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

解答

\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}

求解步骤

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

使用分布积分法

= [\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}

化简

[\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}

带入上下限计算后得:

\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})

化简

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

\frac{d y}{d x} = 2 y - y^{2}

\frac{d}{d x} (2 x^{7} - 3 x^{6} + 3 x^{3} - 4 x^{2} - 7)

\int \frac{7 - 7 x}{1 - x} d x

\int x (x^{2} - 1)^{3} d x

d er i v a t i v e f (x) = 2 - 3 x