What is the integral from 0 to 5 of xe^{2x} ?

The integral from 0 to 5 of xe^{2x} is (9e^{10}+1)/4

Popular >

integral from 0 to 5 of xe^{2x}

\int_{0}^{5} x e^{2 x} d x

\frac{9 e ^{10} + 1}{4}

Decimal Notation

49559.79803 \dots

Solution steps

\int_{0}^{5} x e^{2 x} d x

Apply u-substitution

: \int_{0}^{10} \frac{e ^{u} u}{4} d u

= \int_{0}^{10} \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

= \frac{1}{4} \cdot \int_{0}^{10} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{10}

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{10}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{10}

Compute the boundaries:

9 e^{10} + 1

= \frac{1}{4} (9 e^{10} + 1)

Simplify

= \frac{9 e ^{10} + 1}{4}

\int_{1}^{2} (\frac{ln ( x )}{x ^{2}}) d x

\int_{- x - 2}^{x} x^{2} d x

\int_{0}^{1} x cos (x y) d y

\int_{3}^{5} 3 x^{2} + 2 x d x

\int_{2}^{2} \frac{1}{x} d x

What is the integral from 0 to 5 of xe^{2x} ?
The integral from 0 to 5 of xe^{2x} is (9e^{10}+1)/4