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

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

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

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

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

Decimal Notation

4.19452 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \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}

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

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

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

Simplify

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

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

Compute the boundaries:

e^{2} + 1

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

Simplify

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

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