What is the integral of te^{-(t^2)/2} ?

The integral of te^{-(t^2)/2} is -e^{-(t^2)/2}+C

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integral of te^{-(t^2)/2}

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

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

Solution steps

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

Apply u-substitution

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

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

Add a constant to the solution

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

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