What is the integral from 1 to 2 of x^2e^{-x} ?

The integral from 1 to 2 of x^2e^{-x} is -(10)/(e^2)+5/e

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integral from 1 to 2 of x^2e^{-x}

\int_{1}^{2} x^{2} e^{- x} d x

- \frac{10}{e ^{2}} + \frac{5}{e}

Decimal Notation

0.48604 \dots

Solution steps

\int_{1}^{2} x^{2} e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{1}^{2}

= [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{1}^{2}

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

= [- e^{- x} x^{2} - (- 2 (- e^{- x} x - e^{- x}))]_{1}^{2}

Simplify

= [- e^{- x} x^{2} + 2 (- e^{- x} x - e^{- x})]_{1}^{2}

Compute the boundaries:

- \frac{10}{e ^{2}} + \frac{5}{e}

= - \frac{10}{e ^{2}} + \frac{5}{e}

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What is the integral from 1 to 2 of x^2e^{-x} ?
The integral from 1 to 2 of x^2e^{-x} is -(10)/(e^2)+5/e