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Bonrozzy
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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 xaxis x >= 1 into two equal parts.
 9 months ago
 9 months ago
Bonrozzy Group Title
Find the value of k that the line x=k divides the area of the first quadrant region of y=e^x and the xaxis x >= 1 into two equal parts.
 9 months ago
 9 months ago

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myininaya Group TitleBest ResponseYou've already chosen the best response.1
So you have talked about improper integrals?
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
so we want this: \[1/2 \int\limits_{1}^{k}e^{x} dx=\lim_{a \rightarrow \infty}\int\limits_{1}^{a}e^{x} dx\]
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
we want that one area to be half of that other area
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
did you get this far?
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
I hadn't thought about constructing it that way. It makes sense.
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
so we can try to evaluate both integrals first then we just solve the equation for k
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
That's true haha it seems so simple now thank you :)
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
I get that k=1, but i know that's not the answer.
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
\[1/e^k = 1/e\]
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
sorry it wouldnt be one, but this is the end result i believe
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
that negative is weird killing me because e^(to a positive number) can't be negative
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i was agreeing it isn't 1 because it would give us 0 for that one area
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
the antiderivative of e^x is e^x right?
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
yep you are right
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i got it
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i wrote the equation just a little wrong
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[ \int\limits\limits_{1}^{k}e^{x} dx=1/2 \lim_{a \rightarrow \infty}\int\limits\limits_{1}^{a}e^{x} dx\]
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
you will get the answer you got using your calculator except it will be exact
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i'm idiot sorry
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
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!
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Did you get it?
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
\[1/e^k+1/e=1/2e\]
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
thats the function right?
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
yep good so far
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
Then just solve for k?
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
yep! :) I left it as \[e^{k}+e^{1}=e^{1}/2 \]
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
oops with a negative in front of the e^(k)
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{k}+e^{1}=1/2e^{1} => e^{k}=1/2 e^{1} \]
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
get rid of those negatives then take natural log of both sides
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
Yes I got it! Thank you so much :) it's gonna be: \[\ln (1/2e)\]
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
or you could write it as ln(2e)
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
true haha
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
Yes I noticed that haha. I can't thank you enough.
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I'm going to need to learn that cool calculator trick for myself
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
I'm rather inept at using the calculator, but that's one trick I know ;)
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i know the basics and i know there are really cool tricks in stuff i could learn just never got around to it
 9 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
anyways fun problem we are going over improper integrals today in class :)
 9 months ago

Bonrozzy Group TitleBest ResponseYou've already chosen the best response.1
Well, where i am it's just about dinner time. Thanks again it really saved a lot of hair pulling. See ya
 9 months ago
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