5.6E: Exercises for Section 5.6
- Page ID
- 72042
( \newcommand{\kernel}{\mathrm{null}\,}\)
For exercises 1 - 8, compute each indefinite integral.
1) ∫e2xdx
2) ∫e−3xdx
- Answer
- ∫e−3xdx=−13e−3x+C
3) ∫2xdx
4) ∫3−xdx
- Answer
- ∫3−xdx=−3−xln3+C
5) ∫12xdx
6) ∫2xdx
- Answer
- ∫2xdx=2ln|x|+C=ln(x2)+C
7) ∫1x2dx
8) ∫1√xdx
- Answer
- ∫1√xdx=2√x+C
In exercises 9 - 16, find each indefinite integral by using appropriate substitutions.
9) ∫lnxxdx
10) ∫dxx(lnx)2
- Answer
- ∫dxx(lnx)2=−1lnx+C
11) ∫dxxlnx(x>1)
12) ∫dxxlnxln(lnx)
- Answer
- ∫dxxlnxln(lnx)=ln(ln(lnx))+C
13) ∫tanθdθ
14) ∫cosx−xsinxxcosxdx
- Answer
- ∫cosx−xsinxxcosxdx=ln(xcosx)+C
15) ∫ln(sinx)tanxdx
16) ∫ln(cosx)tanxdx
- Answer
- ∫ln(cosx)tanxdx=−12(ln(cos(x)))2+C
17) ∫xe−x2dx
18) ∫x2e−x3dx
- Answer
- ∫x2e−x3dx=−e−x33+C
19) ∫esinxcosxdx
20) ∫etanxsec2xdx
- Answer
- ∫etanxsec2xdx=etanx+C
21) ∫elnxxdx
22) ∫eln(1−t)1−tdt
- Answer
- ∫eln(1−t)1−tdt=∫1−t1−tdt=∫1dt=t+C
In exercises 23 - 28, verify by differentiation that ∫lnxdx=x(lnx−1)+C, then use appropriate changes of variables to compute the integral.
23) ∫lnxdx (Hint: ∫lnxdx=12∫xln(x2)dx)
24) ∫x2ln2xdx
- Answer
- ∫x2ln2xdx=19x3(ln(x3)−1)+C
25) ∫lnxx2dx (Hint: Set u=1x.)
26) ∫lnx√xdx (Hint: Set u=√x.)
- Answer
- ∫lnx√xdx=2√x(lnx−2)+C
27) Write an integral to express the area under the graph of y=1t from t=1 to ex and evaluate the integral.
28) Write an integral to express the area under the graph of y=et between t=0 and t=lnx, and evaluate the integral.
- Answer
- ∫lnx0etdt=et|lnx0=elnx−e0=x−1
In exercises 29 - 35, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.
29) ∫tan(2x)dx
30) ∫sin(3x)−cos(3x)sin(3x)+cos(3x)dx
- Answer
- ∫sin(3x)−cos(3x)sin(3x)+cos(3x)dx=−13ln|sin(3x)+cos(3x)|+C
31) ∫xsin(x2)cos(x2)dx
32) ∫xcsc(x2)dx
- Answer
- ∫xcsc(x2)dx=−12ln∣csc(x2)+cot(x2)∣+C
33) ∫ln(cosx)tanxdx
34) ∫ln(cscx)cotxdx
- Answer
- ∫ln(cscx)cotxdx=−12(ln(cscx))2+C
35) ∫ex−e−xex+e−xdx
In exercises 36 - 40, evaluate the definite integral.
36) ∫211+2x+x23x+3x2+x3dx
- Answer
- ∫211+2x+x23x+3x2+x3dx=13ln(267)
37) ∫π/40tanxdx
38) ∫π/30sinx−cosxsinx+cosxdx
- Answer
- ∫π/30sinx−cosxsinx+cosxdx=ln(√3−1)
39) ∫π/2π/6cscxdx
40) ∫π/3π/4cotxdx
- Answer
- ∫π/3π/4cotxdx=12ln32
In exercises 41 - 46, integrate using the indicated substitution.
