What is the integral of 1/32 xe^{(-x^2)/(64)} ?

The integral of 1/32 xe^{(-x^2)/(64)} is -e^{-(x^2)/(64)}+C

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integral of 1/32 xe^{(-x^2)/(64)}

\int \frac{1}{32} x e^{\frac{- x ^{2}}{64}} d x

- e^{- \frac{x ^{2}}{64}} + C

Solution steps

\int \frac{1}{32} x e^{\frac{- x ^{2}}{64}} d x

Take the constant out:

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

= \frac{1}{32} \cdot \int x e^{\frac{- x ^{2}}{64}} d x

Simplify

= \frac{1}{32} \cdot \int e^{- \frac{x ^{2}}{64}} x d x

Apply u-substitution

: \int - 32 e^{u} d u

= \frac{1}{32} \cdot \int - 32 e^{u} d u

Take the constant out:

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

= \frac{1}{32} (- 32 \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{32} (- 32 e^{u})

Substitute back

u = - \frac{x ^{2}}{64}

= \frac{1}{32} (- 32 e^{- \frac{x ^{2}}{64}})

Simplify

\frac{1}{32} (- 32 e^{- \frac{x ^{2}}{64}}) : - e^{- \frac{x ^{2}}{64}}

= - e^{- \frac{x ^{2}}{64}}

Add a constant to the solution

= - e^{- \frac{x ^{2}}{64}} + C

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What is the integral of 1/32 xe^{(-x^2)/(64)} ?
The integral of 1/32 xe^{(-x^2)/(64)} is -e^{-(x^2)/(64)}+C