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

The integral from 0 to t of x^2e^{-x} is -t^2e^{-t}+2(-te^{-t}-e^{-t})+2

Popular >

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

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

- t^{2} e^{- t} + 2 (- t e^{- t} - e^{- t}) + 2

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

- t^{2} e^{- t} + 2 (- t e^{- t} - e^{- t}) + 2

= - t^{2} e^{- t} + 2 (- t e^{- t} - e^{- t}) + 2

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

\int_{0}^{2} π (2 - x) d x

\int_{0}^{4} \frac{2}{9 x + 8} d x

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

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

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