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

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

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integral from 0 to 1 of (e^x-xe^x)

\int_{0}^{1} (e^{x} - x e^{x}) d x

e - 2

Decimal

0.71828 \dots

Solution steps

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

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

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

= e - 1 - 1

Simplify

= e - 2

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