What is the integral from 0 to 7 of x^2e^{-x} ?

The integral from 0 to 7 of x^2e^{-x} is -(65)/(e^7)+2

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integral from 0 to 7 of x^2e^{-x}

\int_{0}^{7} x^{2} e^{- x} d x

- \frac{65}{e ^{7}} + 2

Decimal Notation

1.94072 \dots

Solution steps

\int_{0}^{7} x^{2} e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{0}^{7}

= [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{0}^{7}

\int - 2 e^{- x} x d x = - 2 (- e^{- x} x - e^{- x})

= [- e^{- x} x^{2} - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{7}

Simplify

= [- e^{- x} x^{2} + 2 (- e^{- x} x - e^{- x})]_{0}^{7}

Compute the boundaries:

- \frac{65}{e ^{7}} + 2

= - \frac{65}{e ^{7}} + 2

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What is the integral from 0 to 7 of x^2e^{-x} ?
The integral from 0 to 7 of x^2e^{-x} is -(65)/(e^7)+2