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

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

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

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

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

Decimal

0.85914 \dots

Solution steps

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

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

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

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

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