What is the integral from 0 to 2 of 2xe^{-x} ?

The integral from 0 to 2 of 2xe^{-x} is 2(-3/(e^2)+1)

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integral from 0 to 2 of 2xe^{-x}

\int_{0}^{2} 2 x e^{- x} d x

2 (- \frac{3}{e ^{2}} + 1)

Decimal

1.18798 \dots

Solution steps

\int_{0}^{2} 2 x e^{- x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x - \int - e^{- x} d x]_{0}^{2}

= 2 [- e^{- x} x - \int - e^{- x} d x]_{0}^{2}

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

= 2 [- e^{- x} x - e^{- x}]_{0}^{2}

Compute the boundaries:

- \frac{3}{e ^{2}} + 1

= 2 (- \frac{3}{e ^{2}} + 1)

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What is the integral from 0 to 2 of 2xe^{-x} ?
The integral from 0 to 2 of 2xe^{-x} is 2(-3/(e^2)+1)