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

The integral from 1 to 2 of xe^{-2x^2} is (e^6-1)/(4e^8)

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

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

\frac{e ^{6} - 1}{4 e ^{8}}

Decimal

0.03374 \dots

Solution steps

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

Apply u-substitution

= \int_{- 2}^{- 8} - \frac{e ^{u}}{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 8}^{- 2} - \frac{e ^{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_{- 8}^{- 2} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{4} [e^{u}]_{- 8}^{- 2})

Simplify

= \frac{1}{4} [e^{u}]_{- 8}^{- 2}

Compute the boundaries:

\frac{1}{e ^{2}} - \frac{1}{e ^{8}}

= \frac{1}{4} (\frac{1}{e ^{2}} - \frac{1}{e ^{8}})

Simplify

= \frac{e ^{6} - 1}{4 e ^{8}}

What is the integral from 1 to 2 of xe^{-2x^2} ?
The integral from 1 to 2 of xe^{-2x^2} is (e^6-1)/(4e^8)