What is the integral of 2e^{-0.5x} ?

The integral of 2e^{-0.5x} is -4e^{-0.5x}+C

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integral of 2e^{-0.5x}

\int 2 e^{- 0.5 x} d x

- 4 e^{- 0.5 x} + C

Solution steps

\int 2 e^{- 0.5 x} d x

Take the constant out:

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

= 2 \cdot \int e^{- 0.5 x} d x

Apply u-substitution

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

= 2 \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

= 2 (- \frac{1}{0.5} \cdot \int e^{u} d u)

Simplify

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

Use the common integral:

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

= 2 (- 2 e^{u})

Substitute back

u = - 0.5 x

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

Simplify

= - 4 e^{- 0.5 x}

Add a constant to the solution

= - 4 e^{- 0.5 x} + C

\frac{d}{d x} (2 x^{3} - 4 x + 5)

x \to 0 lim (+ arctan (ln (x)))

x \to - 1 lim (11 x + 9)

\frac{t}{t - 1}

(1 + cos (t)) y^{^{'}} = (1 + e - y) sin (t)

What is the integral of 2e^{-0.5x} ?
The integral of 2e^{-0.5x} is -4e^{-0.5x}+C