What is the integral from 0 to 3 of x(e^{x^2}-1) ?

The integral from 0 to 3 of x(e^{x^2}-1) is (e^9-10)/2

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integral from 0 to 3 of x(e^{x^2}-1)

\int_{0}^{3} x (e^{x^{2}} - 1) d x

\frac{e ^{9} - 10}{2}

Decimal

4046.54196 \dots

Solution steps

\int_{0}^{3} x (e^{x^{2}} - 1) d x

Apply u-substitution

: \int_{0}^{9} \frac{e ^{u} - 1}{2} d u

= \int_{0}^{9} \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}^{9} 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}^{9} e^{u} d u - \int_{0}^{9} 1 d u)

\int_{0}^{9} e^{u} d u = e^{9} - 1

\int_{0}^{9} 1 d u = 9

= \frac{1}{2} (e^{9} - 1 - 9)

Simplify

\frac{1}{2} (e^{9} - 1 - 9) : \frac{e ^{9} - 10}{2}

= \frac{e ^{9} - 10}{2}

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What is the integral from 0 to 3 of x(e^{x^2}-1) ?
The integral from 0 to 3 of x(e^{x^2}-1) is (e^9-10)/2