What is the integral from 0 to 1 of xe^{3x} ?

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

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integral from 0 to 1 of xe^{3x}

\int_{0}^{1} x e^{3 x} d x

\frac{2 e ^{3} + 1}{9}

Decimal

4.57456 \dots

Solution steps

\int_{0}^{1} x e^{3 x} d x

Apply u-substitution

= \int_{0}^{3} \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}^{3} e^{u} u d u

Apply Integration By Parts

= \frac{1}{9} [e^{u} u - \int e^{u} d u]_{0}^{3}

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

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

Compute the boundaries:

2 e^{3} + 1

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

Simplify

= \frac{2 e ^{3} + 1}{9}

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