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

The integral of e^{-0.4x} is -2.5e^{-0.4x}+C

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integral of e^{-0.4x}

\int e^{- 0.4 x} d x

- 2.5 e^{- 0.4 x} + C

Solution steps

\int e^{- 0.4 x} d x

Apply u-substitution

: \int - \frac{1}{0.4} e^{u} d u

= \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

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

Simplify

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

Use the common integral:

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

= - 2.5 e^{u}

Substitute back

u = - 0.4 x

= - 2.5 e^{- 0.4 x}

Add a constant to the solution

= - 2.5 e^{- 0.4 x} + C

x \to - 2 lim (\frac{x + 2}{x ^{3} + 8})

z \to 2 lim (6 z - 1)

7 y^{^{''}} + 2 y^{^{'}} = 0

n = 1 \sum \infty \frac{- 1}{n}

\int_{0}^{2} \frac{x ^{2} - 3}{x + 1} d x

What is the integral of e^{-0.4x} ?
The integral of e^{-0.4x} is -2.5e^{-0.4x}+C