41) ∫xx−100dx;u=x−100
42) ∫y−1y+1dy;u=y+1
- Answer
- ∫y−1y+1dy=y−2ln|y+1|+C
43) ∫1−x23x−x3dx;u=3x−x3
44) \displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx;\quad u=\sin x−\cos x
- Answer
- \displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx \quad=\quad \ln|\sin x−\cos x|+C
45) \displaystyle ∫e^{2x}\sqrt{1−e^{2x}}\,dx;\quad u=1−e^{2x}
46) \displaystyle ∫\ln(x)\frac{\sqrt{1−(\ln x)^2}}{x}\,dx;\quad u=1−(\ln x)^2
- Answer
- \displaystyle ∫\ln(x)\frac{\sqrt{1−(\ln x)^2}}{x}\,dx \quad = \quad −\frac{1}{3}(1−(\ln x)^2)^{3/2}+C
47) \displaystyle \int \frac{\sqrt{x}}{\sqrt{x} + 2}\,dx; \quad u = \sqrt{x} + 2
- Answer
- \displaystyle \int \frac{\sqrt{x}}{\sqrt{x} + 2}\,dx \quad = \quad \left( \sqrt{x} + 2 \right)^2 - 8\left( \sqrt{x} + 2 \right) + 8\ln\left( \sqrt{x} + 2 \right) + C
48) \displaystyle \int e^x\sec(e^x+1)\tan(e^x+1)\,dx; \quad u = e^{x} + 1
- Answer
- \displaystyle \int e^x\sec(e^x+1)\tan(e^x+1)\,dx \quad = \quad \sec(e^x+1) + C
In exercises 49 - 54, state whether the right-endpoint approximation overestimates or underestimates the exact area. Then calculate the right endpoint estimate R_{50} and solve for the exact area.
49) [T] y=e^x over [0,1]
50) [T] y=e^{−x} over [0,1]
- Answer
- Since f is decreasing, the right endpoint estimate underestimates the area.
Exact solution: \dfrac{e−1}{e},\quad R_{50}=0.6258.
51) [T] y=\ln(x) over [1,2]
52) [T] y=\dfrac{x+1}{x^2+2x+6} over [0,1]
- Answer
- Since f is increasing, the right endpoint estimate overestimates the area.
Exact solution: \dfrac{2\ln(3)−\ln(6)}{2},\quad R_{50}=0.2033.
53) [T] y=2^x over [−1,0]
54) [T] y=−2^{−x} over [0,1]
- Answer
- Since f is increasing, the right endpoint estimate overestimates the area (the actual area is a larger negative number).
Exact solution: −\dfrac{1}{\ln(4)},\quad R_{50}=−0.7164.
In exercises 55 - 58, f(x)≥0 for a≤x≤b. Find the area under the graph of f(x) between the given values a and b by integrating.
55) f(x)=\dfrac{\log_{10}(x)}{x};\quad a=10,b=100
56) f(x)=\dfrac{\log_2(x)}{x};\quad a=32,b=64
- Answer
- \dfrac{11}{2}\ln 2
57) f(x)=2^{−x};\quad a=1,b=2
58) f(x)=2^{−x};\quad a=3,b=4
- Answer
- \dfrac{1}{\ln(65,536)}
59) Find the area under the graph of the function f(x)=xe^{−x^2} between x=0 and x=5.
60) Compute the integral of f(x)=xe^{−x^2} and find the smallest value of N such that the area under the graph f(x)=xe^{−x^2} between x=N and x=N+1 is, at most, 0.01.
- Answer
- \displaystyle ∫^{N+1}_Nxe^{−x^2}\,dx=\frac{1}{2}(e^{−N^2}−e^{−(N+1)^2}). The quantity is less than 0.01 when N=2.
61) Find the limit, as N tends to infinity, of the area under the graph of f(x)=xe^{−x^2} between x=0 and x=5.
62) Show that \displaystyle ∫^b_a\frac{dt}{t}=∫^{1/a}_{1/b}\frac{dt}{t} when 0<a≤b.
