What is the integral from 0 to 1 of xe^{4x} ?

The integral from 0 to 1 of xe^{4x} is (3e^4+1)/(16)

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integral from 0 to 1 of xe^{4x}

\int_{0}^{1} x e^{4 x} d x

\frac{3 e ^{4} + 1}{16}

Decimal

10.29965 \dots

Solution steps

\int_{0}^{1} x e^{4 x} d x

Apply u-substitution

: \int_{0}^{4} \frac{e ^{u} u}{16} d u

= \int_{0}^{4} \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{0}^{4} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{4}

= \frac{1}{16} [e^{u} u - \int e^{u} d u]_{0}^{4}

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

= \frac{1}{16} [e^{u} u - e^{u}]_{0}^{4}

Compute the boundaries:

3 e^{4} + 1

= \frac{1}{16} (3 e^{4} + 1)

Simplify

= \frac{3 e ^{4} + 1}{16}

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What is the integral from 0 to 1 of xe^{4x} ?
The integral from 0 to 1 of xe^{4x} is (3e^4+1)/(16)