integral from a to b of x(b-x)^2

解

\int_{a}^{b} x (b - x)^{2} d x

\frac{b ^{4} - 3 a ^{4} + 8 b a ^{3} - 6 b ^{2} a ^{2}}{12}

解答ステップ

\int_{a}^{b} x (b - x)^{2} d x

部分積分を適用する

= [\frac{1}{2} x^{2} (b - x)^{2} - \int - x^{2} (b - x) d x]_{a}^{b}

\int - x^{2} (b - x) d x = - \frac{b x ^{3}}{3} + \frac{x ^{4}}{4}

= [\frac{1}{2} x^{2} (b - x)^{2} - (- \frac{b x ^{3}}{3} + \frac{x ^{4}}{4})]_{a}^{b}

簡素化

[\frac{1}{2} x^{2} (b - x)^{2} - (- \frac{b x ^{3}}{3} + \frac{x ^{4}}{4})]_{a}^{b} : [\frac{1}{12} (3 x^{4} - 8 b x^{3} + 6 b^{2} x^{2})]_{a}^{b}

= [\frac{1}{12} (3 x^{4} - 8 b x^{3} + 6 b^{2} x^{2})]_{a}^{b}

限度を計算する:

\frac{b ^{4}}{12} - \frac{1}{12} (3 a^{4} - 8 b a^{3} + 6 b^{2} a^{2})

= \frac{b ^{4}}{12} - \frac{1}{12} (3 a^{4} - 8 b a^{3} + 6 b^{2} a^{2})

簡素化

= \frac{b ^{4} - 3 a ^{4} + 8 b a ^{3} - 6 b ^{2} a ^{2}}{12}