta)\;\longrightarrow\;dx=3\sec(\theta)\tan(\theta)\;d\theta$

From our triangle, $\sqrt{x^2-9}=3\tan(\theta)$

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{x^3}{\sqrt{x^2-9}}\;dx&=& \... ...m] &=&9(x^2-9)^{1/2}+\displaystyle\frac{(x^2-9)^{3/2}}{3}+\;C\\ \end{eqnarray*}$$

5) $$\int e^{2x}\sqrt{1+e^{2x}}\;dx$$

Solution:

Let $u=1+e^{2x}\;\longrightarrow du=2e^{2x}dx\;\longrightarrow \displaystyle\frac{du}{2}=e^{2x}dx$

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

6) $$\int\sin^3(x)\cos^3(x)\;dx$$

Solution:

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

7) $$\int x^2e^x\;dx$$

Solution:

$$u=x^2\;\;\;\;dv=e^x\;dx$$

$$du=2x\;dx\;\;\;\;v=e^x$$

$$\displaystyle\int x^2e^x\;dx=x^2e^x-2\displaystyle\int xe^x\;dx$$

$$u=x\;\;\;\;dv=e^x\;dx$$

$$du=dx\;\;\;\;v=e^x$$

$$\begin{eqnarray*}\displaystyle\int x^2e^x\;dx&=&x^2e^x-2\bigg[xe^x-\displaystyle... ...\bigg[xe^x-e^x\bigg]+\;C\\ \\ [1mm] &=&x^2e^x-2xe^x+2e^x+\;C\\ \end{eqnarray*}$$

8) $$\int\displaystyle\frac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}\;dx$$

Solution:

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{\sin^2(x)+\cos^2(x)}{\sin^2... ...&=&\displaystyle\int\csc^2(x)\;dx\\ \\ [1mm] &=&-\cot(x)+\;C\\ \end{eqnarray*}$$

9) $$\int\displaystyle\frac{1}{x^2-1}\;dx$$

Solution:

$$\frac{1}{x^2-1}=\displaystyle\frac{1}{(x-1)(x+1)}= \displaystyle\f... ...e\frac{A(x+1)+B(x-1)}{(x+1)(x-1)}= \displaystyle\frac{x(A+B)+(A-B)}{(x+1)(x-1)}$$

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

$$A-B=1\longrightarrow A=\displaystyle\frac{1}{2}\;\;B=-\displaystyle\frac{1}{2}$$

$$\begin{eqnarray*}\displaystyle\int\displaystyle\frac{1}{x^2-1}\;dx&=& \displayst... ...ac{1}{2}\ln\displaystyle\frac{\mid x-1\mid}{\mid x+1\mid}+\;C\\ \end{eqnarray*}$$

Section 2

Evaluate the following limits.
10) $$\lim_{x \to 0^+}\displaystyle\frac{e^x-(1+x)}{x^3}$$

Solution:

$$\begin{eqnarray*}\displaystyle\lim_{x \to 0^+}\displaystyle\frac{e^x-(1+x)}{x^3}... ...playstyle\lim_{x \to 0^+}\displaystyle\frac{e^x}{6x}&=&\infty\\ \end{eqnarray*}$$

11) $$\lim_{x \to 0}\displaystyle\frac{\sin(2x)}{\sin(5x)}$$

Solution:

$$\begin{eqnarray*}\displaystyle\lim_{x \to 0}\displaystyle\frac{\sin(2x)}{\sin(5x... ...aystyle\frac{2\cos(2x)}{5\cos(5x)}&=&\displaystyle\frac{2}{5}\\ \end{eqnarray*}$$

12) $$\lim_{x \to \infty}x\sin\left(\displaystyle\frac{1}{x}\right)$$

Solution:

$$\begin{eqnarray*}\displaystyle\lim_{x \to \infty}x\sin\left(\displaystyle\frac{1... ...yle\frac{1}{x}\right)\\ \\ [1mm] &=&\cos(0)\\ \\ [1mm] &=&1\\ \end{eqnarray*}$$

Section 3
Determine the convergence or divergence of the following improper integrals.

