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Mathematics LibreTexts

5.5E: Exercises for Section 5.5

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1) Why is u-substitution referred to as a change of variable?

2) If f=gh, when reversing the chain rule, ddx(gh)(x)=g(h(x))h(x), should you take u=g(x) or u=h(x)?

Answer
u=h(x)

In exercises 3 - 7, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form f(u)du.

3) xx+1dx=215(x+1)3/2(3x2)+C;u=x+1

4) x2x1dx=215x1(3x2+4x+8)+C,(x>1);u=x1

Answer
f(u)=(u+1)2u

5) x4x2+9dx=112(4x2+9)3/2+C;u=4x2+9

6) x4x2+9dx=144x2+9+C;u=4x2+9

Answer
du=8xdx;f(u)=18u

7) x(4x2+9)2dx=18(4x2+9)+C;u=4x2+9

In exercises 8 - 17, find the antiderivative using the indicated substitution.

8) (x+1)4dx;u=x+1

Answer
(x+1)4dx=15(x+1)5+C

9) (x1)5dx;u=x1

10) (2x3)7dx;u=2x3

Answer
(2x3)7dx=112(2x3)6+C

11) (3x2)11dx;u=3x2

12) xx2+1dx;u=x2+1

Answer
xx2+1dx=x2+1+C

13) x1x2dx;u=1x2

14) (x1)(x22x)3dx;u=x22x

Answer
(x1)(x22x)3dx=18(x22x)4+C

15) (x22x)(x33x2)2dx;u=x3=3x2

16) cos3θdθ;u=sinθ (Hint: cos2θ=1sin2θ)

Answer
cos3θdθ=sinθsin3θ3+C

17) sin3θdθ;u=cosθ (Hint: sin2θ=1cos2θ)

In exercises 18 - 34, use a suitable change of variables to determine the indefinite integral.

18) x(1x)99dx

Answer
x(1x)99dx=(1x)101101(1x)100100+C=(1x)10010100[100x+1]+C

19) t(1t2)10dt

20) (11x7)3dx

Answer
(11x7)3dx=122(11x7)2+C

21) (7x11)4dx

22) cos3θsinθdθ

Answer
cos3θsinθdθ=cos4θ4+C

23) sin7θcosθdθ

24) cos2(πt)sin(πt)dt

Answer
cos2(πt)sin(πt)dt=cos3(πt)3π+C

25) sin2xcos3xdx (Hint: sin2x+cos2x=1)

26) tsin(t2)cos(t2)dt

Answer
tsin(t2)cos(t2)dt=14cos2(t2)+C

27) t2cos2(t3)sin(t3)dt

28) x2(x33)2dx

Answer
x2(x33)2dx=13(x33)+C

29) x31x2dx

30) y5(1y3)3/2dy

Answer
y5(1y3)3/2dy=2(y32)31y3+C

31) cosθ(1cosθ)99sinθdθ

32) (1cos3θ)10cos2θsinθdθ

Answer
(1cos3θ)10cos2θsinθdθ=133(1cos3θ)11+C

33) (cosθ1)(cos2θ2cosθ)3sinθdθ

34) (sin2θ2sinθ)(sin3θ3sin2θ)3cosθdθ

Answer
(sin2θ2sinθ)(sin3θ3sin2θ)3cosθdθ=112(sin3θ3sin2θ)4+C

In exercises 35 - 38, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer.

35) [T] y=3(1x)2 over [0,2]

36) [T] y=x(1x2)3 over [1,2]

Answer
L50=8.5779. The exact area is 818 units2.

37) [T] y=sinx(1cosx)2 over [0,π]

38) [T] y=x(x2+1)2 over [1,1]

Answer
L50=0.006399. The exact area is 0.

In exercises 39 - 44, use a change of variables to evaluate the definite integral.

39) 10x1x2dx

40) 10x1+x2dx

Answer
u=1+x2,du=2xdx,10x1+x2dx=1221u1/2du=21

41) 20t5+t2dt

42) 10t21+t3dt

Answer
u=1+t3,du=3t2,10t21+t3dt=1321u1/2du=23(21)

43) π/40sec2θtanθdθ

44) π/40sinθcos4θdθ

Answer
u=cosθ,du=sinθdθ,π/40sinθcos4θdθ=2/21u4du=12/2u4du=13(221)

In exercises 45 - 50, evaluate the indefinite integral f(x)dx with constant C=0 using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral F(x)=xaf(t)dt, with a the left endpoint of the given interval.

45) [T] (2x+1)ex2+x6dx over [3,2]

46) [T] cos(ln(2x))xdx on [0,2]

Answer

Two graphs. The first shows the function f(x) = cos(ln(2x)) / x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.

