Question #9:
\(\displaystyle \int_0^1 (x^7-x^3) \, dx
= \left.\frac{x^8}{8}-\frac{x^4}{4}\right|_0^1
= \frac{1}{8}-\frac{1}{4} = -\frac{1}{8}\).
Answer: (b)
Question #10:
\( \displaystyle F(x) = \int e^{3x}-\sin x\,dx
= \frac{e^{3x}}{3}+\cos x + C\).
To find \(C\), we use
\(\displaystyle \frac{1}{3} = F(0)
= \frac{e^{0}}{3}+\cos(0) + C = \frac{1}{3}+1+C\),
so \(C=-1\). Hence
\( \displaystyle F(x) = \frac{e^{3x}}{3}+\cos x -1\).
Answer: (d)
Question #12:
\(u = {\sqrt{x}}\), so \(x=u^2\) and \(dx=2u\,du\).
\(\displaystyle \int {{\sqrt {x} }\over {{\sqrt{\sqrt{x} - 1}}}} dx
= \int \frac{u}{\sqrt{u-1}} \cdot 2u\,du
= \int \frac{2u^2}{\sqrt{u-1}}du\).
Answer: (a)
Question #13:
Use integration by parts \(\displaystyle \int u\,dv=uv-\int v\,du\).
Set \(u=x\) and \(dv=e^{-x}dx\), so \(du=dx\) and \(v=-e^{-x}\). Then
\(\displaystyle \int x e^{-x} dx
= x(-e^{-x}) + \int e^{-x}dx
= -xe^{-x} - e^{-x} + C\).
Answer: (c)
Question #14:
\(\displaystyle \int^\infty_1 {1 \over {x^3}}
= \lim_{N\to\infty} \int^N_1 {1 \over {x^3}}
= \lim_{N\to\infty}\left.-\frac{1}{2x^2}\right|_1^N
= \lim_{N\to\infty}-\frac{1}{2N^2}+\frac{1}{2}
= \frac{1}{2}\).
Answer: (b)
Question #15:
\(\displaystyle \int^{\pi/2}_0 \cos^5 x \sin x \,dx
= \left.-\frac{1}{6}\cos^6x\right|_0^{\pi/2}
=\frac{1}{6}\).
Answer: (a)
Question #16:
Use partial fractions,
\(\displaystyle {1 \over {x^2 - 4}}=\frac{1}{4}\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\).
\(\displaystyle \int^1_0 {1 \over {x^2 - 4}} dx
= \frac{1}{4}\int_0^1\frac{1}{x-2}-\frac{1}{x+2} dx
= \left.\left(\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\right)\right|_0^1
=\frac{1}{4}\left(\ln\left(\frac{1}{3}\right)-\ln(1)\right)
=-\frac{1}{4}\ln 3\).
Answer: (b)