integral from 0 to 1 of (1+e^x)/(e^x)

解答

\int_{0}^{1} \frac{1 + e ^{x}}{e ^{x}} d x

- \frac{1}{e} + 2

十进制

1.63212 \dots

求解步骤

\int_{0}^{1} \frac{1 + e ^{x}}{e ^{x}} d x

乘开

\frac{1 + e ^{x}}{e ^{x}} : \frac{1}{e ^{x}} + 1

= \int_{0}^{1} \frac{1}{e ^{x}} + 1 d x

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} \frac{1}{e ^{x}} d x + \int_{0}^{1} 1 d x

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

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

= - \frac{1}{e} + 1 + 1

化简

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

\frac{\partial}{\partial x} (\frac{- x - 6 y}{y ^{2} + 5 x ^{2}})

\frac{\partial}{\partial x} (2 - x^{2})

in v erse l a pl a ce \frac{1}{s ^{2} + 4}

\int \frac{1}{x 8 x} d x

7 t \cdot \frac{d y}{d t} - 4 y = t