integral from 1 to 2 of xsqrt(2x-x^2)

解答

\int_{1}^{2} x 2 x - x^{2} d x

\frac{1}{3} + \frac{π}{4}

十进制

1.11873 \dots

求解步骤

\int_{1}^{2} x 2 x - x^{2} d x

对

2 x - x^{2}

配方:

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

= \int_{1}^{2} x - (x - 1)^{2} + 1 d x

使用换元积分法

= \int_{0}^{1} (u + 1) - u^{2} + 1 d u

乘开

(u + 1) - u^{2} + 1 : u - u^{2} + 1 + - u^{2} + 1

= \int_{0}^{1} u - u^{2} + 1 + - u^{2} + 1 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 - u^{2} + 1 d u + \int_{0}^{1} - u^{2} + 1 d u

\int_{0}^{1} u - u^{2} + 1 d u = \frac{1}{3}

\int_{0}^{1} - u^{2} + 1 d u = \frac{π}{4}

= \frac{1}{3} + \frac{π}{4}