What is the integral from 0 to 1 of (x+e)/(e^x) ?

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

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

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

e - \frac{2}{e}

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

= e - \frac{2}{e}

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What is the integral from 0 to 1 of (x+e)/(e^x) ?
The integral from 0 to 1 of (x+e)/(e^x) is e-2/e