What is the integral of pe^{-0.1p} ?

The integral of pe^{-0.1p} is 100(-0.1e^{-0.1p}p-e^{-0.1p})+C

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integral of pe^{-0.1p}

\int p e^{- 0.1 p} d p

100 (- 0.1 e^{- 0.1 p} p - e^{- 0.1 p}) + C

Solution steps

\int p e^{- 0.1 p} d p

Apply u-substitution

: \int \frac{10}{0.1} e^{u} u d u

= \int \frac{10}{0.1} e^{u} u d u

Take the constant out:

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

= \frac{10}{0.1} \cdot \int e^{u} u d u

Simplify

= 100 \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 100 (e^{u} u - \int e^{u} d u)

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

= 100 (e^{u} u - e^{u})

Substitute back

u = - 0.1 p

= 100 (e^{- 0.1 p} (- 0.1 p) - e^{- 0.1 p})

Simplify

= 100 (- 0.1 e^{- 0.1 p} p - e^{- 0.1 p})

Add a constant to the solution

= 100 (- 0.1 e^{- 0.1 p} p - e^{- 0.1 p}) + C

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What is the integral of pe^{-0.1p} ?
The integral of pe^{-0.1p} is 100(-0.1e^{-0.1p}p-e^{-0.1p})+C