What is the integral from 0 to 1 of xe^{8x} ?

The integral from 0 to 1 of xe^{8x} is (7e^8+1)/(64)

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integral from 0 to 1 of xe^{8x}

\int_{0}^{1} x e^{8 x} d x

\frac{7 e ^{8} + 1}{64}

Decimal

326.05790 \dots

Solution steps

\int_{0}^{1} x e^{8 x} d x

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u} u}{64} d u

= \int_{0}^{8} \frac{e ^{u} u}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int_{0}^{8} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{8}

= \frac{1}{64} [e^{u} u - \int e^{u} d u]_{0}^{8}

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

= \frac{1}{64} [e^{u} u - e^{u}]_{0}^{8}

Compute the boundaries:

7 e^{8} + 1

= \frac{1}{64} (7 e^{8} + 1)

Simplify

= \frac{7 e ^{8} + 1}{64}

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What is the integral from 0 to 1 of xe^{8x} ?
The integral from 0 to 1 of xe^{8x} is (7e^8+1)/(64)