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

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

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

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

\frac{e - 2}{2}

Decimal

0.35914 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{0}^{1} e^{u} d u - \int_{0}^{1} 1 d u)

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

\int_{0}^{1} 1 d u = 1

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

Simplify

\frac{1}{2} (e - 1 - 1) : \frac{e - 2}{2}

= \frac{e - 2}{2}

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