integral from 2 to 3 of 2xsqrt(x^2-1)

解答

\int_{2}^{3} 2 x x^{2} - 1 d x

2 (\frac{16 2}{3} - 3)

十进制

11.62084 \dots

求解步骤

\int_{2}^{3} 2 x x^{2} - 1 d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{3} x x^{2} - 1 d x

使用换元积分法

= 2 \cdot \int_{3}^{22} u^{2} d u

使用幂法则

= 2 [\frac{u ^{3}}{3}]_{3}^{22}

带入上下限计算后得:

\frac{16 2}{3} - 3

= 2 (\frac{16 2}{3} - 3)

\int_{1}^{3} x^{2} - 3 x d x

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

\int_{0}^{1} x^{3} - x^{2} - 2 x d x

\int_{- 3}^{4} 25 - x^{2} d x

\int_{- 4}^{- 1} x^{2} d x