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

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

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

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

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

Decimal

0.13212 \dots

Solution steps

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

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

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

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

\int_{1}^{2} 2 x (x^{2} - 2)^{2} d x

\int_{- 1}^{3} 2 x + 4 d x

\int_{0}^{3} (y - 3) (2 y + 3) d y

\int_{5}^{6} - x^{3} + 11 x^{2} - 30 x d x

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

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