What is the integral of xe^{-0.5x} ?

The integral of xe^{-0.5x} is 4(-0.5e^{-0.5x}x-e^{-0.5x})+C

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integral of xe^{-0.5x}

\int x e^{- 0.5 x} d x

4 (- 0.5 e^{- 0.5 x} x - e^{- 0.5 x}) + C

Solution steps

\int x e^{- 0.5 x} d x

Apply u-substitution

: \int \frac{2}{0.5} e^{u} u d u

= \int \frac{2}{0.5} e^{u} u d u

Take the constant out:

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

= \frac{2}{0.5} \cdot \int e^{u} u d u

Simplify

= 4 \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= 4 (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= 4 (e^{u} u - e^{u})

Substitute back

u = - 0.5 x

= 4 (e^{- 0.5 x} (- 0.5 x) - e^{- 0.5 x})

Simplify

= 4 (- 0.5 e^{- 0.5 x} x - e^{- 0.5 x})

Add a constant to the solution

= 4 (- 0.5 e^{- 0.5 x} x - e^{- 0.5 x}) + C

n = 1 \sum \infty n e^{- 3 n^{2}}

\frac{\partial}{\partial x} (e^{yz})

y^{^{'}} = a \cdot e^{- 2 x} + b \cdot e^{2 x} + \frac{3 x}{4} - \frac{1}{4}

f (x) = e^{x} sin (x) + cos (x) e^{x}

y = arccos (x)

What is the integral of xe^{-0.5x} ?
The integral of xe^{-0.5x} is 4(-0.5e^{-0.5x}x-e^{-0.5x})+C