How to solve the definite integral $\int_{-4}^{-2}e^{-x}\,dx$?

Question

I'm trying to find the value of the integral $\int_{-4}^{-2}e^{-x}\,dx$ but I just couldn't solve it.

Actually I found in a List of integrals that $\int e^x\,dx=e^x+C$ so I concluded:

$$ \int e^{-x}\,dx=\int\frac{1}{e^x}\,dx=\ln|e^x|$$

$$\int\limits_{-4}^{-2}e^{-x}\,dx=\left(\ln|e^{-2}|\right)-\left(\ln|e^{-4}|\right)=-2+4=2$$

I know the solution is wrong, but how can I solve this integral or any another integral like this?

Answer 1

$$ \int_{-4}^{-2} e^{-x} dx = \left . \left ( -e^{-x}\right ) \right |_{-4}^{-2} = -e^2+e^4 = e^2 \left ( e^2-1\right ) $$

Answer 2

Hint:

$$u=-x$$

$$dx=-du$$

Can you figure out what to do from here?

Answer 3

Isn't easy to see (think to the derivative) that if $\alpha\neq 0$ $$\int e^{\alpha x}=\frac{1}{\alpha} e^{\alpha x}+C \ ?$$