integral from 1 to 0 of ze^{-1}

解答

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

- \frac{1}{2 e}

十进制

- 0.18393 \dots

求解步骤

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

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

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

提出常数:

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

= - e^{- 1} \cdot \int_{0}^{1} z d z

使用指数法则:

a^{- 1} = \frac{1}{a}

= - \frac{1}{e} \cdot \int_{0}^{1} z d z

使用幂法则

= - \frac{1}{e} [\frac{z ^{2}}{2}]_{0}^{1}

带入上下限计算后得:

\frac{1}{2}

= - \frac{1}{e} \cdot \frac{1}{2}

化简

= - \frac{1}{2 e}

\int_{0}^{π} e^{- s t} d t

\int_{1}^{3} \frac{6}{x} d x

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

\int_{- l n (2)}^{l n (2)} x sinh (x) d x

\int_{- 2}^{0} 2 x + 4 d x