- Answer
- \displaystyle ∫^b_a\frac{dx}{x}=\ln(b)−\ln(a)=\ln(\frac{1}{a})−\ln(\frac{1}{b})=∫^{1/a}_{1/b}\frac{dx}{x}
63) Suppose that f(x)>0 for all x and that f and g are differentiable. Use the identity f^g=e^{g\ln f} and the chain rule to find the derivative of f^g.
64) Use the previous exercise to find the antiderivative of h(x)=x^x(1+\ln x) and evaluate \displaystyle ∫^3_2x^x(1+\ln x)\,dx.
- Answer
- 23
65) Show that if c>0, then the integral of \frac{1}{x} from ac to bc (\text{for}\,0<a<b) is the same as the integral of \frac{1}{x} from a to b.
The following exercises are intended to derive the fundamental properties of the natural log starting from the definition \displaystyle \ln(x)=∫^x_1\frac{dt}{t}, using properties of the definite integral and making no further assumptions.
66) Use the identity \displaystyle \ln(x)=∫^x_1\frac{dt}{t} to derive the identity \ln\left(\dfrac{1}{x}\right)=−\ln x.
- Answer
- We may assume that x>1,so \dfrac{1}{x}<1. Then, \displaystyle ∫^{1/x}_{1}\frac{dt}{t}. Now make the substitution u=\dfrac{1}{t}, so du=−\dfrac{dt}{t^2} and \dfrac{du}{u}=−\dfrac{dt}{t}, and change endpoints: \displaystyle ∫^{1/x}_1\frac{dt}{t}=−∫^x_1\frac{du}{u}=−\ln x.
67) Use a change of variable in the integral \displaystyle ∫^{xy}_1\frac{1}{t}\,dt to show that \ln xy=\ln x+\ln y for x,y>0.
68) Use the identity \displaystyle \ln x=∫^x_1\frac{dt}{t} to show that \ln(x) is an increasing function of x on [0,∞), and use the previous exercises to show that the range of \ln(x) is (−∞,∞). Without any further assumptions, conclude that \ln(x) has an inverse function defined on (−∞,∞).
69) Pretend, for the moment, that we do not know that e^x is the inverse function of \ln(x), but keep in mind that \ln(x) has an inverse function defined on (−∞,∞). Call it E. Use the identity \ln xy=\ln x+\ln y to deduce that E(a+b)=E(a)E(b) for any real numbers a, b.
70) Pretend, for the moment, that we do not know that e^x is the inverse function of \ln x, but keep in mind that \ln x has an inverse function defined on (−∞,∞). Call it E. Show that E'(t)=E(t).
- Answer
- x=E(\ln(x)). Then, 1=\dfrac{E'(\ln x)}{x} or x=E'(\ln x). Since any number t can be written t=\ln x for some x, and for such t we have x=E(t), it follows that for any t,\,E'(t)=E(t).
71) The sine integral, defined as \displaystyle S(x)=∫^x_0\frac{\sin t}{t}\,dt is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x. Show that for k≥1,\quad |S(2πk)−S(2π(k+1))|≤\dfrac{1}{k(2k+1)π}. (Hint: \sin(t+π)=−\sin t)
72) [T] The normal distribution in probability is given by p(x)=\dfrac{1}{σ\sqrt{2π}}e^{−(x−μ)^2/2σ^2}, where σ is the standard deviation and μ is the average. The standard normal distribution in probability, p_s, corresponds to μ=0 and σ=1. Compute the left endpoint estimates R_{10} and R_{100} of \displaystyle ∫^1_{−1}\frac{1}{\sqrt{2π}}e^{−x^{2/2}}\,dx.
- Answer
- R_{10}=0.6811,\quad R_{100}=0.6827
73) [T] Compute the right endpoint estimates R_{50} and R_{100} of \displaystyle ∫^5_{−3}\frac{1}{2\sqrt{2π}}e^{−(x−1)^2/8}.
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
- Paul Seeburger (Monroe Community College) added problems 47-48 to Section 5.6 exercises.