定積分$\dint_{-1}^{1} \left| x^2-\bunsuu{1}{2}x-\bunsuu{1}{2}\right| dx$を求めよ．
基本的に絶対値の積分は，絶対値を外してから場合分けをして積分することになります．
\[\left| x^2-\bunsuu{1}{2}x-\bunsuu{1}{2}\right| = \begin{cases} \left(x+\bunsuu{1}{2}\right)(x-1) &\left(-1\leq x\leq -\bunsuu{1}{2}\right)\\ -\left(x+\bunsuu{1}{2}\right)(x-1) &\left(-\bunsuu{1}{2}\leq x\leq 1\right)\end{cases}\]
であるから，
\begin{eqnarray*}
&&\dint_{-1}^{1} \left| x^2-\bunsuu{1}{2}x-\bunsuu{1}{2}\right| dx\\
&=& \dint_{-1}^{-\frac{1}{2}} \left(x+\bunsuu{1}{2}\right)(x-1) dx – \int_{-\frac{1}{2}}^1 \left(x+\bunsuu{1}{2}\right)(x-1)dx \\
&=& \dint_{-1}^{-\frac{1}{2}} \left(x+\bunsuu{1}{2}\right)\left\{\left(x+\bunsuu{1}{2}\right)-\bunsuu{3}{2}\right\} dx -\left(-\bunsuu{1}{6}\left(1+\bunsuu{1}{2}\right)^3\right)\\
&=& \dint_{-1}^{-\frac{1}{2}} \left\{\left(x+\bunsuu{1}{2}\right)^2-\bunsuu{3}{2}\left(x+\bunsuu{1}{2}\right)\right\} dx +\bunsuu{9}{16}\\
&=& \left[ \bunsuu{1}{3}\left(x+\bunsuu{1}{2}\right)^3 – \bunsuu{3}{4}\left(x+\bunsuu{1}{2}\right)^2\right]_{-1}^{-\frac{1}{2}} +\bunsuu{9}{16}\\
&=& \bunsuu{19}{24}
\end{eqnarray*}