Answer
If $f^{\prime}$ is continuous on $[a, b]$, show that
$$
2 \int_{a}^{b} f(x) f^{\prime}(x) d x=[f(b)]^{2}-[f(a)]^{2}
$$
Let $u=f(x)$ and $d u=f^{\prime}(x) d x$. When $x = a, u = f(a)$; when $x = b, u =f(b)$. Thus, the Substitution Rule gives
$$
\begin{aligned}
R.H.S.=2 \int_{a}^{b} f(x) f^{\prime}(x) d x& =2 \int_{f(a)}^{f(b)} u d u \\
&=\left[u^{2}\right]_{f(a)}^{f(b)} \\
&=[f(b)]^{2}-[f(a)]^{2}\\
& =L.H.S.
\end{aligned}
$$
Work Step by Step
If $f^{\prime}$ is continuous on $[a, b]$, show that
$$
2 \int_{a}^{b} f(x) f^{\prime}(x) d x=[f(b)]^{2}-[f(a)]^{2}
$$
Let $u=f(x)$ and $d u=f^{\prime}(x) d x$. When $x = a, u = f(a)$; when $x = b, u =f(b)$. Thus, the Substitution Rule gives
$$
\begin{aligned}
R.H.S.=2 \int_{a}^{b} f(x) f^{\prime}(x) d x& =2 \int_{f(a)}^{f(b)} u d u \\
&=\left[u^{2}\right]_{f(a)}^{f(b)} \\
&=[f(b)]^{2}-[f(a)]^{2}\\
& =L.H.S.
\end{aligned}
$$
