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

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

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

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

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

Decimal

0.34956 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 1}^{- 4} - \frac{e ^{u}}{2} d u

= 2 \cdot \int_{- 1}^{- 4} - \frac{e ^{u}}{2} d u

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

= 2 (- \int_{- 4}^{- 1} - \frac{e ^{u}}{2} d u)

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{- 4}^{- 1} e^{u} d u))

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

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