integral from 0 to-1 of x*e^x

解答

\int_{0}^{- 1} x \cdot e^{x} d x

1 - \frac{2}{e}

十进制

0.26424 \dots

求解步骤

\int_{0}^{- 1} x e^{x} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} x e^{x} d x

使用分布积分法

= - [e^{x} x - \int e^{x} d x]_{- 1}^{0}

\int e^{x} d x = e^{x}

= - [e^{x} x - e^{x}]_{- 1}^{0}

带入上下限计算后得:

- 1 + \frac{2}{e}

= - (- 1 + \frac{2}{e})

化简

= 1 - \frac{2}{e}

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

\int_{5}^{10} 1

\int_{- 2}^{0} \frac{- 2 x}{1 + x ^{2}} d x

\int_{0}^{\frac{t}{3}} e^{t^{2}} d t

\int_{0}^{\frac{π}{4}} 18 cos (2 t) d t