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

The integral from 0 to 2 of 2x^2e^{-2x} is 2(-(13)/(4e^4)+1/4)

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

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

2 (- \frac{13}{4 e ^{4}} + \frac{1}{4})

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{2} 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}^{2}

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

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

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

Simplify

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

Compute the boundaries:

- \frac{13}{4 e ^{4}} + \frac{1}{4}

= 2 (- \frac{13}{4 e ^{4}} + \frac{1}{4})

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What is the integral from 0 to 2 of 2x^2e^{-2x} ?
The integral from 0 to 2 of 2x^2e^{-2x} is 2(-(13)/(4e^4)+1/4)