What is the integral from 0 to 4 of xe^{3x} ?

The integral from 0 to 4 of xe^{3x} is (11e^{12}+1)/9

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integral from 0 to 4 of xe^{3x}

\int_{0}^{4} x e^{3 x} d x

\frac{11 e ^{12} + 1}{9}

Decimal

198922.63395 \dots

Solution steps

\int_{0}^{4} x e^{3 x} d x

Apply u-substitution

: \int_{0}^{12} \frac{e ^{u} u}{9} d u

= \int_{0}^{12} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{12} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{12}

= \frac{1}{9} [e^{u} u - \int e^{u} d u]_{0}^{12}

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

= \frac{1}{9} [e^{u} u - e^{u}]_{0}^{12}

Compute the boundaries:

11 e^{12} + 1

= \frac{1}{9} (11 e^{12} + 1)

Simplify

= \frac{11 e ^{12} + 1}{9}

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What is the integral from 0 to 4 of xe^{3x} ?
The integral from 0 to 4 of xe^{3x} is (11e^{12}+1)/9