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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - 2 e^{u} d u

= \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{1}{2} x

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

Simplify

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

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

Add a constant to the solution

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

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

\int_{- 1}^{0} \frac{4}{1 + ( 2 x + 1 ) ^{2}} d x

x^{3} + 9 x

\frac{d}{d x} (\frac{2 x}{2 x + 1})

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