What is the integral of xe^{-0.5x^2} ?

The integral of xe^{-0.5x^2} is -e^{-0.5x^2}+C

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integral of xe^{-0.5x^2}

\int x e^{- 0.5 x^{2}} d x

- e^{- 0.5 x^{2}} + C

Solution steps

\int x e^{- 0.5 x^{2}} d x

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = - 0.5 x^{2}

= - e^{- 0.5 x^{2}}

Add a constant to the solution

= - e^{- 0.5 x^{2}} + C

\int_{0}^{2 π} sin^{2} (\frac{1}{3} x) d x

x \to 0 lim ((1 + x)^{\frac{1}{x}})

\frac{d}{d x} (\frac{( x - 1 ) ^{2}}{x ^{2} + 2})

f (x) = (x - 1) (x^{2} - 8 x + 7)

f (x) = \frac{- cos ( x )}{4 x ^{2} + 2 x - 3}

What is the integral of xe^{-0.5x^2} ?
The integral of xe^{-0.5x^2} is -e^{-0.5x^2}+C