integral from 0 to 1-x of 24y(1-x-y)

解

\int_{0}^{1 - x} 24 y (1 - x - y) d y

4 (- x^{3} + 3 x^{2} - 3 x + 1)

解答ステップ

\int_{0}^{1 - x} 24 y (1 - x - y) d y

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 \cdot \int_{0}^{1 - x} (1 - x - y) y d y

部分積分を適用する

= 24 [\frac{1}{2} y^{2} (1 - x - y) - \int - \frac{y ^{2}}{2} d y]_{0}^{1 - x}

\int - \frac{y ^{2}}{2} d y = - \frac{y ^{3}}{6}

= 24 [\frac{1}{2} y^{2} (1 - x - y) - (- \frac{y ^{3}}{6})]_{0}^{1 - x}

簡素化

24 [\frac{1}{2} y^{2} (1 - x - y) - (- \frac{y ^{3}}{6})]_{0}^{1 - x} : 24 [\frac{1}{6} (3 y^{2} - 3 x y^{2} - 2 y^{3})]_{0}^{1 - x}

= 24 [\frac{1}{6} (3 y^{2} - 3 x y^{2} - 2 y^{3})]_{0}^{1 - x}

限度を計算する:

\frac{- 3 x ^{3} + 9 x ^{2} - 9 x - 2 ( - x + 1 ) ^{3} + 3}{6}

= 24 \cdot \frac{- 3 x ^{3} + 9 x ^{2} - 9 x - 2 ( - x + 1 ) ^{3} + 3}{6}

簡素化

= 4 (- x^{3} + 3 x^{2} - 3 x + 1)