What is the integral from 3 to 5 of e^{-x^2}x ?

The integral from 3 to 5 of e^{-x^2}x is (1/(e^9)-1/(e^{25)})/2

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integral from 3 to 5 of e^{-x^2}x

\int_{3}^{5} e^{- x^{2}} x d x

\frac{\frac{1}{e ^{9}} - \frac{1}{e ^{25}}}{2}

Decimal

0.00006 \dots

Solution steps

\int_{3}^{5} e^{- x^{2}} x d x

Apply u-substitution

: \int_{- 9}^{- 25} - \frac{e ^{u}}{2} d u

= \int_{- 9}^{- 25} - \frac{e ^{u}}{2} d u

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

= - \int_{- 25}^{- 9} - \frac{e ^{u}}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{- 25}^{- 9} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 25}^{- 9})

Simplify

= \frac{1}{2} [e^{u}]_{- 25}^{- 9}

Compute the boundaries:

\frac{1}{e ^{9}} - \frac{1}{e ^{25}}

= \frac{1}{2} (\frac{1}{e ^{9}} - \frac{1}{e ^{25}})

Simplify

= \frac{\frac{1}{e ^{9}} - \frac{1}{e ^{25}}}{2}

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What is the integral from 3 to 5 of e^{-x^2}x ?
The integral from 3 to 5 of e^{-x^2}x is (1/(e^9)-1/(e^{25)})/2