What is the integral of e^{-t^2}t ?

The integral of e^{-t^2}t is -1/2 e^{-t^2}+C

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integral of e^{-t^2}t

\int e^{- t^{2}} t d t

- \frac{1}{2} e^{- t^{2}} + C

Solution steps

\int e^{- t^{2}} t d t

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = - t^{2}

= - \frac{1}{2} e^{- t^{2}}

Add a constant to the solution

= - \frac{1}{2} e^{- t^{2}} + C

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What is the integral of e^{-t^2}t ?
The integral of e^{-t^2}t is -1/2 e^{-t^2}+C