integral from 1 to 2 of e^{x-1}-e

解答

\int_{1}^{2} e^{x - 1} - e d x

\frac{e ^{2} - e}{e} - e

十进制

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求解步骤

\int_{1}^{2} e^{x - 1} - e 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_{1}^{2} e^{x - 1} d x - \int_{1}^{2} e d x

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

\int_{1}^{2} e d x = e

= \frac{e ^{2} - e}{e} - e

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