7.R: Chapter 7 Review Exercises
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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
In exercises 21 - 22, consider the gamma function given by \displaystyle Γ(a)=∫^∞_0e^{−y}y^{a−1}\,dy.
21) Show that \displaystyle Γ(a)=(a−1)Γ(a−1).
22) Extend to show that \displaystyle Γ(a)=(a−1)!, assuming a is a positive integer.
The fastest car in the world, the Bugati Veyron, can reach a top speed of 408 km/h. The graph represents its velocity.
23) [T] Use the graph to estimate the velocity every 20 sec and fit to a graph of the form v(t)=ae^{bx}\sin(cx)+d. (Hint: Consider the time units.)
24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- Answer
- Answers may vary. Ex: 9.405 km