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