部分積分の問題

■問題

次の問題を積分せよ(定積分)．

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

■答

${\displaystyle \frac{1}{9}{e}^3\left(4e^9-1\right)}$

■ヒント

部分積分法

${\displaystyle \int f^{\prime }\left(x\right)g\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f\left(x\right)g^{\prime }\left(x\right)dx}$

を用いる．

■解説

${\displaystyle f\left(x\right)=e^{3x}}$ より，${\displaystyle f^{\prime }\left(x\right)=\frac{1}{3}{e}^{3x}}$

${\displaystyle g\left(x\right)=x}$ より，${\displaystyle g{\left(x\right)}^{\prime }=1}$

として部分積分を行う．

${\displaystyle {\displaystyle \underset{1}{\overset{4}{\int }}x{e}^{3x}}dx}$ ${\displaystyle ={\displaystyle \underset{1}{\overset{4}{\int }}x{\left(\frac{1}{3}{e}^{3e}\right)}^{\prime }dx}}$

${\displaystyle ={\left[x\cdot \frac{1}{3}{e}^{3x}\right]}_1^4-{\displaystyle \underset{1}{\overset{4}{\int }}{\left(x\right)}^{\prime }}\cdot \frac{1}{3}{e}^{3x}dx}$

${\displaystyle =\frac{4}{3}{e}^{12}-\frac{1}{3}{e}^3-\frac{1}{3}{\displaystyle \underset{1}{\overset{4}{\int }}{e}^{3x}}dx}$

${\displaystyle =\frac{4}{3}{e}^{12}-\frac{1}{3}{e}^3-\frac{1}{3}{\left[\frac{1}{3}{e}^{3x}\right]}_1^4}$

${\displaystyle =\frac{4}{3}{e}^{12}-\frac{1}{3}{e}^3-\frac{1}{3}\left(\frac{1}{3}{e}^{12}-\frac{1}{3}{e}^3\right)}$

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

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

${\displaystyle =\frac{1}{9}{e}^3\left(11e^9-2\right)}$

 

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最終更新日： 2023年11月23日