【2011京都大学・理系・第１問[２]】
定積分 \(\displaystyle\int^{\frac{1}{2}}_{0}(x+1)\sqrt{1-2x^2} dx\) を求めよ．
根号(ルート)を含む積分についての有名・基本例題は⏬で確認を！
解答・解説
(与式) を \(I\) とおく.
\(I=\displaystyle\int^{\frac{1}{2}}_{0}x\sqrt{1-2x^2} dx+\displaystyle\int^{\frac{1}{2}}_{0}\sqrt{1-2x^2} dx\) より
\(I_{1}=\displaystyle\int^{\frac{1}{2}}_{0}x\sqrt{1-2x^2} dx\) , \(I_{2}=\displaystyle\int^{\frac{1}{2}}_{0}\sqrt{1-2x^2} dx\) とおく．
\(I_{1}\) について
\(t=1-2x^2\) とおくと
\(\displaystyle\frac{dt}{dx}=-4x\) ,
\(x\) : \(0 \rightarrow \displaystyle\frac{1}{2}\) のとき \(t\) : \(1 \rightarrow \displaystyle\frac{1}{2}\) より
\(I_{1}=-\displaystyle\frac{1}{4}\displaystyle\int^{\frac{1}{2}}_{1} t^{\frac{1}{2}}dt\)
\(=-\displaystyle\frac{1}{4}\cdot\displaystyle\frac{2}{3}\Bigl[t^{\frac{3}{2}}\Bigr]^{\frac{1}{2}}_{1}\)
\(=\displaystyle\frac{1}{6}-\displaystyle\frac{\sqrt{2}}{24}\)
\(I_{2}\) について
\(I_{2}=\sqrt{2}\displaystyle\int^{\frac{1}{2}}_{0}\sqrt{\displaystyle\frac{1}{2}-x^2} dx\)
\(x=\displaystyle\frac{1}{\sqrt{2}}\sin\theta\) とおくと
\(\displaystyle\frac{dx}{d \theta}=\displaystyle\frac{1}{\sqrt{2}}\cos \theta\) ，
\(x\) : \(0 \rightarrow \displaystyle\frac{1}{2}\) のとき \(\theta\) : \(0 \rightarrow \displaystyle\frac{\pi}{4}\) より
\(I_{2}=\sqrt{2}\displaystyle\int^{\frac{\pi}{4}}_{0}\sqrt{\displaystyle\frac{1}{2}-\displaystyle\frac{1}{2}\sin^2\theta}\cdot\displaystyle\frac{1}{\sqrt{2}}\cos \theta d \theta\)
\(=\displaystyle\frac{1}{\sqrt{2}}\displaystyle\int^{\frac{\pi}{4}}_{0}\cos^2 \theta d \theta\)
\(=\displaystyle\frac{1}{\sqrt{2}}\displaystyle\int^{\frac{\pi}{4}}_{0}\displaystyle\frac{1+\cos 2 \theta}{2} d \theta\)
\(=\displaystyle\frac{1}{2\sqrt{2}}\Bigl[ \theta+\displaystyle\frac{1}{2}\sin 2 \theta\Bigr]^{\frac{\pi}{4}}_{0}\)
\(=\displaystyle\frac{\sqrt{2}}{16}\pi+\displaystyle\frac{\sqrt{2}}{8}\)
したがって求める定積分の値は，
\(I=I_{1}+I_{2}\)
\(=\displaystyle\frac{1}{6}-\displaystyle\frac{\sqrt{2}}{24}+\displaystyle\frac{\sqrt{2}}{16}\pi+\displaystyle\frac{\sqrt{2}}{8}\)
\(=\displaystyle\frac{1}{6}+\displaystyle\frac{\sqrt{2}}{12}+\displaystyle\frac{\sqrt{2}}{16}\pi\)
コメント