What is the integral of 5e^{-0.5x} ?

The integral of 5e^{-0.5x} is -10e^{-0.5x}+C

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integral of 5e^{-0.5x}

\int 5 e^{- 0.5 x} d x

- 10 e^{- 0.5 x} + C

Solution steps

\int 5 e^{- 0.5 x} d x

Take the constant out:

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

= 5 \cdot \int e^{- 0.5 x} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 (- \frac{1}{0.5} \cdot \int e^{u} d u)

Simplify

= 5 (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

= 5 (- 2 e^{u})

Substitute back

u = - 0.5 x

= 5 (- 2 e^{- 0.5 x})

Simplify

= - 10 e^{- 0.5 x}

Add a constant to the solution

= - 10 e^{- 0.5 x} + C

\frac{d}{d x} (65 - 55 cos (\frac{π x}{15}))

\frac{\partial}{\partial x} (\frac{x ^{2} + 4}{y - 3})

y^{^{'}} = \frac{y ^{2} - y}{x}

x + 2, x^{2}, - 1, 2

y = \frac{x}{3 + x}

What is the integral of 5e^{-0.5x} ?
The integral of 5e^{-0.5x} is -10e^{-0.5x}+C