integral from 0 to 1 of (2x+1)/(e^x)

解答

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

- \frac{5}{e} + 3

十进制

1.16060 \dots

求解步骤

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

乘开

\frac{2 x + 1}{e ^{x}} : \frac{2 x}{e ^{x}} + \frac{1}{e ^{x}}

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

\int_{0}^{1} \frac{2 x}{e ^{x}} d x = 2 (- \frac{2}{e} + 1)

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

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

化简

2 (- \frac{2}{e} + 1) - \frac{1}{e} + 1 : - \frac{5}{e} + 3

= - \frac{5}{e} + 3

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