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

The integral from 0 to 2 of xe^{3x} is (5e^6+1)/9

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

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

\frac{5 e ^{6} + 1}{9}

Decimal

224.23821 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{6} \frac{e ^{u} u}{9} d u

= \int_{0}^{6} \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}^{6} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{6}

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

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

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

Compute the boundaries:

5 e^{6} + 1

= \frac{1}{9} (5 e^{6} + 1)

Simplify

= \frac{5 e ^{6} + 1}{9}

\int_{9}^{10} (x - 10)^{9} d x

\int_{3}^{5} \frac{x}{x ^{2} - 4} d x

\int_{0}^{1} \frac{x ^{2}}{2 x + 1} d x

\int_{3}^{10} \frac{1}{x} d x

\int_{0}^{1} \frac{e ^{x} + 2}{e ^{x}} d x

What is the integral from 0 to 2 of xe^{3x} ?
The integral from 0 to 2 of xe^{3x} is (5e^6+1)/9