What is the integral from 1 to 3 of xe^{3x} ?

The integral from 1 to 3 of xe^{3x} is (8e^9-2e^3)/9

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integral from 1 to 3 of xe^{3x}

\int_{1}^{3} x e^{3 x} d x

\frac{8 e ^{9} - 2 e ^{3}}{9}

Solution steps

\int_{1}^{3} x e^{3 x} d x

Apply u-substitution

: \int_{3}^{9} \frac{e ^{u} u}{9} d u

= \int_{3}^{9} \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_{3}^{9} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{3}^{9}

= \frac{1}{9} [e^{u} u - \int e^{u} d u]_{3}^{9}

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

= \frac{1}{9} [e^{u} u - e^{u}]_{3}^{9}

Compute the boundaries:

8 e^{9} - 2 e^{3}

= \frac{1}{9} (8 e^{9} - 2 e^{3})

Simplify

= \frac{8 e ^{9} - 2 e ^{3}}{9}

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What is the integral from 1 to 3 of xe^{3x} ?
The integral from 1 to 3 of xe^{3x} is (8e^9-2e^3)/9