What is the integral of 10(20-x)e^{-x/(20)} ?

The integral of 10(20-x)e^{-x/(20)} is 200e^{-x/(20)}x+C

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integral of 10(20-x)e^{-x/(20)}

\int 10 (20 - x) e^{- \frac{x}{20}} d x

200 e^{- \frac{x}{20}} x + C

Solution steps

\int 10 (20 - x) e^{- \frac{x}{20}} d x

Take the constant out:

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

= 10 \cdot \int (20 - x) e^{- \frac{x}{20}} d x

Expand

(20 - x) e^{- \frac{x}{20}} : 20 e^{- \frac{x}{20}} - e^{- \frac{x}{20}} x

= 10 \cdot \int 20 e^{- \frac{x}{20}} - e^{- \frac{x}{20}} x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 10 (\int 20 e^{- \frac{x}{20}} d x - \int e^{- \frac{x}{20}} x d x)

\int 20 e^{- \frac{x}{20}} d x = - 400 e^{- \frac{x}{20}}

\int e^{- \frac{x}{20}} x d x = - 20 e^{- \frac{x}{20}} x - 400 e^{- \frac{x}{20}}

= 10 (- 400 e^{- \frac{x}{20}} - (- 20 e^{- \frac{x}{20}} x - 400 e^{- \frac{x}{20}}))

Simplify

10 (- 400 e^{- \frac{x}{20}} - (- 20 e^{- \frac{x}{20}} x - 400 e^{- \frac{x}{20}})) : 200 e^{- \frac{x}{20}} x

= 200 e^{- \frac{x}{20}} x

Add a constant to the solution

= 200 e^{- \frac{x}{20}} x + C

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What is the integral of 10(20-x)e^{-x/(20)} ?
The integral of 10(20-x)e^{-x/(20)} is 200e^{-x/(20)}x+C