# How do you integrate int x^2(x^3+8)^2dx from [-2,4]?

May 24, 2018

$41472$

#### Explanation:

the Integrand is given by
${x}^{2} \left({x}^{6} + 16 {x}^{3} + 64\right) = {x}^{8} + 16 {x}^{5} + 64 {x}^{2}$
and a primitive is ${x}^{9} / 9 + \frac{16}{6} {x}^{6} + \frac{64}{3} {x}^{3}$

May 24, 2018

$41472$.

#### Explanation:

Suppose that, $I = {\int}_{-} {2}^{4} {x}^{2} {\left({x}^{3} + 8\right)}^{2} \mathrm{dx}$.

Let, $\left({x}^{3} + 8\right) = t . \therefore 3 {x}^{2} \mathrm{dx} = \mathrm{dt} , \mathmr{and} , {x}^{2} \mathrm{dx} = \frac{1}{3} \mathrm{dt}$.

Also, when $x = - 2 , t = {x}^{3} + 8 = {\left(- 2\right)}^{3} + 8 = 0$ and,

similarly when, $x = 4 , t = 72$.

$\therefore I = {\int}_{-} {2}^{4} {\left({x}^{3} + 8\right)}^{2} {x}^{2} \mathrm{dx}$,

$= {\int}_{0}^{72} {t}^{2} \cdot \frac{1}{3} \mathrm{dt}$,

$= \frac{1}{3} {\left[\frac{1}{3} {t}^{3}\right]}_{0}^{72}$,

$= \frac{1}{9} \left[{72}^{3} - 0\right]$,

$= \frac{1}{9} \cdot 72 \cdot {72}^{2}$,

$= 8 \times 5184$.

$\Rightarrow I = 41472$.