# How do you find the integral int_0^1x^2*e^(x^3)dx ?

Sep 22, 2014

We have to use a substitution technique to solve this problem. The strategy is to find an expression that when then differentiated can be substituted back into the original integral.

Let $u = {x}^{3}$

${\int}_{0}^{1} {x}^{2} {e}^{u} \mathrm{dx}$

$\mathrm{du} = 3 {x}^{2} \mathrm{dx}$

$\frac{\mathrm{du}}{3} = {x}^{2} \mathrm{dx}$

$\frac{1}{3} \cdot \mathrm{du} = {x}^{2} \mathrm{dx}$

$\int {e}^{u} {x}^{2} \mathrm{dx}$, notice that ${x}^{2} \mathrm{dx}$ can be replaced by $\frac{1}{3} \cdot \mathrm{du}$

$\int {e}^{u} \frac{1}{3} \cdot \mathrm{du}$

$\frac{1}{3} \int {e}^{u} \cdot \mathrm{du}$, Constants can be moved outside of the integral

Now lets evaluate the boundaries. Look back to the original $u$ substitution: $u = {x}^{3}$

Substitute in the current high and low boundaries.

$u = {\left(1\right)}^{3} = 1 \to$ upper boundary
$u = {\left(0\right)}^{3} = 0 \to$ lower boundary

In this problem the boundaries did not change

$\frac{1}{3} {\int}_{0}^{1} {e}^{u} \cdot \mathrm{du}$

$= \frac{1}{3} {\left[{e}^{u}\right]}_{0}^{1} = \frac{1}{3} \left[{e}^{1} - {e}^{0}\right] = \frac{1}{3} \left[{e}^{1} - 1\right] = \frac{{e}^{1} - 1}{3} = \frac{e - 1}{3}$

$= 0.5728$