What is the integral from 0 to 3 of ((xe^{x^2})/3) ?

The integral from 0 to 3 of ((xe^{x^2})/3) is (e^9-1)/6

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integral from 0 to 3 of ((xe^{x^2})/3)

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

\frac{e ^{9} - 1}{6}

Decimal

1350.34732 \dots

Solution steps

\int_{0}^{3} \frac{x e ^{x^{2}}}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{3} x e^{x^{2}} d x

Apply u-substitution

= \frac{1}{3} \cdot \int_{0}^{9} \frac{e ^{u}}{2} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int_{0}^{9} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{3} \cdot \frac{1}{2} [e^{u}]_{0}^{9}

Simplify

\frac{1}{3} \cdot \frac{1}{2} [e^{u}]_{0}^{9} : \frac{1}{6} [e^{u}]_{0}^{9}

= \frac{1}{6} [e^{u}]_{0}^{9}

Compute the boundaries:

e^{9} - 1

= \frac{1}{6} (e^{9} - 1)

Simplify

= \frac{e ^{9} - 1}{6}

What is the integral from 0 to 3 of ((xe^{x^2})/3) ?
The integral from 0 to 3 of ((xe^{x^2})/3) is (e^9-1)/6