问题及解答

求

\[\int_{-1}^{\frac{1}{2}}\sqrt{x^2-x^4}dx.\]

原积分等于

\[
\begin{split}
&\int_{-1}^{0}\sqrt{x^2-x^4}dx+\int_{0}^{\frac{1}{2}}\sqrt{x^2-x^4}dx\\
=&\int_{0}^{1}\sqrt{x^2-x^4}dx+\int_{0}^{\frac{1}{2}}\sqrt{x^2-x^4}dx\\
\end{split}
\]

其中

\[
\begin{split}
\int\sqrt{x^2-x^4}dx&=\int x\sqrt{1-x^2}dx=\frac{1}{2}\int\sqrt{1-x^2}dx^2\\
&\stackrel{t=x^2}{=}\frac{1}{2}\int\sqrt{1-t}dt\\
&\stackrel{y=\sqrt{1-t}}{=}-\int y^2dy\\
&=-\frac{1}{3}y^3+C\\
&=-\frac{1}{3}(\sqrt{1-x^2})^3+C.
\end{split}
\]

因此

\[
\begin{split}
&\int_{0}^{1}\sqrt{x^2-x^4}dx+\int_{0}^{\frac{1}{2}}\sqrt{x^2-x^4}dx\\
=&-\frac{1}{3}(\sqrt{1-x^2})^3\biggr|_{1}^{0}-\frac{1}{3}(\sqrt{1-x^2})^3\biggr|_{0}^{\frac{1}{2}}\\
=&\frac{1}{3}-\frac{1}{3}\biggl[(\sqrt{1-(\frac{1}{2})^2})^3-1\biggr]\\
=&\frac{2}{3}-\frac{\sqrt{3}}{8}.
\end{split}
\]