What is the integral from 0 to 3 of 4xe^{-x} ?

The integral from 0 to 3 of 4xe^{-x} is 4(-4/(e^3)+1)

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integral from 0 to 3 of 4xe^{-x}

\int_{0}^{3} 4 x e^{- x} d x

4 (- \frac{4}{e ^{3}} + 1)

Solution steps

\int_{0}^{3} 4 x e^{- x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{3} x e^{- x} d x

Apply Integration By Parts

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

= 4 [- e^{- x} x - \int - e^{- x} d x]_{0}^{3}

\int - e^{- x} d x = e^{- x}

= 4 [- e^{- x} x - e^{- x}]_{0}^{3}

Compute the boundaries:

- \frac{4}{e ^{3}} + 1

= 4 (- \frac{4}{e ^{3}} + 1)

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What is the integral from 0 to 3 of 4xe^{-x} ?
The integral from 0 to 3 of 4xe^{-x} is 4(-4/(e^3)+1)