What is the integral from 0 to 9 of xe^{-x/3} ?

The integral from 0 to 9 of xe^{-x/3} is -(36)/(e^3)+9

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integral from 0 to 9 of xe^{-x/3}

\int_{0}^{9} x e^{- \frac{x}{3}} d x

- \frac{36}{e ^{3}} + 9

Solution steps

\int_{0}^{9} x e^{- \frac{x}{3}} d x

Apply Integration By Parts

: [- 3 e^{- \frac{x}{3}} x - \int - 3 e^{- \frac{x}{3}} d x]_{0}^{9}

= [- 3 e^{- \frac{x}{3}} x - \int - 3 e^{- \frac{x}{3}} d x]_{0}^{9}

\int - 3 e^{- \frac{x}{3}} d x = 9 e^{- \frac{x}{3}}

= [- 3 e^{- \frac{x}{3}} x - 9 e^{- \frac{x}{3}}]_{0}^{9}

Compute the boundaries:

- \frac{36}{e ^{3}} + 9

= - \frac{36}{e ^{3}} + 9

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What is the integral from 0 to 9 of xe^{-x/3} ?
The integral from 0 to 9 of xe^{-x/3} is -(36)/(e^3)+9