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 Notation

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}

= [- 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}

\int_{0}^{3} \frac{e ^{\frac{3}{x}}}{x ^{2}} d x

\int_{0}^{2} \frac{2}{x ^{2}} d x

\int_{0}^{2} (4 x - x^{2} - x^{2}) d x

\int_{4}^{\infty} \frac{3}{x ^{5}} d x

\int_{1}^{3} 2 x^{2} + 3 d x

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