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

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

Popular >

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

\int_{0}^{1} (x e^{2 x}) d x

\frac{e ^{2} + 1}{4}

Decimal

2.09726 \dots

Solution steps

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

Apply u-substitution

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

= \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

= \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}

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

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

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

Compute the boundaries:

e^{2} + 1

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

Simplify

= \frac{e ^{2} + 1}{4}

\int_{1}^{4} \frac{5 e ^{- x}}{2} d x

\int_{x}^{0} e^{- 3 t} d t

\int_{2}^{4} \frac{3 x - 6}{2} d x

\int_{0}^{2} (4 - x^{2}) d x

\int_{- 2}^{6} x^{3} d x

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