integral from-1 to 0 of x(x+1)^{100}

解答

\int_{- 1}^{0} x (x + 1)^{100} d x

- \frac{1}{10302}

十进制

- 0.00009 \dots

求解步骤

\int_{- 1}^{0} x (x + 1)^{100} d x

使用换元积分法

= \int_{0}^{1} (u - 1) u^{100} d u

乘开

(u - 1) u^{100} : u^{101} - u^{100}

= \int_{0}^{1} u^{101} - u^{100} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} u^{101} d u - \int_{0}^{1} u^{100} d u

\int_{0}^{1} u^{101} d u = \frac{1}{102}

\int_{0}^{1} u^{100} d u = \frac{1}{101}

= \frac{1}{102} - \frac{1}{101}

化简

\frac{1}{102} - \frac{1}{101} : - \frac{1}{10302}

= - \frac{1}{10302}

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