5.6E: Exercises for Section 5.6
- Page ID
- 53724
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1) \(\displaystyle ∫e^{2x}\,dx\)
2) \(\displaystyle ∫e^{−3x}\,dx\)
- Answer
- \(\displaystyle ∫e^{−3x}\,dx \quad = \quad \frac{−1}{3}e^{−3x}+C\)
3) \(\displaystyle ∫2^x\,dx\)
4) \(\displaystyle ∫3^{−x}\,dx\)
- Answer
- \(\displaystyle ∫3^{−x}\,dx \quad = \quad −\frac{3^{−x}}{\ln 3}+C\)
5) \(\displaystyle ∫\frac{1}{2x}\,dx\)
6) \(\displaystyle ∫\frac{2}{x}\,dx\)
- Answer
- \(\displaystyle ∫\frac{2}{x}\,dx \quad = \quad 2\ln |x|+C \quad = \quad \ln(x^2)+C\)
7) \(\displaystyle ∫\frac{1}{x^2}\,dx\)
8) \(\displaystyle ∫\frac{1}{\sqrt{x}}\,dx\)
- Answer
- \(\displaystyle ∫\frac{1}{\sqrt{x}}\,dx \quad = \quad 2\sqrt{x}+C\)
In exercises 9 - 16, find each indefinite integral by using appropriate substitutions.
9) \(\displaystyle ∫\frac{\ln x}{x}\,dx\)
10) \(\displaystyle ∫\frac{dx}{x(\ln x)^2}\)
- Answer
- \(\displaystyle ∫\frac{dx}{x(\ln x)^2} \quad = \quad −\frac{1}{\ln x}+C\)
11) \(\displaystyle ∫\frac{dx}{x\ln x}\quad (x>1)\)
12) \(\displaystyle ∫\frac{dx}{x\ln x\ln(\ln x)}\)
- Answer
- \(\displaystyle ∫\frac{dx}{x\ln x\ln(\ln x)} \quad = \quad \ln(\ln(\ln x))+C\)
13) \(\displaystyle ∫\tan θ\,dθ\)
14) \(\displaystyle ∫\frac{\cos x−x\sin x}{x\cos x}\,dx\)
- Answer
- \(\displaystyle ∫\frac{\cos x−x\sin x}{x\cos x}\,dx \quad = \quad \ln(x\cos x)+C\)
15) \(\displaystyle ∫\frac{\ln(\sin x)}{\tan x}\,dx\)
16) \(\displaystyle ∫\ln(\cos x)\tan x\,dx\)
- Answer
- \(\displaystyle ∫\ln(\cos x)\tan x\,dx \quad = \quad −\dfrac{1}{2}(\ln(\cos(x)))^2+C\)
17) \(\displaystyle ∫xe^{−x^2}\,dx\)
18) \(\displaystyle ∫x^2e^{−x^3}\,dx\)
- Answer
- \(\displaystyle ∫x^2e^{−x^3}\,dx \quad = \quad \dfrac{−e^{−x^3}}{3}+C\)
19) \(\displaystyle ∫e^{\sin x}\cos x\,dx\)
20) \(\displaystyle ∫e^{\tan x}\sec^2 x\,dx\)
- Answer
- \(\displaystyle ∫e^{\tan x}\sec^2 x\,dx\quad = \quad e^{\tan x}+C\)
21) \(\displaystyle ∫\frac{e^{\ln x}}{x}\,dx \)
22) \(\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt\)
- Answer
- \(\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt = \int \frac{1-t}{1-t}\,dt = \int 1\, dt \quad = \quad t+C\)
In exercises 23 - 28, verify by differentiation that \(\displaystyle ∫\ln x\,dx=x(\ln x−1)+C\), then use appropriate changes of variables to compute the integral.
