4.6E: Exercises for Section 4.6

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

2) If \( f=g∘h\), when reversing the chain rule, \(\dfrac{d}{dx}(g∘h)(x)=g′(h(x))h′(x)\), 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 \(\displaystyle∫f(u)\,du.\)

3) \(\displaystyle ∫\frac{1}{\sqrt{2x+1}}\,dx=\sqrt{2x+1}+C;\quad u=2x+1\)

4) \(\displaystyle∫xe^{x^2-3}\,dx=\frac{1}{2}e^{x^2-3}+C,\quad u=x^2-3\)

Answer

\(du=2x\,dx;\quad f(u)=\frac{1}{2}e^u\)

5) \(\displaystyle∫x\sqrt{4x^2+9}\,dx=\frac{1}{12}(4x^2+9)^{3/2}+C;\quad u=4x^2+9\)

6) \(\displaystyle∫\frac{x}{\sqrt{4x^2+9}}\,dx=\frac{1}{4}\sqrt{4x^2+9}+C;\quad u=4x^2+9\)

Answer

\( du=8x\,dx;\quad f(u)=\frac{1}{8\sqrt{u}}\)

7) \(\displaystyle∫\frac{x}{(4x^2+9)^2}\,dx=−\frac{1}{8(4x^2+9)} + C;\quad u=4x^2+9\)

In exercises 8 - 15, find the antiderivative using the indicated substitution.

8) \(\displaystyle∫3\left(3x+4\right)^4\,dx;\quad u=3x+4\)

Answer

\(\displaystyle∫3\left(3x+4\right)^4\,dx = \frac{1}{5}\left(3x+4\right)^5+C\)

9) \(\displaystyle∫e^{7x+1}\,dx;\quad u=7x+1\)

10) \(\displaystyle∫(2x−3)^{−7}\,dx;\quad u=2x−3\)

Answer

\(\displaystyle∫(2x−3)^{−7}\,dx = −\frac{1}{12(2x−3)^6}+C\)

11) \(\displaystyle∫(3x−2)^{−1}\,dx;\quad u=3x−2\)

12) \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx;\quad u=x^2+1\)

Answer

\(\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx = \sqrt{x^2+1}+C\)

13) \(\displaystyle∫x\sqrt{1−x^2}\,dx;\quad u=1−x^2\)

14) \(\displaystyle∫(x−1)(x^2−2x)^3\,dx;\quad u=x^2−2x\)

Answer

\(\displaystyle∫(x−1)(x^2−2x)^3\,dx = \frac{1}{8}(x^2−2x)^4+C\)

15) \(\displaystyle∫(x^2−2x)(x^3−3x^2)^2\,dx;\quad u=x^3=3x^2\)

In exercises 16-20, use a suitable change of variables to determine the indefinite integral.

16) \(\displaystyle∫\frac{1}{1-2x}\,dx\)

Answer

\(\displaystyle∫\frac{1}{1-2x}\,dx= -\frac{1}{2}\ln{|1-2x|}+C \)

17) \(\displaystyle∫t(1−t^2)^{10}dt\)

18) \(\displaystyle∫(11x−7)^{−3}\,dx\)

Answer

\(\displaystyle∫(11x−7)^{−3}\,dx = −\frac{1}{22(11x−7)^2}+C\)

19) \(\displaystyle∫(7x−11)^4\,dx\)

20) \(\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx\)

Answer

\(\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx = −\frac{1}{3(x^3−3)}+C\)

In exercises 21-26, use a change of variables to evaluate the definite integral.

21) \(\displaystyle∫^1_0x\sqrt{1−x^2}\,dx\)

22) \(\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx\)

Answer

\(\displaystyle u=1+x^2,\quad du=2x\,dx,\quad ∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1\)

23) \(\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}\,dt\)

24) \(\displaystyle∫^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt\)

Answer

\(\displaystyle u=1+t^3,\quad du=3t^2,\quad ∫^1_0\frac{t^2}{\sqrt{1+t^3}}\,dt = \frac{1}{3}∫^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)\)

25) \(\displaystyle∫^{0}_{-1}e^{-2x}\,dx\)

26) \(\displaystyle∫^{2}_0\frac{2}{4x+1}\,dx\)

Answer

\(\displaystyle u=4x+1,\quad du=4dx,\quad \displaystyle∫^{2}_0\frac{2}{4x+1}\,dx =\frac{1}{2}∫^{9}_{1}\frac{1}{u}du = \frac{1}{2}\ln{9}\)

27) If \(h(a)=h(b)\) in \(\displaystyle ∫^b_ag'(h(x))h(x)\,dx,\) what can you say about the value of the integral?

28) Is the substitution \(u=1−x^2\) in the definite integral \(\displaystyle ∫^2_0\frac{x}{1−x^2}\,dx\) okay? If not, why not?

Answer

No, because the integrand is discontinuous at \(x=1\).

In exercises 29-30, use a change of variables to show that each definite integral is equal to zero.

29) \(\displaystyle ∫^1_0(1−2t)\,dt\)

30) \(\displaystyle ∫^1_0\frac{1−2t}{1+(t−\frac{1}{2})^2}\,dt\)

Answer

\(u=1+(t−\frac{1}{2})^2;\) the integral becomes \(\displaystyle −∫^{5/4}_{5/4}\frac{1}{u}\,du\).

31) 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 \(\left[\frac{a}{c},\frac{b}{c}\right]\) for \(c>0.\)

Answer

Setting \(u=cx\) and \(du=c\,dx\) gets you \(\displaystyle \frac{1}{\frac{b}{c}−\frac{a}{c}}∫^{b/c}_{a/c}f(cx)\,dx=\frac{c}{b−a}∫^{u=b}_{u=a}f(u)\frac{du}{c}=\frac{1}{b−a}∫^b_af(u)\,du.\)

32) Find the area under the graph of \(f(t)=\dfrac{t}{(1+t^2)^a}\) between \(t=0\) and \(t=x\) where \(a>0\) and \(a≠1\) is fixed, and evaluate the limit as \(x→∞\).

33) Find the area under the graph of \(g(t)=\dfrac{t}{(1−t^2)^a}\) between \(t=0\) and \(t=x\), where \(0<x<1\) and \(a>0\) is fixed. Evaluate the limit as \(x→1\).

Answer

\(\displaystyle ∫^x_0g(t)\,dt=\frac{1}{2}∫^1_{u=1−x^2} \frac{du}{u^a}=\frac{1}{2(1−a)}u^{1−a}∣1u=\frac{1}{2(1−a)}(1−(1−x^2)^{1−a})\) As \(x→1\) the limit is \(\dfrac{1}{2(1−a)}\) if \(a<1\), and the limit diverges to \(+∞\) if \(a>1\).