What is the integral of 4e^{-0.5x} ?

The integral of 4e^{-0.5x} is -8e^{-0.5x}+C

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integral of 4e^{-0.5x}

\int 4 e^{- 0.5 x} d x

- 8 e^{- 0.5 x} + C

Solution steps

\int 4 e^{- 0.5 x} d x

Take the constant out:

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

= 4 \cdot \int e^{- 0.5 x} d x

Apply u-substitution

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

= 4 \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

= 4 (- \frac{1}{0.5} \cdot \int e^{u} d u)

Simplify

= 4 (- 2 \cdot \int e^{u} d u)

Use the common integral:

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

= 4 (- 2 e^{u})

Substitute back

u = - 0.5 x

= 4 (- 2 e^{- 0.5 x})

Simplify

= - 8 e^{- 0.5 x}

Add a constant to the solution

= - 8 e^{- 0.5 x} + C

f^{^{'}} (x) = x \cdot 1 5^{x}

\frac{x}{2 x + 3}

\int_{1}^{16} \frac{6}{x} d x

\int_{2}^{5} \frac{1}{2 - 5 t} d t

9 y^{^{''}} - 6 y^{^{'}} + 4 y = 0

What is the integral of 4e^{-0.5x} ?
The integral of 4e^{-0.5x} is -8e^{-0.5x}+C