Question

Integration Procedures(Itevrations by Parts)

∫ x³dx/cuberoot of (8-x²) from 0 to √7

Answer 1

we evaluate the integral

$\int^{\sqrt7}_{0} \frac {x^3}{\sqrt[3]{8-x^2}}dx$

letting $u=8-x^3 ...(i)$

then $du= -2xdx$ or$\frac {-1}{2}du=xdx ...(ii)$

for the limits; when $x=0, u= 8 ...(iii), x=\sqrt{7}, u=1 ...(iv)$

$\therefore \int^{\sqrt7}_{0} \frac{x^3}{\sqrt[3]{8-x^2}dx}=\int^{\sqrt7}_{0} \frac{x^2}{\sqrt[3]{8-x^2}}xdx$

$=\frac{-1}{2}\int^{1}_{8}\frac{(8-u)}{\sqrt[3]u}du$ from (i) and (iv)

$=\frac{-1}{2}\int^{1}_{8} (8u^{-1/3}-u^{2/3})du$

$=\frac{-1}{2}[12u^{2/3}-\frac{3}{5}u^{5/3}]^{1}_{8}$

$=\frac {-1}{2} [\frac{57}{5}-\frac{144}{5}]=\frac{87}{10}=8.7$