What is the integral from 0 to 1 of xe^x-(e^{2x})/2 ?

The integral from 0 to 1 of xe^x-(e^{2x})/2 is 1-(e^2-1)/4

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integral from 0 to 1 of xe^x-(e^{2x})/2

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

1 - \frac{e ^{2} - 1}{4}

Decimal

- 0.59726 \dots

Solution steps

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

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

\int_{0}^{1} \frac{e ^{2 x}}{2} d x = \frac{e ^{2} - 1}{4}

= 1 - \frac{e ^{2} - 1}{4}

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What is the integral from 0 to 1 of xe^x-(e^{2x})/2 ?
The integral from 0 to 1 of xe^x-(e^{2x})/2 is 1-(e^2-1)/4