Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 58

Suppose that $f$ is continuous and $\displaystyle \int^a_0 f(x) dx = 4$, find $\displaystyle \int^3_0 x f(x^2) dx$.

Let $u = x^2$, then $du = 2x dx$, so $\displaystyle xdx = \frac{du}{2}$. When $x = 0, u = 0$ and when $x = 3, u = 9$. Thus,



$
\begin{equation}
\begin{aligned}

\int^3_0 x f(x) dx =& \int^9_0 f(u) \frac{du}{2}
\\
\\
\int^3_0 x f(x) dx =& \frac{1}{2} \int^9_0 f(u) du

\end{aligned}
\end{equation}
$



We know that $\displaystyle \int^9_0 f(x) dx = 4$ or $\displaystyle \int^9_0 f(u) du = 4$, so



$
\begin{equation}
\begin{aligned}

\int^3_0 x f(x) dx =& \frac{1}{2} (4)
\\
\\
\int^3_0 x f(x) dx =& 2

\end{aligned}
\end{equation}
$