Hi everyone...I am having a brain pain...I seem to have forgotten how to do this specific integral. Integrate e^(-3x)...Can someone take a look and explain how to do it? I know that the integral of e^anything = e^anything but I can't seem to remember why I also need to divide by -1/3...please help

Question

anonymous · Answer

attachment

anonymous · Answer

Could someone take a look at the 13.bmp and tell me how to get the -1/3?

anonymous · Answer

Think about what happens when you derivate  
$$(e^{-3\cdot x})' = -3\cdot e^{-3\cdot x}$$ Divide both sides by (-1/3). What can you deduce from that? Remember that the following rule holds for differentiating:
$$(c\cdot f(x))' = c * f'(x)$$

anonymous · Answer

so am I setting -3x = u or something?

anonymous · Answer

No, not really. Do you understand the first equation I posted?

anonymous · Answer

OKay I remember now! I set u=-3x then dx=-1/3...yada yada yada...I got it...Thanks guys!