integral from 1 to 2 of x*sqrt(x-1)

解答

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

\frac{16}{15}

十进制

1.06666 \dots

求解步骤

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

使用换元积分法

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

乘开

(u + 1) u : u^{\frac{3}{2}} + u

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

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

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

= \frac{2}{5} + \frac{2}{3}

化简

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

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