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

The integral from-1 to 1 of xe^{2x} is (e^4+3)/(4e^2)

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

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

\frac{e ^{4} + 3}{4 e ^{2}}

Decimal

1.94876 \dots

Solution steps

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

Apply u-substitution

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

= \int_{- 2}^{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_{- 2}^{2} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 2}^{2}

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

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

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

Compute the boundaries:

e^{2} + \frac{3}{e ^{2}}

= \frac{1}{4} (e^{2} + \frac{3}{e ^{2}})

Simplify

= \frac{e ^{4} + 3}{4 e ^{2}}

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What is the integral from-1 to 1 of xe^{2x} ?
The integral from-1 to 1 of xe^{2x} is (e^4+3)/(4e^2)