What is the integral from 0 to 1 of 2xe^{3x} ?

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

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integral from 0 to 1 of 2xe^{3x}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{3} \frac{e ^{u} u}{9} d u

= 2 \cdot \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

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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{3}

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

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

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

Simplify

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

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

Compute the boundaries:

2 e^{3} + 1

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

Simplify

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

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