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

The integral from 0 to 1 of 3xe^{2x} is (3(e^2+1))/4

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

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

\frac{3 ( e ^{2} + 1 )}{4}

Decimal Notation

6.29179 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{2} \frac{e ^{u} u}{4} d u

= 3 \cdot \int_{0}^{2} \frac{e ^{u} u}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int_{0}^{2} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{2}

= 3 \cdot \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{2}

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

= 3 \cdot \frac{1}{4} [e^{u} u - e^{u}]_{0}^{2}

Simplify

3 \cdot \frac{1}{4} [e^{u} u - e^{u}]_{0}^{2} : \frac{3}{4} [e^{u} u - e^{u}]_{0}^{2}

= \frac{3}{4} [e^{u} u - e^{u}]_{0}^{2}

Compute the boundaries:

e^{2} + 1

= \frac{3}{4} (e^{2} + 1)

Simplify

= \frac{3 ( e ^{2} + 1 )}{4}

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What is the integral from 0 to 1 of 3xe^{2x} ?
The integral from 0 to 1 of 3xe^{2x} is (3(e^2+1))/4