3.R: Chapter 3 Review Exercises
- Page ID
- 196037
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In exercises 1 - 4, determine whether the statement is true or false. Justify your answer with a proof or a counterexample.
1) \(\displaystyle ∫e^x\sin(x)\,dx\) cannot be integrated by parts.
2) \(\displaystyle ∫\frac{1}{x^4+1}\,dx\) cannot be integrated using partial fractions.
- Answer
- False
3) In numerical integration, increasing the number of points decreases the error.
4) Integration by parts can always yield the integral.
- Answer
- False
In exercises 5 - 10, evaluate the integral using the specified method.
5) \(\displaystyle ∫x^2\sin(4x)\,dx,\) using integration by parts
6) \(\displaystyle ∫\frac{1}{x^2\sqrt{x^2+16}}\,dx,\) using trigonometric substitution
- Answer
- \(\displaystyle ∫\frac{1}{x^2\sqrt{x^2+16}}\,dx = −\frac{\sqrt{x^2+16}}{16x}+C\)
7) \(\displaystyle ∫\sqrt{x}\ln x\,dx,\) using integration by parts
8) \(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx,\) using partial fractions
- Answer
- \(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\)
9) \(\displaystyle ∫\frac{x^5}{(4x^2+4)^{5/2}}\,dx,\) using trigonometric substitution
10) \(\displaystyle ∫\frac{\sqrt{4−\sin^2(x)}}{\sin^2(x)}\cos(x)\,dx,\) using a table of integrals or a CAS
- Answer
- \(\displaystyle ∫\frac{\sqrt{4−\sin^2(x)}}{\sin^2(x)}\cos(x)\,dx = −\frac{\sqrt{4−\sin^2(x)}}{\sin(x)}−\frac{x}{2}+C\)
In exercises 11 - 15, integrate using whatever method you choose.
11) \(\displaystyle ∫\sin^2 x\cos^2 x\,dx\)
12) \(\displaystyle ∫x^3\sqrt{x^2+2}\,dx\)
- Answer
- \(\displaystyle ∫x^3\sqrt{x^2+2}\,dx = \frac{1}{15}(x^2+2)^{3/2}(3x^2−4)+C\)
13) \(\displaystyle ∫\frac{3x^2+1}{x^4−2x^3−x^2+2x}\,dx\)
14) \(\displaystyle ∫\frac{1}{x^4+4}\,dx\)
- Answer
- \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\)
15) \(\displaystyle ∫\frac{\sqrt{3+16x^4}}{x^4}\,dx\)
In exercises 16 - 18, approximate the integrals using the midpoint rule, trapezoidal rule, and Simpson’s rule using four subintervals, rounding to three decimals.
16) [T] \(\displaystyle ∫^2_1\sqrt{x^5+2}\,dx\)
- Answer
- \(M_4=3.312,\)
\(T_4=3.354,\)
\(S_4=3.326\)
17) [T] \(\displaystyle ∫^{\sqrt{π}}_0e^{−\sin(x^2)}\,dx\)
18) [T] \(\displaystyle ∫^4_1\frac{\ln(1/x)}{x}\,dx\)
- Answer
- \(M_4=−0.982,\)
\(T_4=−0.917,\)
\(S_4=−0.952\)
In exercises 19 - 20, evaluate the integrals, if possible.
19) \(\displaystyle ∫^∞_1\frac{1}{x^n}\,dx,\) for what values of \(n\) does this integral converge or diverge?
20) \(\displaystyle ∫^∞_1\frac{e^{−x}}{x}\,dx\)
- Answer
- approximately 0.2194
21) We have now covered many methods for evaluating integrals. Use ChatGPT (or another LLM) to analyze how to decide which method to use. Submit your prompt(s), a summary of the output from ChatGPT, and your analysis of how accurate this output is based on the classwork and homework.