The antiderivative is y=sin(ln(2x)). Since the antiderivative is not continuous at x=0, one cannot find a value of C that would make y=sin(ln(2x))C work as a definite integral.

47) [T] 3x2+2x+1x3+x2+x+4dx over [1,2]

48) [T] sinxcos3xdx over [π3,π3]

Answer

Two graphs. The first is the function f(x) = sin(x) / cos(x)^3 over [-5pi/16, 5pi/16]. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = ½ sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.

The antiderivative is y=12sec2x. You should take C=2 so that F(π3)=0.

49) [T] (x+2)ex24x+3dx over [5,1]

50) [T] 3x22x3+1dx over [0,1]

Answer

Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1/3 * (2x^3 + 1)^(1/3). It is an increasing concave up curve starting at about 0.3.

The antiderivative is y=13(2x3+1)3/2. One should take C=13.

51) If h(a)=h(b) in bag(h(x))h(x)dx, what can you say about the value of the integral?

52) Is the substitution u=1x2 in the definite integral 20x1x2dx okay? If not, why not?

Answer
No, because the integrand is discontinuous at x=1.

In exercises 53 - 59, use a change of variables to show that each definite integral is equal to zero.

53) π0cos2(2θ)sin(2θ)dθ

54) π0tcos(t2)sin(t2)dt

Answer
u=sin(t2); the integral becomes 1200udu.

55) 10(12t)dt

56) 1012t1+(t12)2dt

Answer
u=1+(t12)2; the integral becomes 5/45/41udu.

57) π0sin((tπ2)3)cos(tπ2)dt

58) 20(1t)cos(πt)dt

Answer
u=1t; Since the integrand is odd, the integral becomes
11ucos(π(1u))du=11u[cosπcosusinπsinu]du=11ucosudu=11ucosudu=0

59) 3π/4π/4sin2tcostdt

60) Show that the average value of f(x) over an interval [a,b] is the same as the average value of f(cx) over the interval [ac,bc] for c>0.

Answer
Setting u=cx and du=cdx gets you 1bcacb/ca/cf(cx)dx=cbau=bu=af(u)duc=1babaf(u)du.

61) Find the area under the graph of f(t)=t(1+t2)a between t=0 and t=x where a>0 and a1 is fixed, and evaluate the limit as x.

62) Find the area under the graph of g(t)=t(1t2)a between t=0 and t=x, where 0<x<1 and a>0 is fixed. Evaluate the limit as x1.

Answer
x0g(t)dt=121u=1x2duua=12(1a)u1a1u=12(1a)(1(1x2)1a) As x1 the limit is 12(1a) if a<1, and the limit diverges to + if a>1.

63) The area of a semicircle of radius 1 can be expressed as 111x2dx. Use the substitution x=cost to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

64) The area of the top half of an ellipse with a major axis that is the x-axis from x=1 to a and with a minor axis that is the y-axis from y=b to y=b can be written as aab1x2a2dx. Use the substitution x=acost to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

Answer
t=0t=πb1cos2t×(asint)dt=t=πt=0absin2tdt

65) [T] The following graph is of a function of the form f(t)=asin(nt)+bsin(mt). Estimate the coefficients a and b and the frequency parameters n and m. Use these estimates to approximate π0f(t)dt.

A graph of a function of the given form over [0, 2pi], which has six turning points. They are located at just before pi/4, just after pi/2, between 3pi/4 and pi, between pi and 5pi/4, just before 3pi/2, and just after 7pi/4 at about 3, -2, 1, -1, 2, and -3. It begins at the origin and ends at (2pi, 0). It crosses the x axis between pi/4 and pi/2, just before 3pi/4, pi, just after 5pi/4, and between 3pi/2 and 4pi/4.

66) [T] The following graph is of a function of the form f(x)=acos(nt)+bcos(mt). Estimate the coefficients a and b and the frequency parameters n and m. Use these estimates to approximate π0f(t)dt.

The graph of a function of the given form over [0, 2pi]. It begins at (0,1) and ends at (2pi, 1). It has five turning points, located just after pi/4, between pi/2 and 3pi/4, pi, between 5pi/4 and 3pi/2, and just before 7pi/4 at about -1.5, 2.5, -3, 2.5, and -1. It crosses the x axis between 0 and pi/4, just before pi/2, just after 3pi/4, just before 5pi/4, just after 3pi/2, and between 7pi/4 and 2pi.

Answer
f(t)=2cos(3t)cos(2t);π/20(2cos(3t)cos(2t))dt=23

5.5E: Exercises for Section 5.5 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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