重積分の基礎

■問題

次の重積分の値を求めよ．

$3$\displaystyle{ \displaystyle{ \hspace{1}{\int\int }_{D} {e}^{x}\sin ydxdy } }$ $3$\displaystyle{ $ D:0\leq x\leq 1,0\leq y\leq {\pi}x $ }$

■答

$3$\displaystyle{\frac{\hspace{2} e+1 \hspace{2}}{\hspace{2} 1+{{\pi}}^{2} \hspace{2}}+e- 1}$

■ヒント

領域 $3$\displaystyle{D\text{\hspace{1} }}$ から変数 $3$\displaystyle{x\text{\hspace{1} }}$ と変数 $3$\displaystyle{y\text{\hspace{1} }}$ の積分範囲を決定する．

次に， $3$\displaystyle{x\text{\hspace{1} }}$ を定数とみなして $3$\displaystyle{y\text{\hspace{1} }}$ について積分し，その結果を更に $3$\displaystyle{x\text{\hspace{1} }}$ で積分する．

$3$\displaystyle{I=\displaystyle{ \int {e}^{x}\sin xdx }}$ や $3$\displaystyle{I=\displaystyle{ \int {e}^{x}\cos xdx }}$ は部分積分を使う．

■解き方

$3$\displaystyle{\displaystyle{ \hspace{1}{\int\int }_{D} {e}^{x}\sin ydxdy }}$

$3$\displaystyle{=\displaystyle{\int }_{0}^{1}\left({ \displaystyle{\int }_{0}^{ {\pi}x }{e}^{x}\sin ydy }\right)dx}$

$3$\displaystyle{=\displaystyle{ \int _{0}^{1} {e}^{x} } \[- \cos y \] _{0}^{ {\pi}x }dx}$

$3$\displaystyle{=- \displaystyle{ \int _{0}^{1} {e}^{x}\cos {\pi}xdx }+\displaystyle{ \int _{0}^{1} {e}^{x} }dx}$

$3$\displaystyle{=- \displaystyle{ \int _{0}^{1} {e}^{x}\cos {\pi}xdx }+ \[ {e}^{x} \] _{0}^{1}}$

$3$\displaystyle{=- \displaystyle{ \int _{0}^{1} {e}^{x}\cos {\pi}xdx }+e- 1}$ 　･･････(1)

$3$\displaystyle{=- \displaystyle{ \int _{0}^{1} {e}^{x} }{ $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}}\sin {\pi}x $ }^{\prime }dx+e- 1}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}} \{ \[ {e}^{x}\sin {\pi}x \] _{0}^{1}- \displaystyle{ \int _{0}^{1} {e}^{x}\sin {\pi}xdx } \} +e- 1}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}} $ - \displaystyle{ \int _{0}^{1} {e}^{x}\sin {\pi}xdx } $ +e- 1}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}}\displaystyle{ \int _{0}^{1} {e}^{x} }\sin {\pi}xdx+e- 1}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}}\displaystyle{ \int _{0}^{1} {e}^{x} }{ $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}{\pi}\hspace{2}}\cos {\pi}x $ }^{\prime }dx+e- 1}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}} \{ \[ {e}^{x}cox{\pi}x \] _{0}^{1}- \displaystyle{ \int _{0}^{1} {e}^{x} }\cos {\pi}xdx \} +e- 1}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}} \{ $ - e- 1 $ - \displaystyle{ \int _{0}^{1} {e}^{x} }\cos {\pi}xdx \} +e- 1}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}} $ e+1 $ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}}\displaystyle{ \int _{0}^{1} {e}^{x} }\cos {\pi}xdx+e- 1}$ 　･･････(2)

ここで， $3$\displaystyle{I=\displaystyle{ \int _{0}^{1} {e}^{x} }\cos {\pi}xdx}$ とおくと，(1)，(2)より

$3$\displaystyle{- I+e- 1}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}} $ e+1 $ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}}I+e- 1}$ 　･･････(3)

が得られる．(3)を $3$I$ について解く

$3$\displaystyle{ $ \frac{\hspace{2} 1+{{\pi}}^{2} \hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}} $ I=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {{\pi}}^{2} \hspace{2}} $ e+1 $ }$

$3$\displaystyle{I=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{{\pi}}^{2} \hspace{2}} $ e+1 $ }$

よって

$3$\displaystyle{ \displaystyle{ \hspace{1}{\int\int }_{D} {e}^{x}\sin ydxdy } }$ $3$\displaystyle{=- I+e- 1}$

$3$\displaystyle{=\frac{\hspace{2} e+1 \hspace{2}}{\hspace{2} 1+{{\pi}}^{2} \hspace{2}}+e- 1}$

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最終更新日： 2023年10月19日