What is the integral from 1 to 3 of x^2e^{-x} ?

The integral from 1 to 3 of x^2e^{-x} is -(17)/(e^3)+5/e

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integral from 1 to 3 of x^2e^{-x}

\int_{1}^{3} x^{2} e^{- x} d x

- \frac{17}{e ^{3}} + \frac{5}{e}

Decimal

0.99301 \dots

Solution steps

\int_{1}^{3} x^{2} e^{- x} d x

Apply Integration By Parts

= [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{1}^{3}

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

= [- e^{- x} x^{2} - (- 2 (- e^{- x} x - e^{- x}))]_{1}^{3}

Simplify

= [- e^{- x} x^{2} + 2 (- e^{- x} x - e^{- x})]_{1}^{3}

Compute the boundaries:

- \frac{17}{e ^{3}} + \frac{5}{e}

= - \frac{17}{e ^{3}} + \frac{5}{e}

What is the integral from 1 to 3 of x^2e^{-x} ?
The integral from 1 to 3 of x^2e^{-x} is -(17)/(e^3)+5/e