# Evaluate the integral  I = int_0^3 \ xf(x^2) \ dx ?

Jan 26, 2018

${\int}_{0}^{3} \setminus x f \left({x}^{2}\right) \setminus \mathrm{dx} = \frac{3}{2}$

#### Explanation:

We seek:

$I = {\int}_{0}^{3} \setminus x f \left({x}^{2}\right) \setminus \mathrm{dx}$

We can perform a substitution:

Let $u = {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

And the substitution will require a change in limits:

When $x = \left\{\begin{matrix}0 \\ 3\end{matrix}\right. \implies u = \left\{\begin{matrix}0 \\ 9\end{matrix}\right.$

Substituting into the integral, changing the variable of integration from $x$ to $u$ we get:

$I = \frac{1}{2} \setminus {\int}_{0}^{3} \setminus 2 x f \left({x}^{2}\right) \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{9} \setminus f \left(u\right) \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{2} \setminus \left(3\right)$, as ${\int}_{0}^{9} \setminus f \left(u\right) \setminus \mathrm{du} = {\int}_{0}^{9} \setminus f \left(x\right) \setminus \mathrm{dx}$

$\setminus \setminus = \frac{3}{2}$