23) \(\displaystyle ∫\ln x\,dx\) (Hint: \(\displaystyle ∫\ln x\,dx=\frac{1}{2}∫x\ln(x^2)\,dx\))
24) \(\displaystyle ∫x^2\ln^2 x\,dx\)
- Answer
- \(\displaystyle ∫x^2\ln^2 x\,dx \quad = \quad \dfrac{1}{9}x^3(\ln(x^3)−1)+C\)
25) \(\displaystyle ∫\frac{\ln x}{x^2}\,dx\) (Hint: Set \(u=\dfrac{1}{x}.)\)
26) \(\displaystyle ∫\frac{\ln x}{\sqrt{x}}\,dx\) (Hint: Set \(u=\sqrt{x}.)\)
- Answer
- \( \displaystyle ∫\frac{\ln x}{\sqrt{x}}\,dx \quad = \quad 2\sqrt{x}(\ln x−2)+C\)
27) Write an integral to express the area under the graph of \(y=\dfrac{1}{t}\) from \( t=1\) to \(e^x\) and evaluate the integral.
28) Write an integral to express the area under the graph of \(y=e^t\) between \(t=0\) and \(t=\ln x\), and evaluate the integral.
- Answer
- \(\displaystyle ∫^{\ln x}_0e^t\,dt=e^t\bigg|^{\ln x}_0=e^{\ln x}−e^0=x−1\)
In exercises 29 - 35, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.
29) \(\displaystyle ∫\tan(2x)\,dx\)
30) \(\displaystyle ∫\frac{\sin(3x)−\cos(3x)}{\sin(3x)+\cos(3x)}\,dx\)
- Answer
- \( \displaystyle ∫\frac{\sin(3x)−\cos(3x)}{\sin(3x)+\cos(3x)}\,dx \quad = \quad −\frac{1}{3}\ln|\sin(3x)+\cos(3x)| + C\)
31) \(\displaystyle ∫\frac{x\sin(x^2)}{\cos(x^2)}\,dx\)
32) \(\displaystyle ∫x\csc(x^2)\,dx\)
- Answer
- \( \displaystyle ∫x\csc(x^2)\,dx \quad = \quad −\frac{1}{2}\ln∣\csc(x^2)+\cot(x^2)∣+C\)
33) \(\displaystyle ∫\ln(\cos x)\tan x\,dx\)
34) \(\displaystyle ∫\ln(\csc x)\cot x\,dx\)
- Answer
- \( \displaystyle ∫\ln(\csc x)\cot x\,dx \quad = \quad −\frac{1}{2}(\ln(\csc x))^2+C\)
35) \(\displaystyle ∫\frac{e^x−e^{−x}}{e^x+e^{−x}}\,dx\)
In exercises 36 - 40, evaluate the definite integral.
36) \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx\)
- Answer
- \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx \quad = \quad \frac{1}{3}\ln\left(\tfrac{26}{7}\right)\)
37) \(\displaystyle ∫^{π/4}_0\tan x\,dx\)
38) \(\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx\)
- Answer
- \(\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx \quad = \quad \ln(\sqrt{3}−1)\)
39) \(\displaystyle ∫^{π/2}_{π/6}\csc x\,dx\)
40) \(\displaystyle ∫^{π/3}_{π/4}\cot x\,dx\)
- Answer
- \(\displaystyle ∫^{π/3}_{π/4}\cot x\,dx \quad = \quad \frac{1}{2}\ln\frac{3}{2}\)
In exercises 41 - 46, integrate using the indicated substitution.
41) \(\displaystyle ∫\frac{x}{x−100}\,dx;\quad u=x−100\)
42) \(\displaystyle ∫\frac{y−1}{y+1}\,dy;\quad u=y+1\)
- Answer
- \( \displaystyle ∫\frac{y−1}{y+1}\,dy \quad = \quad y−2\ln|y+1|+C\)
43) \(\displaystyle ∫\frac{1−x^2}{3x−x^3}\,dx;\quad u=3x−x^3\)
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 \(x=N+10\) 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 \(\displaystyle ∫^5_{−3}\frac{1}{2\sqrt{2π}}e^{−(x−1)^2/8}\).
Contributors and Attributions
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.