部分積分の問題

■問題

次の問題を積分せよ(定積分)．

$3$\displaystyle{ \displaystyle{ {\int }\limits_{1}^{4} x{e}^{ 3x } }dx }$

■答

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}{e}^{3} $ 11{e}^{9}- 2 $ }$

■ヒント

部分積分法

$3$\displaystyle{\int {f}^{\prime } $x$ g $x$ dx=f $x$ g $x$ - \int f $x$ {g}^{\prime } $x$ dx }$

を用いる．

■解説

$3$\displaystyle{ f $x$ ={e}^{ 3x } }$ より， $3$\displaystyle{ {f}^{\prime } $x$ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 3x } }$

$3$\displaystyle{ g $x$ =x }$ より， $3$\displaystyle{g{ $x$ }^{\prime }=1}$

として部分積分を行う．

$3$\displaystyle{ \displaystyle{ {\int }\limits_{1}^{4} x{e}^{ 3x } }dx }$ $3$\displaystyle{ = \displaystyle{ {\int }\limits_{1}^{4} x{ $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 3x } $ }^{\prime } dx} }$

$3$\displaystyle{ = \[ x\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 3x } \] _{1}^{4}- \displaystyle{ {\int }\limits_{1}^{4}{ $x$ }^{\prime } } \cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 3x }dx }$

$3$\displaystyle{ =\frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 12 }- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{3}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\displaystyle{ {\int }\limits_{1}^{4} {e}^{ 3x } }dx }$

$3$\displaystyle{ =\frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 12 }- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{3}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} \[ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 3x } \] _{1}^{4} }$

$3$\displaystyle{ =\frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 12 }- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{3}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 12 }- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{3} $ }$

$3$\displaystyle{ =\frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 12 }- \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{3}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}{e}^{ 12 }+\frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}{e}^{3} }$

$3$\displaystyle{ =\frac{\hspace{2}11\hspace{2}}{\hspace{2}9\hspace{2}}{e}^{ 12 }- \frac{\hspace{2}2\hspace{2}}{\hspace{2}9\hspace{2}}{e}^{3} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}{e}^{3} $ 11{e}^{9}- 2 $ }$

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最終更新日： 2025年3月8日