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.

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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.

Mathematics
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So you have talked about improper integrals?
yes
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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Other answers:

we want that one area to be half of that other area
did you get this far?
i see
I hadn't thought about constructing it that way. It makes sense.
so we can try to evaluate both integrals first then we just solve the equation for k
That's true haha it seems so simple now thank you :)
I get that k=1, but i know that's not the answer.
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
\[-1/e^k = 1/e\]
sorry it wouldnt be one, but this is the end result i believe
that negative is weird killing me because e^(to a positive number) can't be negative
i was agreeing it isn't 1 because it would give us 0 for that one area
the antiderivative of e^-x is -e^x right?
-E^-x*
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
yep you are right
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
i got it
i wrote the equation just a little wrong
oh
\[ \int\limits\limits_{1}^{k}e^{-x} dx=1/2 \lim_{a \rightarrow \infty}\int\limits\limits_{1}^{a}e^{-x} dx\]
you will get the answer you got using your calculator except it will be exact
i'm idiot sorry
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!
Did you get it?
\[-1/e^k+1/e=1/2e\]
thats the function right?
yep good so far
Then just solve for k?
yep! :) I left it as \[e^{-k}+e^{-1}=e^{-1}/2 \]
oops with a negative in front of the e^(-k)
\[-e^{-k}+e^{-1}=1/2e^{-1} => -e^{-k}=-1/2 e^{-1} \]
get rid of those negatives then take natural log of both sides
Yes I got it! Thank you so much :) it's gonna be: \[-\ln (1/2e)\]
or you could write it as ln(2e)
true haha
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
Yes I noticed that haha. I can't thank you enough.
I'm going to need to learn that cool calculator trick for myself
I'm rather inept at using the calculator, but that's one trick I know ;)
i know the basics and i know there are really cool tricks in stuff i could learn just never got around to it
anyways fun problem we are going over improper integrals today in class :)
cool.
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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