What is the integral of 1/4 x^2e^{-x/2} ?

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

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integral of 1/4 x^2e^{-x/2}

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: - 2 e^{- \frac{x}{2}} x^{2} - \int - 4 e^{- \frac{x}{2}} x d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

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