What is the integral of 10e^{-x/2} ?

The integral of 10e^{-x/2} is -20e^{-x/2}+C

integral of 10e^{-x/2}

\int 10 e^{- \frac{x}{2}} d x

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

Solution steps

\int 10 e^{- \frac{x}{2}} d x

Take the constant out:

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

= 10 \cdot \int e^{- \frac{x}{2}} d x

Apply u-substitution

= 10 \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

= 10 (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

= 10 (- 2 e^{u})

Substitute back

u = - \frac{x}{2}

= 10 (- 2 e^{- \frac{x}{2}})

Simplify

= - 20 e^{- \frac{x}{2}}

Add a constant to the solution

= - 20 e^{- \frac{x}{2}} + C

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What is the integral of 10e^{-x/2} ?
The integral of 10e^{-x/2} is -20e^{-x/2}+C