What is the integral of t^3e^{-t^2} ?

The integral of t^3e^{-t^2} is 1/2 (-e^{-t^2}t^2-e^{-t^2})+C

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integral of t^3e^{-t^2}

\int t^{3} e^{- t^{2}} d t

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

Solution steps

\int t^{3} e^{- t^{2}} d t

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - t^{2}

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

Simplify

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

Add a constant to the solution

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

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What is the integral of t^3e^{-t^2} ?
The integral of t^3e^{-t^2} is 1/2 (-e^{-t^2}t^2-e^{-t^2})+C