3. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function with $\int_{0}^{1} f(x) f^{\prime}(x) d x=0$ and $\int_{0}^{1} f(x)^{2} f^{\prime}(x) d x=$ 18. What is $\int_{0}^{1} f(x)^{4} f^{\prime}(x) d x$ ?
Solution:
$$
\begin{array}{l}
0=\int_{0}^{1} f(x) f^{\prime}(x) d x=\int_{f(0)}^{f(1)} u d u=\frac{1}{2}\left(f(1)^{2}-f(0)^{2}\right), \text { and } \\
18=\int_{0}^{1} f(x)^{2} f^{\prime}(x) d x=\int_{f(0)}^{f(1)} u^{2} d u=\frac{1}{3}\left(f(1)^{3}-f(0)^{3}\right) .
\end{array}
$$

The first equation implies $f(0)= \pm f(1)$. The second equation shows that $f(0) \neq f(1)$, and in fact $54=f(1)^{3}-f(0)^{3}=2 f(1)^{3}$, so $f(1)=3$ and $f(0)=-3$. Then
$$
\int_{0}^{1} f(x)^{4} f^{\prime}(x) d x=\int_{f(0)}^{f(1)} u^{4} d u=\frac{1}{5}\left(f(1)^{5}-f(0)^{5}\right)=\frac{1}{5}(243+243)=\frac{486}{5} .
$$