What is the integral of 16t^7e^{-t^8} ?

The integral of 16t^7e^{-t^8} is -2e^{-t^8}+C

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integral of 16t^7e^{-t^8}

\int 16 t^{7} e^{- t^{8}} d t

- 2 e^{- t^{8}} + C

Solution steps

\int 16 t^{7} e^{- t^{8}} d t

Take the constant out:

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

= 16 \cdot \int t^{7} e^{- t^{8}} d t

Apply u-substitution

= 16 \cdot \int - \frac{e ^{u}}{8} d u

Take the constant out:

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

= 16 (- \frac{1}{8} \cdot \int e^{u} d u)

Use the common integral:

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

= 16 (- \frac{1}{8} e^{u})

Substitute back

u = - t^{8}

= 16 (- \frac{1}{8} e^{- t^{8}})

Simplify

16 (- \frac{1}{8} e^{- t^{8}}) : - 2 e^{- t^{8}}

= - 2 e^{- t^{8}}

Add a constant to the solution

= - 2 e^{- t^{8}} + C

What is the integral of 16t^7e^{-t^8} ?
The integral of 16t^7e^{-t^8} is -2e^{-t^8}+C