1. ## improper integrals

Hi everyone,

Could someone please tell me if this is correct?

∫-infinity to -1 e^-2xdx

lim t goes to -infinity ∫t to -1 e^2x dx
lim t goes to -infinity e^-2x/-2|t to -1

lim t goes to -infinity e^2/-2 -e^-2t/-2

Would I have to use L'hopitals rule for both of these?

Thank you very much

2. Originally Posted by chocolatelover
Hi everyone,

Could someone please tell me if this is correct?

∫-infinity to -1 e^-2xdx

lim t goes to -infinity ∫t to -1 e^2x dx
lim t goes to -infinity e^-2x/-2|t to -1

lim t goes to -infinity e^2/-2 -e^-2t/-2

Would I have to use L'hopitals rule for both of these?

Thank you very much
the conditions for L'Hopital's rule are not fulfilled here, and it is not needed to begin with. these limits are easy enough to take directly

3. Could you show me what to do, please? How would you do it without L'Hopital's rule? Would you just look at the graph?

Thank you

4. Originally Posted by chocolatelover
Could you show me what to do, please? How would you do it without L'Hopital's rule? Would you just look at the graph?

Thank you
it is quite obvious that $\displaystyle \lim_{x \to \infty}e^x = \infty$, yes, you can look at the graph

5. Would the final anwer be e^2/-2-infinity???? or would it just be infinity?

Thank you

6. Originally Posted by chocolatelover
Would the final anwer be e^2/-2-infinity???? or would it just be infinity?

Thank you
you should have $\displaystyle \lim_{t \to - \infty} \left( - \frac 12 e^2 + \frac 12 e^{-2t}\right)$ and the answer would just be $\displaystyle \infty$ of course. you have infinity minus a very small constant, it's still infinity