5.5E: Exercises for Section 5.5
( \newcommand{\kernel}{\mathrm{null}\,}\)
1) Why is u-substitution referred to as a change of variable?
2) If f=g∘h, when reversing the chain rule, ddx(g∘h)(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) ∫x√x+1dx=215(x+1)3/2(3x−2)+C;u=x+1
4) ∫x2√x−1dx=215√x−1(3x2+4x+8)+C,(x>1);u=x−1
- Answer
- f(u)=(u+1)2√u
5) ∫x√4x2+9dx=112(4x2+9)3/2+C;u=4x2+9
6) ∫x√4x2+9dx=14√4x2+9+C;u=4x2+9
- Answer
- du=8xdx;f(u)=18√u
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) ∫(x−1)5dx;u=x−1
10) ∫(2x−3)−7dx;u=2x−3
- Answer
- ∫(2x−3)−7dx=−112(2x−3)6+C
11) ∫(3x−2)−11dx;u=3x−2
12) ∫x√x2+1dx;u=x2+1
- Answer
- ∫x√x2+1dx=√x2+1+C
13) ∫x√1−x2dx;u=1−x2
14) ∫(x−1)(x2−2x)3dx;u=x2−2x
- Answer
- ∫(x−1)(x2−2x)3dx=18(x2−2x)4+C
15) ∫(x2−2x)(x3−3x2)2dx;u=x3=3x2
16) ∫cos3θdθ;u=sinθ (Hint: cos2θ=1−sin2θ)
- Answer
- ∫cos3θdθ=sinθ−sin3θ3+C
17) ∫sin3θdθ;u=cosθ (Hint: sin2θ=1−cos2θ)
In exercises 18 - 34, use a suitable change of variables to determine the indefinite integral.
18) ∫x(1−x)99dx
- Answer
- ∫x(1−x)99dx=(1−x)101101−(1−x)100100+C=−(1−x)10010100[100x+1]+C
19) ∫t(1−t2)10dt
20) ∫(11x−7)−3dx
- Answer
- ∫(11x−7)−3dx=−122(11x−7)2+C
21) ∫(7x−11)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(x3−3)2dx
- Answer
- ∫x2(x3−3)2dx=−13(x3−3)+C
29) ∫x3√1−x2dx
30) ∫y5(1−y3)3/2dy
- Answer
- ∫y5(1−y3)3/2dy=−2(y3−2)3√1−y3+C
31) ∫cosθ(1−cosθ)99sinθdθ
32) ∫(1−cos3θ)10cos2θsinθdθ
- Answer
- ∫(1−cos3θ)10cos2θsinθdθ=133(1−cos3θ)11+C
33) ∫(cosθ−1)(cos2θ−2cosθ)3sinθdθ
34) ∫(sin2θ−2sinθ)(sin3θ−3sin2θ)3cosθdθ
- Answer
- ∫(sin2θ−2sinθ)(sin3θ−3sin2θ)3cosθdθ=112(sin3θ−3sin2θ)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(1−x)2 over [0,2]
36) [T] y=x(1−x2)3 over [−1,2]
- Answer
- L50=−8.5779. The exact area is −818 units2.
37) [T] y=sinx(1−cosx)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) ∫10x√1−x2dx
40) ∫10x√1+x2dx
- Answer
- u=1+x2,du=2xdx,∫10x√1+x2dx=12∫21u−1/2du=√2−1
41) ∫20t√5+t2dt
42) ∫10t2√1+t3dt
- Answer
- u=1+t3,du=3t2,∫10t2√1+t3dt=13∫21u−1/2du=23(√2−1)
43) ∫π/40sec2θtanθdθ
44) ∫π/40sinθcos4θdθ
- Answer
- u=cosθ,du=−sinθdθ,∫π/40sinθcos4θdθ=−∫√2/21u−4du=∫1√2/2u−4du=13(2√2−1)
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+x−6dx over [−3,2]
46) [T] ∫cos(ln(2x))xdx on [0,2]
- Answer
-
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+1√x3+x2+x+4dx over [−1,2]
48) [T] ∫sinxcos3xdx over [−π3,π3]
- Answer
-
The antiderivative is y=12sec2x. You should take C=−2 so that F(−π3)=0.
49) [T] ∫(x+2)e−x2−4x+3dx over [−5,1]
50) [T] ∫3x2√2x3+1dx over [0,1]
- Answer
-
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=1−x2 in the definite integral ∫20x1−x2dx 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 12∫00udu.
55) ∫10(1−2t)dt
56) ∫101−2t1+(t−12)2dt
- Answer
- u=1+(t−12)2; the integral becomes −∫5/45/41udu.
57) ∫π0sin((t−π2)3)cos(t−π2)dt
58) ∫20(1−t)cos(πt)dt
- Answer
- u=1−t; Since the integrand is odd, the integral becomes
∫−11ucos(π(1−u))du=∫−11u[cosπcosu−sinπsinu]du=−∫−11ucosudu=∫1−1ucosudu=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 1bc−ac∫b/ca/cf(cx)dx=cb−a∫u=bu=af(u)duc=1b−a∫baf(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 a≠1 is fixed, and evaluate the limit as x→∞.
62) Find the area under the graph of g(t)=t(1−t2)a between t=0 and t=x, where 0<x<1 and a>0 is fixed. Evaluate the limit as x→1.
- Answer
- ∫x0g(t)dt=12∫1u=1−x2duua=12(1−a)u1−a∣1u=12(1−a)(1−(1−x2)1−a) As x→1 the limit is 12(1−a) if a<1, and the limit diverges to +∞ if a>1.
63) The area of a semicircle of radius 1 can be expressed as ∫1−1√1−x2dx. 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 ∫a−ab√1−x2a2dx. 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=πb√1−cos2t×(−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.
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.
- Answer
- f(t)=2cos(3t)−cos(2t);∫π/20(2cos(3t)−cos(2t))dt=−23