13) $$\int^{\infty}_{0}\displaystyle\frac{e^{-x}}{1+e^{-x}}\;dx$$

Solution:

$$\begin{eqnarray*}\displaystyle\int^{\infty}_{0}\displaystyle\frac{e^{-x}}{1+e^{-... ...(1+e^{-b})+\ln(1+e^0)\bigg]\\ &=&-\ln(1)+\ln(2)\\ &=&\ln(2)\\ \end{eqnarray*}$$

Therefore, the improper integral converges.

14) $$\int^{\infty}_{0}\displaystyle\frac{2e^x}{e^x+e^{-x}}\;dx$$

Solution:

$$\begin{eqnarray*}\displaystyle\int^{\infty}_{0}\displaystyle\frac{2e^x}{e^x+e^{-... ...ty}\bigg[\ln(e^{2b}+1)-\ln(e^0+1)\bigg]\\ \\ [1mm] &=&\infty\\ \end{eqnarray*}$$

Therefore, the improper integral diverges.

Section 4

What is wrong with the following statements?

15) $$\int\displaystyle\frac{1}{x^2(x^2-1)}\;dx=\displaystyle\int\displaystyle\frac{A}{x}+ \displaystyle\frac{B}{x^2}+\displaystyle\frac{Cx+D}{x^2-1}\;dx$$

Solution:
The denominator term x²-1=(x+1)(x-1) and so is not irreducible. Remember that we must factor until we have irreducible factors in the denominator before we may start a partial fraction decomposition.

16) $$\int\displaystyle\frac{e^x}{x^2}\;dx=x^2e^x-\displaystyle\int 2xe^x\;dx$$

hint: What are u and dv?

Solution:

$$\displaystyle\frac{e^x}{x^2}=\left(\displaystyle\frac{1}{x^2}\right)e^x$$

This solution chose u=x² and $dv=e^x\;dx$. But it should have been $u=\displaystyle\frac{1}{x^2}$ and $dv=e^x\;dx$.

Section 5

17) List the first 4 terms for the sequence $\{a_n\}=\displaystyle\frac{(n-1)!}{2^{n+1}+1}$, where n=1,2,3,...

Solution:

$$\begin{eqnarray*}a_1&=&\displaystyle\frac{0!}{2^2+1}\\ &=&\displaystyle\frac{1}... ... &=&\displaystyle\frac{6}{33}\\ &=&\displaystyle\frac{2}{11}\\ \end{eqnarray*}$$

18) Evaluate $$\sum^4_{n=1}\displaystyle\frac{(2n+2)!}{(2n-1)!}$$

Solution:

$$\begin{eqnarray*}\displaystyle\sum^4_{n=1}\displaystyle\frac{(2n+2)!}{(2n-1)!}&=... ...+(10)(9)(8)\\ \\ [1mm] &=&24+120+336+720\\ \\ [1mm] &=&1200\\ \end{eqnarray*}$$

19) Might the following series converge?

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

Solution:

$$\begin{eqnarray*}\displaystyle\lim_{n \to \infty}\displaystyle\frac{e^n}{n^3}&=&... ...m_{n \to \infty}\displaystyle\frac{e^n}{6}&=&\infty\\ &\ne&0\\ \end{eqnarray*}$$

Therefore, the series must diverge.

20) The $$\lim_{n \to \infty}\displaystyle\frac{1}{n}=0$$, therefore $$\sum^{\infty}_{n=1}\displaystyle\frac{1}{n}$$ converges. (True/False)

Solution:

False. We have shown that this series is divergent with the integral test.

$$\displaystyle\lim_{n \to \infty}a_n=0$$

doesn't tell us anything about the convergence or divergence of a series.