What is the integral from 0 to 4 of x^2e^{-x} ?

The integral from 0 to 4 of x^2e^{-x} is -(26)/(e^4)+2

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integral from 0 to 4 of x^2e^{-x}

\int_{0}^{4} x^{2} e^{- x} d x

- \frac{26}{e ^{4}} + 2

Solution steps

\int_{0}^{4} x^{2} e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{0}^{4}

= [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{0}^{4}

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

= [- e^{- x} x^{2} - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{4}

Simplify

= [- e^{- x} x^{2} + 2 (- e^{- x} x - e^{- x})]_{0}^{4}

Compute the boundaries:

- \frac{26}{e ^{4}} + 2

= - \frac{26}{e ^{4}} + 2

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

\int_{1}^{2} (x^{2} - 4 x + 3) d x

\int_{- 1}^{1} t^{7} (1 + t^{8})^{7} d t

\int_{0}^{1} x arctan (x^{2}) d x

\int_{0}^{1} - x^{2} + 2 x d x

What is the integral from 0 to 4 of x^2e^{-x} ?
The integral from 0 to 4 of x^2e^{-x} is -(26)/(e^4)+2