## anonymous 2 years ago Find the value of k that the line x=k divides the area of the first quadrant region of y=e^-x and the x-axis x >= 1 into two equal parts.

1. myininaya

So you have talked about improper integrals?

2. anonymous

yes

3. myininaya

so we want this: $1/2 \int\limits_{1}^{k}e^{-x} dx=\lim_{a \rightarrow \infty}\int\limits_{1}^{a}e^{-x} dx$

4. myininaya

we want that one area to be half of that other area

5. myininaya

did you get this far?

6. anonymous

i see

7. anonymous

8. myininaya

so we can try to evaluate both integrals first then we just solve the equation for k

9. anonymous

That's true haha it seems so simple now thank you :)

10. anonymous

I get that k=1, but i know that's not the answer.

11. myininaya

what is the answer? also yep if k is one, then that one area is 0 and we know 0 is not half the other area

12. anonymous

$-1/e^k = 1/e$

13. anonymous

sorry it wouldnt be one, but this is the end result i believe

14. myininaya

that negative is weird killing me because e^(to a positive number) can't be negative

15. myininaya

i was agreeing it isn't 1 because it would give us 0 for that one area

16. anonymous

the antiderivative of e^-x is -e^x right?

17. anonymous

-E^-x*

18. myininaya

maybe since we can't solve that equation for k there is no way possible to divide the area with a vertical line so we have two equal areas on both sides

19. myininaya

yep you are right

20. anonymous

Okay, well here's some help. plugging the e^-x in my calculator and then taking the integral of that I find the area, i divided it by 2 and then checked the x y table to find the x that gives half the area. It is about 1.693

21. myininaya

i got it

22. myininaya

i wrote the equation just a little wrong

23. anonymous

oh

24. myininaya

$\int\limits\limits_{1}^{k}e^{-x} dx=1/2 \lim_{a \rightarrow \infty}\int\limits\limits_{1}^{a}e^{-x} dx$

25. myininaya

you will get the answer you got using your calculator except it will be exact

26. myininaya

i'm idiot sorry

27. anonymous

oh such an easily overseen mistake! Ill check this out right now hahhaaha it's perfectly fine :D thanks for taking the time to figure it out!

28. myininaya

Did you get it?

29. anonymous

$-1/e^k+1/e=1/2e$

30. anonymous

thats the function right?

31. myininaya

yep good so far

32. anonymous

Then just solve for k?

33. myininaya

yep! :) I left it as $e^{-k}+e^{-1}=e^{-1}/2$

34. myininaya

oops with a negative in front of the e^(-k)

35. myininaya

$-e^{-k}+e^{-1}=1/2e^{-1} => -e^{-k}=-1/2 e^{-1}$

36. myininaya

get rid of those negatives then take natural log of both sides

37. anonymous

Yes I got it! Thank you so much :) it's gonna be: $-\ln (1/2e)$

38. myininaya

or you could write it as ln(2e)

39. anonymous

true haha

40. myininaya

but yep if you put that in your calculator you will see that its approximation is the same as the approximation you got using just the calculator to find what was it 1.6 something

41. anonymous

Yes I noticed that haha. I can't thank you enough.

42. myininaya

I'm going to need to learn that cool calculator trick for myself

43. anonymous

I'm rather inept at using the calculator, but that's one trick I know ;)

44. myininaya

i know the basics and i know there are really cool tricks in stuff i could learn just never got around to it

45. myininaya

anyways fun problem we are going over improper integrals today in class :)

46. anonymous

cool.

47. anonymous

Well, where i am it's just about dinner time. Thanks again it really saved a lot of hair pulling. See ya