anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
So you have talked about improper integrals?
anonymous
  • anonymous
yes
myininaya
  • 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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myininaya
  • myininaya
we want that one area to be half of that other area
myininaya
  • myininaya
did you get this far?
anonymous
  • anonymous
i see
anonymous
  • anonymous
I hadn't thought about constructing it that way. It makes sense.
myininaya
  • myininaya
so we can try to evaluate both integrals first then we just solve the equation for k
anonymous
  • anonymous
That's true haha it seems so simple now thank you :)
anonymous
  • anonymous
I get that k=1, but i know that's not the answer.
myininaya
  • 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
  • anonymous
\[-1/e^k = 1/e\]
anonymous
  • anonymous
sorry it wouldnt be one, but this is the end result i believe
myininaya
  • myininaya
that negative is weird killing me because e^(to a positive number) can't be negative
myininaya
  • myininaya
i was agreeing it isn't 1 because it would give us 0 for that one area
anonymous
  • anonymous
the antiderivative of e^-x is -e^x right?
anonymous
  • anonymous
-E^-x*
myininaya
  • 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
  • myininaya
yep you are right
anonymous
  • 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
  • myininaya
i got it
myininaya
  • myininaya
i wrote the equation just a little wrong
anonymous
  • anonymous
oh
myininaya
  • myininaya
\[ \int\limits\limits_{1}^{k}e^{-x} dx=1/2 \lim_{a \rightarrow \infty}\int\limits\limits_{1}^{a}e^{-x} dx\]
myininaya
  • myininaya
you will get the answer you got using your calculator except it will be exact
myininaya
  • myininaya
i'm idiot sorry
anonymous
  • 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
  • myininaya
Did you get it?
anonymous
  • anonymous
\[-1/e^k+1/e=1/2e\]
anonymous
  • anonymous
thats the function right?
myininaya
  • myininaya
yep good so far
anonymous
  • anonymous
Then just solve for k?
myininaya
  • myininaya
yep! :) I left it as \[e^{-k}+e^{-1}=e^{-1}/2 \]
myininaya
  • myininaya
oops with a negative in front of the e^(-k)
myininaya
  • myininaya
\[-e^{-k}+e^{-1}=1/2e^{-1} => -e^{-k}=-1/2 e^{-1} \]
myininaya
  • myininaya
get rid of those negatives then take natural log of both sides
anonymous
  • anonymous
Yes I got it! Thank you so much :) it's gonna be: \[-\ln (1/2e)\]
myininaya
  • myininaya
or you could write it as ln(2e)
anonymous
  • anonymous
true haha
myininaya
  • 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
  • anonymous
Yes I noticed that haha. I can't thank you enough.
myininaya
  • myininaya
I'm going to need to learn that cool calculator trick for myself
anonymous
  • anonymous
I'm rather inept at using the calculator, but that's one trick I know ;)
myininaya
  • 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
  • myininaya
anyways fun problem we are going over improper integrals today in class :)
anonymous
  • anonymous
cool.
anonymous
  • 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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