What is the integral of t^3e^{-t^4} ?

The integral of t^3e^{-t^4} is -1/4 e^{-t^4}+C

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integral of t^3e^{-t^4}

\int t^{3} e^{- t^{4}} d t

- \frac{1}{4} e^{- t^{4}} + C

Solution steps

\int t^{3} e^{- t^{4}} d t

Apply u-substitution

: \int - \frac{e ^{u}}{4} d u

= \int - \frac{e ^{u}}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{4} e^{u}

Substitute back

u = - t^{4}

= - \frac{1}{4} e^{- t^{4}}

Add a constant to the solution

= - \frac{1}{4} e^{- t^{4}} + C

x \to 0 lim (x^{2} cos (\frac{5}{x}))

8 x^{2} y + y^{^{'}} = - 3 x^{2}

\frac{d}{d x} (x^{\frac{1}{2}} y)

r \cdot \frac{d q}{d r} + 3 q = 0

x \to 0 + lim (5 x + 6)

What is the integral of t^3e^{-t^4} ?
The integral of t^3e^{-t^4} is -1/4 e^{-t^4}+C