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

The integral from 0 to x of x^2e^{-2x} is (-e^{-2x}x^2-e^{-2x}x)/2-1/4 e^{-2x}+1/4

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

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

\frac{- e ^{- 2 x} x ^{2} - e ^{- 2 x} x}{2} - \frac{1}{4} e^{- 2 x} + \frac{1}{4}

Solution steps

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

Apply Integration By Parts

: [- \frac{1}{2} e^{- 2 x} x^{2} - \int - e^{- 2 x} x d x]_{0}^{x}

= [- \frac{1}{2} e^{- 2 x} x^{2} - \int - e^{- 2 x} x d x]_{0}^{x}

\int - e^{- 2 x} x d x = - \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x})

= [- \frac{1}{2} e^{- 2 x} x^{2} - (- \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}))]_{0}^{x}

Simplify

= [- \frac{1}{2} e^{- 2 x} x^{2} + \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x})]_{0}^{x}

Compute the boundaries:

- \frac{1}{2} e^{- 2 x} x^{2} + \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + \frac{1}{4}

= - \frac{1}{2} e^{- 2 x} x^{2} + \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + \frac{1}{4}

Simplify

= \frac{- e ^{- 2 x} x ^{2} - e ^{- 2 x} x}{2} - \frac{1}{4} e^{- 2 x} + \frac{1}{4}

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What is the integral from 0 to x of x^2e^{-2x} ?
The integral from 0 to x of x^2e^{-2x} is (-e^{-2x}x^2-e^{-2x}x)/2-1/4 e^{-2x}+1/4