integral from 1 to 4 of xsqrt(x-1)

解答

\int_{1}^{4} x x - 1 d x

\frac{18 3}{5} + 23

十进制

9.69948 \dots

求解步骤

\int_{1}^{4} x x - 1 d x

使用换元积分法

= \int_{0}^{3} (u + 1) u d u

乘开

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

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

\int_{0}^{3} u^{\frac{3}{2}} d u = \frac{18 3}{5}

\int_{0}^{3} u d u = 23

= \frac{18 3}{5} + 23

\int_{0}^{x} 1 - t^{2} d t

\int_{0}^{2} 2 x + x^{2} - x^{3} d x

\int_{0}^{5} ∣ 1 - x ∣ d x

\int_{0}^{2} (t - 1) e^{(t - 1)^{2}} d t

\int_{- 4}^{4} - x^{2} - 3 x d x