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

解答

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

- \frac{1}{e} + 2

十进制

1.63212 \dots

求解步骤

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

乘开

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

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

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

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

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

化简

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

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