# How do you integrate int x^2e^(x^3/2)dx from [-2,3]?

Jan 14, 2017

$\frac{2 {e}^{\frac{35}{2}} - 2}{3 {e}^{4}}$

#### Explanation:

$I = {\int}_{- 2}^{3} {x}^{2} {e}^{{x}^{3} / 2} \mathrm{dx}$

Use the substitution:

$u = {x}^{3} / 2 \text{ "=>" } \mathrm{du} = \frac{3}{2} {x}^{2} \mathrm{dx}$

We already have ${x}^{2} \mathrm{dx}$ in the integral, but are off by a factor of $\frac{3}{2}$. Also note that changing from $x$ to $u$ will change our bounds:

$x = 3 \text{ "=>" } u = {3}^{3} / 2 = \frac{27}{2}$

$x = - 2 \text{ "=>" } u = {\left(- 2\right)}^{3} / 2 = - 4$

Thus:

$I = \frac{2}{3} \int {e}^{{x}^{3} / 2} \left(\frac{3}{2} {x}^{2} \mathrm{dx}\right) = \frac{2}{3} {\int}_{- 4}^{\frac{27}{2}} {e}^{u} \mathrm{du}$

The integral of ${e}^{u}$ is itself:

$= \frac{2}{3} {\left[{e}^{u}\right]}_{- 4}^{\frac{27}{2}} = \frac{2}{3} \left({e}^{\frac{27}{2}} - {e}^{- 4}\right) = \frac{2}{3} \left(\frac{{e}^{\frac{35}{2}} - 1}{e} ^ 4\right) = \frac{2 {e}^{\frac{35}{2}} - 2}{3 {e}^{4}}$