What is the integral from 0 to 5 of xe^{5x} ?

The integral from 0 to 5 of xe^{5x} is (24e^{25}+1)/(25)

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integral from 0 to 5 of xe^{5x}

\int_{0}^{5} x e^{5 x} d x

\frac{24 e ^{25} + 1}{25}

Decimal Notation

69124703363.93034

Solution steps

\int_{0}^{5} x e^{5 x} d x

Apply u-substitution

: \int_{0}^{25} \frac{e ^{u} u}{25} d u

= \int_{0}^{25} \frac{e ^{u} u}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int_{0}^{25} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{25}

= \frac{1}{25} [e^{u} u - \int e^{u} d u]_{0}^{25}

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

= \frac{1}{25} [e^{u} u - e^{u}]_{0}^{25}

Compute the boundaries:

24 e^{25} + 1

= \frac{1}{25} (24 e^{25} + 1)

Simplify

= \frac{24 e ^{25} + 1}{25}

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What is the integral from 0 to 5 of xe^{5x} ?
The integral from 0 to 5 of xe^{5x} is (24e^{25}+1)/(25)