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

The integral of 0.5e^{-x} is -0.5e^{-x}+C

integral of 0.5e^{-x}

\int 0.5 e^{- x} d x

- 0.5 e^{- x} + C

Solution steps

\int 0.5 e^{- x} d x

Take the constant out:

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

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

Apply u-substitution

= 0.5 \cdot \int - e^{u} d u

Take the constant out:

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

= 0.5 (- \int e^{u} d u)

Use the common integral:

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

= 0.5 (- e^{u})

Substitute back

u = - x

= 0.5 (- e^{- x})

Simplify

= - 0.5 e^{- x}

Add a constant to the solution

= - 0.5 e^{- x} + C

\int \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (\frac{x ^{5}}{x ^{3}})

x \to 9 lim ([f (18) - g (7)])

n = 2 \sum \infty ln (\frac{1}{2}) ln (n)

t a y l or \frac{e ^{x} + e ^{- x}}{2}