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

The integral from 0 to x of x^2e^{-x/4} is -4e^{-x/4}x^2+8(-4e^{-x/4}x-16e^{-x/4})+128

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

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

- 4 e^{- \frac{x}{4}} x^{2} + 8 (- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}}) + 128

Solution steps

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

Apply Integration By Parts

: [- 4 e^{- \frac{x}{4}} x^{2} - \int - 8 e^{- \frac{x}{4}} x d x]_{0}^{x}

= [- 4 e^{- \frac{x}{4}} x^{2} - \int - 8 e^{- \frac{x}{4}} x d x]_{0}^{x}

\int - 8 e^{- \frac{x}{4}} x d x = - 8 (- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}})

= [- 4 e^{- \frac{x}{4}} x^{2} - (- 8 (- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}}))]_{0}^{x}

Simplify

= [- 4 e^{- \frac{x}{4}} x^{2} + 8 (- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}})]_{0}^{x}

Compute the boundaries:

- 4 e^{- \frac{x}{4}} x^{2} + 8 (- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}}) + 128

= - 4 e^{- \frac{x}{4}} x^{2} + 8 (- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}}) + 128

x \to \infty lim (x - e^{x})

- 10 (I^{2} + x^{2} - I)

(3, - 4), (5, 2)

\int \frac{x ^{2}}{1 - x ^{2}} d x

\frac{\partial}{\partial x} (2 g (x) \cdot (x^{- 2}) \cdot (\frac{D}{L}))

What is the integral from 0 to x of x^2e^{-x/4} ?
The integral from 0 to x of x^2e^{-x/4} is -4e^{-x/4}x^2+8(-4e^{-x/4}x-16e^{-x/4})+128