What is the integral of 5e^{-0.4x} ?

The integral of 5e^{-0.4x} is -12.5e^{-0.4x}+C

integral of 5e^{-0.4x}

\int 5 e^{- 0.4 x} d x

- 12.5 e^{- 0.4 x} + C

Solution steps

\int 5 e^{- 0.4 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^{- 0.4 x} d x

Apply u-substitution

= 5 \cdot \int - \frac{1}{0.4} 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}{0.4} \cdot \int e^{u} d u)

Simplify

= 5 (- 2.5 \cdot \int e^{u} d u)

Use the common integral:

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

= 5 (- 2.5 e^{u})

Substitute back

u = - 0.4 x

= 5 (- 2.5 e^{- 0.4 x})

Simplify

= - 12.5 e^{- 0.4 x}

Add a constant to the solution

= - 12.5 e^{- 0.4 x} + C

t an g e n t f (x) = 7 sec (x), at x = \frac{π}{6}

x \to a lim (\frac{1}{0})

\int_{0}^{7} x^{2} + 49 d x

\int (\frac{2}{x} + \frac{x}{2}) d x

\frac{\partial}{\partial y} (x^{2} y^{2})

What is the integral of 5e^{-0.4x} ?
The integral of 5e^{-0.4x} is -12.5e^{-0.4x}+C