integral from 0 to pi/2 of x(-cos(x)+1)

解

\int_{0}^{\frac{π}{2}} x (- cos (x) + 1) d x

\frac{π ^{2} - 4 π}{8} + 1

十進法表記

0.66290 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} x (- cos (x) + 1) d x

部分積分を適用する

= [x (- sin (x) + x) - \int - sin (x) + x d x]_{0}^{\frac{π}{2}}

\int - sin (x) + x d x = cos (x) + \frac{x ^{2}}{2}

= [x (- sin (x) + x) - (cos (x) + \frac{x ^{2}}{2})]_{0}^{\frac{π}{2}}

簡素化

[x (- sin (x) + x) - (cos (x) + \frac{x ^{2}}{2})]_{0}^{\frac{π}{2}} : [\frac{1}{2} (x^{2} - 2 x sin (x) - 2 cos (x))]_{0}^{\frac{π}{2}}

= [\frac{1}{2} (x^{2} - 2 x sin (x) - 2 cos (x))]_{0}^{\frac{π}{2}}

限度を計算する:

\frac{π ^{2} - 4 π}{8} + 1

= \frac{π ^{2} - 4 π}{8} + 1