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Mathematics LibreTexts

5.6E: Exercises for Section 5.6

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For exercises 1 - 8, compute each indefinite integral.

1) e2xdx

2) e3xdx

Answer
e3xdx=13e3x+C

3) 2xdx

4) 3xdx

Answer
3xdx=3xln3+C

5) 12xdx

6) 2xdx

Answer
2xdx=2ln|x|+C=ln(x2)+C

7) 1x2dx

8) 1xdx

Answer
1xdx=2x+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) cosxxsinxxcosxdx

Answer
cosxxsinxxcosxdx=ln(xcosx)+C

15) ln(sinx)tanxdx

16) ln(cosx)tanxdx

Answer
ln(cosx)tanxdx=12(ln(cos(x)))2+C

17) xex2dx

18) x2ex3dx

Answer
x2ex3dx=ex33+C

19) esinxcosxdx

20) etanxsec2xdx

Answer
etanxsec2xdx=etanx+C

21) elnxxdx

22) eln(1t)1tdt

Answer
eln(1t)1tdt=1t1tdt=1dt=t+C

In exercises 23 - 28, verify by differentiation that lnxdx=x(lnx1)+C, then use appropriate changes of variables to compute the integral.

23) lnxdx (Hint: lnxdx=12xln(x2)dx)

24) x2ln2xdx

Answer
x2ln2xdx=19x3(ln(x3)1)+C

25) lnxx2dx (Hint: Set u=1x.)

26) lnxxdx (Hint: Set u=x.)

Answer
lnxxdx=2x(lnx2)+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=elnxe0=x1

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=12lncsc(x2)+cot(x2)+C

33) ln(cosx)tanxdx

34) ln(cscx)cotxdx

Answer
ln(cscx)cotxdx=12(ln(cscx))2+C

35) exexex+exdx

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/30sinxcosxsinx+cosxdx

Answer
π/30sinxcosxsinx+cosxdx=ln(31)

39) π/2π/6cscxdx

40) π/3π/4cotxdx

Answer
π/3π/4cotxdx=12ln32

In exercises 41 - 46, integrate using the indicated substitution.

41) xx100dx;u=x100

42) y1y+1dy;u=y+1

Answer
y1y+1dy=y2ln|y+1|+C

43) 1x23xx3dx;u=3xx3

44) sinx+cosxsinxcosxdx;u=sinxcosx

Answer
sinx+cosxsinxcosxdx=ln|sinxcosx|+C

45) e2x1e2xdx;u=1e2x

46) ln(x)1(lnx)2xdx;u=1(lnx)2

Answer
ln(x)1(lnx)2xdx=13(1(lnx)2)3/2+C

47) xx+2dx;u=x+2

Answer
xx+2dx=(x+2)28(x+2)+8ln(x+2)+C

48) exsec(ex+1)tan(ex+1)dx;u=ex+1

Answer
exsec(ex+1)tan(ex+1)dx=sec(ex+1)+C

In exercises 49 - 54, state whether the right-endpoint approximation overestimates or underestimates the exact area. Then calculate the right endpoint estimate R50 and solve for the exact area.

49) [T] y=ex over [0,1]

50) [T] y=ex over [0,1]

Answer
Since f is decreasing, the right endpoint estimate underestimates the area.
Exact solution: e1e,R50=0.6258.

51) [T] y=ln(x) over [1,2]

52) [T] y=x+1x2+2x+6 over [0,1]

Answer
Since f is increasing, the right endpoint estimate overestimates the area.
Exact solution: 2ln(3)ln(6)2,R50=0.2033.

53) [T] y=2x over [1,0]

54) [T] y=2x 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: 1ln(4),R50=0.7164.

In exercises 55 - 58, f(x)0 for axb. Find the area under the graph of f(x) between the given values a and b by integrating.

55) f(x)=log10(x)x;a=10,b=100

56) f(x)=log2(x)x;a=32,b=64

Answer
112ln2

57) f(x)=2x;a=1,b=2

58) f(x)=2x;a=3,b=4

Answer
1ln(65,536)

59) Find the area under the graph of the function f(x)=xex2 between x=0 and x=5.

60) Compute the integral of f(x)=xex2 and find the smallest value of N such that the area under the graph f(x)=xex2 between x=N and x=N+1 is, at most, 0.01.

Answer
N+1Nxex2dx=12(eN2e(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)=xex2 between x=0 and x=5.

62) Show that badtt=1/a1/bdtt when 0<ab.

Answer
badxx=ln(b)ln(a)=ln(1a)ln(1b)=1/a1/bdxx

63) Suppose that f(x)>0 for all x and that f and g are differentiable. Use the identity fg=eglnf and the chain rule to find the derivative of fg.

64) Use the previous exercise to find the antiderivative of h(x)=xx(1+lnx) and evaluate 32xx(1+lnx)dx.

Answer
23

65) Show that if c>0, then the integral of 1x from ac to bc (for0<a<b) is the same as the integral of 1x from a to b.

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition ln(x)=x1dtt, using properties of the definite integral and making no further assumptions.

66) Use the identity ln(x)=x1dtt to derive the identity ln(1x)=lnx.

Answer
We may assume that x>1,so 1x<1. Then, 1/x1dtt. Now make the substitution u=1t, so du=dtt2 and duu=dtt, and change endpoints: 1/x1dtt=x1duu=lnx.

67) Use a change of variable in the integral xy11tdt to show that lnxy=lnx+lny for x,y>0.

68) Use the identity lnx=x1dtt 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 ex 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 lnxy=lnx+lny 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 ex is the inverse function of lnx, but keep in mind that lnx has an inverse function defined on (,). Call it E. Show that E(t)=E(t).

Answer
x=E(ln(x)). Then, 1=E(lnx)x or x=E(lnx). Since any number t can be written t=lnx 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 S(x)=x0sinttdt 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 k1,|S(2πk)S(2π(k+1))|1k(2k+1)π. (Hint: sin(t+π)=sint)

72) [T] The normal distribution in probability is given by p(x)=1σ2πe(xμ)2/2σ2, where σ is the standard deviation and μ is the average. The standard normal distribution in probability, ps, corresponds to μ=0 and σ=1. Compute the left endpoint estimates R10 and R100 of 1112πex2/2dx.

Answer
R10=0.6811,R100=0.6827

A graph of the function f(x) = .5 * ( sqrt(2)*e^(-.5x^2)) / sqrt(pi). It is a downward opening curve that is symmetric across the y axis, crossing at about (0, .4). It approaches 0 as x goes to positive and negative infinity. Between 1 and -1, ten rectangles are drawn for a right endpoint estimate of the area under the curve.

73) [T] Compute the right endpoint estimates R50 and R100 of 53122πe(x1)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.

5.6E: Exercises for Section 5.6 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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