What is the integral of (x^0e^{-x/2})/2 ?

The integral of (x^0e^{-x/2})/2 is -e^{-x/2}+C

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integral of (x^0e^{-x/2})/2

\int \frac{x ^{0} e ^{- \frac{x}{2}}}{2} d x

- e^{- \frac{x}{2}} + C

Solution steps

\int \frac{x ^{0} e ^{- \frac{x}{2}}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{0} e^{- \frac{x}{2}} d x

Simplify

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

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

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

Substitute back

u = - \frac{x}{2}

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

Simplify

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

= - e^{- \frac{x}{2}}

Add a constant to the solution

= - e^{- \frac{x}{2}} + C

What is the integral of (x^0e^{-x/2})/2 ?
The integral of (x^0e^{-x/2})/2 is -e^{-x/2}+C