Answer
$$\int^{1}_{0}\frac{10\sqrt v}{(1+v^{3/2})^2}dv=\frac{10}{3}$$ $$\int^{4}_{1}\frac{10\sqrt v}{(1+v^{3/2})^2}dv=\frac{70}{27}$$
Work Step by Step
Part a) $$A=\int^{1}_{0}\frac{10\sqrt v}{(1+v^{3/2})^2}dv$$
Set $u=1+v^{3/2}$, we then have $$du=\frac{3}{2}v^{1/2}dv=\frac{3}{2}\sqrt vdv$$ $$\sqrt vdv=\frac{2}{3}du$$
For $v=0$, we have $u=1$ and for $v=1$, we have $u=2$.
Therefore, $$A=\frac{2}{3}\int^{2}_1\frac{10}{u^2}du$$ $$A=\frac{20}{3}\Big(-\frac{1}{u}\Big)\Big]^2_1$$ $$A=-\frac{20}{3}\Big(\frac{1}{2}-1\Big)$$ $$A=-\frac{20}{3}\Big(-\frac{1}{2}\Big)=\frac{10}{3}$$
Part b) $$A=\int^{4}_{1}\frac{10\sqrt v}{(1+v^{3/2})^2}dv$$
We carry out a similar substitution as in part a).
For $v=4$, we have $u=9$ and for $v=1$, we have $u=2$.
Therefore, $$A=\frac{2}{3}\int^{9}_2\frac{10}{u^2}du=\frac{20}{3}\Big(-\frac{1}{u}\Big)\Big]^9_2$$ $$A=-\frac{20}{3}\Big(\frac{1}{9}-\frac{1}{2}\Big)$$ $$A=-\frac{20}{3}\Big(-\frac{7}{18}\Big)=\frac{70}{27}$$
