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

The integral of e^{-0.5x} is -2e^{-0.5x}+C

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integral of e^{-0.5x}

\int e^{- 0.5 x} d x

- 2 e^{- 0.5 x} + C

Solution steps

\int e^{- 0.5 x} d x

Apply u-substitution

: \int - \frac{1}{0.5} e^{u} d u

= \int - \frac{1}{0.5} e^{u} d u

Take the constant out:

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

= - \frac{1}{0.5} \cdot \int e^{u} d u

Simplify

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

Use the common integral:

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

= - 2 e^{u}

Substitute back

u = - 0.5 x

= - 2 e^{- 0.5 x}

Add a constant to the solution

= - 2 e^{- 0.5 x} + C

\int (x^{2} + 4)^{7} d x

f (x) = x^{2} arctan (3 x)

\int y^{2} e^{x y} d x

f (x) = \frac{4 x}{x - 5}, at x = 7

\frac{\partial}{\partial x} (ln (\frac{x}{y - z}))