What is the integral of-0.0008xe^{x^2} ?

The integral of-0.0008xe^{x^2} is -0.0004e^{x^2}+C

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integral of-0.0008xe^{x^2}

\int - 0.0008 x e^{x^{2}} d x

- 0.0004 e^{x^{2}} + C

Solution steps

\int - 0.0008 x e^{x^{2}} d x

Take the constant out:

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

= - 0.0008 \cdot \int x e^{x^{2}} d x

Apply u-substitution

: \int \frac{e ^{u}}{2} d u

= - 0.0008 \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= - 0.0008 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= - 0.0008 \cdot \frac{1}{2} e^{u}

Substitute back

u = x^{2}

= - 0.0008 \cdot \frac{1}{2} e^{x^{2}}

Simplify

- 0.0008 \cdot \frac{1}{2} e^{x^{2}} : - 0.0004 e^{x^{2}}

= - 0.0004 e^{x^{2}}

Add a constant to the solution

= - 0.0004 e^{x^{2}} + C

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What is the integral of-0.0008xe^{x^2} ?
The integral of-0.0008xe^{x^2} is -0.0004e^{x^2}+C