解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{-2}^{0} |x^2 - 2| dx & = & \displaystyle \int_{-2}^{-\sqrt{2}} x^2 - 2 dx + \displaystyle \int_{-\sqrt{2}}^{0} -(x^2 - 2) dx \\ & = & \left[ \dfrac{1}{3}x^3 - 2x \right]_{-2}^{-\sqrt{2}} - \left[ \dfrac{1}{3}x^3 - 2x \right]_{-\sqrt{2}}^{0} \\ & = & (\dfrac{4\sqrt{2}}{3} - \dfrac{4}{3}) - (-\dfrac{4\sqrt{2}}{3}) \\ & = & \dfrac{8\sqrt{2} - 4}{3} \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{\frac{1}{2e}}^{\frac{e}{2}} |\log 2x| dx & = & \displaystyle \int_{\frac{1}{2e}}^{\frac{1}{2}} -(\log 2x) dx + \displaystyle \int_{\frac{1}{2}}^{\frac{e}{2}} \log 2x dx \\ & = & -\dfrac{1}{2}\left[ 2x\log 2x - 2x \right]_{\frac{1}{2e}}^{\frac{1}{2}} + \dfrac{1}{2}\left[ 2x\log 2x - 2x \right]_{\frac{1}{2}}^{\frac{e}{2}} \\ & = & -\dfrac{1}{2}(-1 + \dfrac{1}{e} + \dfrac{1}{e}) + \dfrac{1}{2}(e - e + 1) \\ & = & 1 - \dfrac{1}{e} \\ \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{3} |e^x - 3| dx & = & \displaystyle \int_{0}^{\log 3} -(e^x - 3) dx + \displaystyle \int_{\log 3}^{3} e^x - 3 dx \\ & = & -\left[ e^x - 3x \right]_{0}^{\log 3} + \left[ e^x - 3x \right]_{\log 3}^{3} \\ & = & -(3 - 3\log 3 - 1) + (e^3 - 9 - 3 + 3\log 3) \\ & = & 6\log 3 + e^3 - 14 \end{eqnarray} \)解説：
\( \cos^2 x\sin x = (1 - \sin^2 x)\sin x = \sin x - \sin^3 x \)より\( -\sin^3 + \sin x = 0 \)を解くと \( \sin x = -1, 0, 1 \) 、\( \therefore x = -\pi, \dfrac{\pi}{2}, 0, \dfrac{\pi}{2} (-\pi \leqq x \leqq \dfrac{\pi}{2}) \)
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{-\pi}^{\frac{\pi}{2}} |\cos^2 x\sin x| dx & = & \displaystyle \int_{-\pi}^{0} -(\cos^2 x\sin x) dx + \displaystyle \int_{0}^{\frac{\pi}{2}} \cos^2 x\sin x dx \\ & = & -\left[ -\dfrac{1}{3}\cos^3 x \right]_{-\pi}^{0} + \left[ -\frac{1}{3}\cos^3 x \right]_{0}^{\frac{\pi}{2}} \\ & = & -(-\dfrac{1}{3} - \dfrac{1}{3}) + (\dfrac{1}{3}) \\ & = & 1 \end{eqnarray} \)解説：
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{3} \sqrt{9^x - 2 \cdot 3^{x + 2} + 81} dx & = & \displaystyle \int_{0}^{3} \sqrt{(3^x - 9)^2} dx \\ & = & \displaystyle \int_{0}^{3} |3^x - 9| dx \\ & = & \displaystyle \int_{0}^{2} -(3^x - 9) dx + \displaystyle \int_{2}^{3} 3^x - 9 dx \\ & = & -\left[ \dfrac{3^x}{\log 3} - 9x \right]_{0}^{2} + \left[ \dfrac{3^x}{\log 3} - 9x \right]_{2}^{3} \\ & = & -(\dfrac{9}{\log 3} - 18 - \dfrac{1}{\log 3}) + (\dfrac{27}{\log 3} - 27 - \dfrac{9}{\log 3} + 18) \\ & = & \dfrac{10}{\log 3} + 9 \end{eqnarray} \)