Question

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.

Answer 1

So you have talked about improper integrals?

Answer 2

yes

Answer 3

so we want this:
\[1/2 \int\limits_{1}^{k}e^{-x} dx=\lim_{a \rightarrow \infty}\int\limits_{1}^{a}e^{-x} dx\]

Answer 4

we want that one area to be half of that other area

Answer 5

did you get this far?

Answer 6

i see

Answer 7

I hadn't thought about constructing it that way. It makes sense.

Answer 8

so we can try to evaluate both integrals first
then we just solve the equation for k

Answer 9

That's true haha it seems so simple now thank you :)

Answer 10

I get that k=1, but i know that's not the answer.

Answer 11

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

Answer 12

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

Answer 13

sorry it wouldnt be one, but this is the end result i believe

Answer 14

that negative is weird killing me because e^(to a positive number) can't be negative

Answer 15

i was agreeing it isn't 1 because it would give us 0 for that one area

Answer 16

the antiderivative of e^-x is -e^x right?

Answer 17

-E^-x*

Answer 18

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

Answer 19

yep you are right

Answer 20

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

Answer 21

i got it

Answer 22

i wrote the equation just a little wrong

Answer 23

oh

Answer 24

\[ \int\limits\limits_{1}^{k}e^{-x} dx=1/2 \lim_{a \rightarrow \infty}\int\limits\limits_{1}^{a}e^{-x} dx\]

Answer 25

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

Answer 26

i'm idiot
sorry

Answer 27

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!

Answer 28

Did you get it?

Answer 29

\[-1/e^k+1/e=1/2e\]

Answer 30

thats the function right?

Answer 31

yep good so far

Answer 32

Then just solve for k?

Answer 33

yep! :)
I left it as \[e^{-k}+e^{-1}=e^{-1}/2 \]

Answer 34

oops with a negative in front of the e^(-k)

Answer 35

\[-e^{-k}+e^{-1}=1/2e^{-1} => -e^{-k}=-1/2 e^{-1} \]

Answer 36

get rid of those negatives then take natural log of both sides

Answer 37

Yes I got it! Thank you so much :) it's gonna be: \[-\ln (1/2e)\]

Answer 38

or you could write it as ln(2e)

Answer 39

true haha

Answer 40

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

Answer 41

Yes I noticed that haha. I can't thank you enough.

Answer 42

I'm going to need to learn that cool calculator trick for myself

Answer 43

I'm rather inept at using the calculator, but that's one trick I know ;)

Answer 44

i know the basics
and i know there are really cool tricks in stuff i could learn
just never got around to it

Answer 45

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

Answer 46

cool.

Answer 47

Well, where i am it's just about dinner time. Thanks again it really saved a lot of hair pulling. See ya