What is the integral from 1 to 3 of xe^{-x^2} ?

The integral from 1 to 3 of xe^{-x^2} is (e^8-1)/(2e^9)

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integral from 1 to 3 of xe^{-x^2}

\int_{1}^{3} x e^{- x^{2}} d x

\frac{e ^{8} - 1}{2 e ^{9}}

Decimal Notation

0.18387 \dots

Solution steps

\int_{1}^{3} x e^{- x^{2}} d x

Apply u-substitution

: \int_{- 1}^{- 9} - \frac{e ^{u}}{2} d u

= \int_{- 1}^{- 9} - \frac{e ^{u}}{2} d u

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

= - \int_{- 9}^{- 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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

= \frac{e ^{8} - 1}{2 e ^{9}}

\frac{d}{d x} (\frac{x - 3}{x ^{2} + 4})

(0.2) (6.6)

x \to 3 lim (\frac{x ^{2}}{9 - x ^{2}})

\frac{d}{d x} (\frac{( e ^{x} )}{x ^{2}})

\frac{2}{x - 1}

What is the integral from 1 to 3 of xe^{-x^2} ?
The integral from 1 to 3 of xe^{-x^2} is (e^8-1)/(2e^9)