What is the integral of 2e^{-0.8x} ?

The integral of 2e^{-0.8x} is -2.5e^{-0.8x}+C

integral of 2e^{-0.8x}

\int 2 e^{- 0.8 x} d x

- 2.5 e^{- 0.8 x} + C

Solution steps

\int 2 e^{- 0.8 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^{- 0.8 x} d x

Apply u-substitution

= 2 \cdot \int - \frac{1}{0.8} 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}{0.8} \cdot \int e^{u} d u)

Simplify

= 2 (- 1.25 \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- 1.25 e^{u})

Substitute back

u = - 0.8 x

= 2 (- 1.25 e^{- 0.8 x})

Simplify

= - 2.5 e^{- 0.8 x}

Add a constant to the solution

= - 2.5 e^{- 0.8 x} + C

x y^{^{'}} + 3 x - 2 y = 0

e^{2 x} y^{^{'}} + 2 e^{2 x} y = 2 x

x \to 6 + lim (x^{3} + c x)

t a y l or ln (x), 3

in v erse l a pl a ce \frac{1}{( s + 2 ) ^{2}}

What is the integral of 2e^{-0.8x} ?
The integral of 2e^{-0.8x} is -2.5e^{-0.8x}+C