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.

- anonymous

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- myininaya

So you have talked about improper integrals?

- anonymous

yes

- myininaya

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

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## More answers

- myininaya

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

- myininaya

did you get this far?

- anonymous

i see

- anonymous

I hadn't thought about constructing it that way. It makes sense.

- myininaya

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

- anonymous

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

- anonymous

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

- 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

- anonymous

\[-1/e^k = 1/e\]

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

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

- anonymous

-E^-x*

- 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

- myininaya

yep you are right

- 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

- myininaya

i got it

- myininaya

i wrote the equation just a little wrong

- anonymous

oh

- myininaya

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

- myininaya

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

- myininaya

i'm idiot
sorry

- 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!

- myininaya

Did you get it?

- anonymous

\[-1/e^k+1/e=1/2e\]

- anonymous

thats the function right?

- myininaya

yep good so far

- anonymous

Then just solve for k?

- myininaya

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

- myininaya

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

- myininaya

\[-e^{-k}+e^{-1}=1/2e^{-1} => -e^{-k}=-1/2 e^{-1} \]

- myininaya

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

- anonymous

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

- myininaya

or you could write it as ln(2e)

- anonymous

true haha

- 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

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

cool.

- anonymous

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

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