# How do you integrate e^(-1.5x)?

##### 1 Answer
Jun 27, 2015

Use substitution with $u = - 1.5 x$

#### Explanation:

Any integral of the form $\int {e}^{k x} \mathrm{dx}$ where $k$ is a constant, may be evaluated by using $u = k x$, so $\mathrm{du} = k \mathrm{dx}$

$\int {e}^{k x} \mathrm{dx} = \frac{1}{k} \int {e}^{k x} k \mathrm{dx}$

$= \frac{1}{k} \int {e}^{u} \mathrm{du} = \frac{1}{k} {e}^{u} + C$

$= \frac{1}{k} {e}^{k x} + C$

This answer makes sense, because the derivative of ${e}^{k x}$ is ${e}^{k x} \cdot k$
That is: we know the integral of ${e}^{- 1.5 x}$ must involve ${e}^{- 1.5 x}$, but the derivative of that is $\frac{d}{\mathrm{dx}} \left({e}^{- 1.5 x}\right) = - 1.5 {e}^{- 1.5 x}$. We'll remove the $- 1.5$ by dividing:

$\int {e}^{- 1.5 x} \mathrm{dx} = \frac{1}{- 1.5} {e}^{- 1.5 x} + C$

Note: fraction with a decimal in the numerator or denominator looks strange to me. (I think: "Make up you mind! One or the other!")

$1.5 = \frac{3}{2}$ so $\frac{1}{1.5} = \frac{2}{3}$ and we can write our answer:

$- \frac{2}{3} {e}^{- 1.5 x} + C$