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So you have talked about improper integrals?

yes

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.

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

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*

yep you are right

i got it

i wrote the equation just a little wrong

oh

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

i'm idiot
sorry

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

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 ;)

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

cool.