Stuck on an integration problem

[CocoatheSeaLion]

Hi all,

For some reason, integrals are one aspect of calculus that I have always struggled with. Even more so now that I am back in school after taking ~3 years off without having to even think about them. Anyway, so I am rather stuck on this problem. I imagine the answer is pretty simple, and maybe I'm just tired and need a fresh perspective.

So here is the question:

So this is what I have for part a. I feel like I am missing something here, or doing something stupid, but I can't figure out what.

1 = ∫f(x)dx
1 = ∫ (a/(2a+x)^2)dx + ∫(Ka/(a+2x)^2)dx
1 = -a/(2a+x) - Ka/2(a+2x)
1 + a/(2a+x) = - Ka/2(a+2x)
(-2a-4x)(1 + a/(2a+x)) = Ka
((-2a-4x)(1 + a/(2a+x)))/a = K

So this gives me the awfully complicated looking term
K = -2a-4x + -2a-4x/(2a+x))

What am I missing, here?

As for part b of this question, I am utterly stumped.

I know if you want the median of f(x), you do:

f(x) dx = 1/2.
But I don't understand how to solve it out. What do I plug in for f(x), here? I have two different functions depending on the value of a, but I don't know the value of a. I don't even know where to begin, here. Could somebody just do the first step of this one for me? I think once I see the first step I can figure it out.

 

[HallsofIvy]

Hi all,

For some reason, integrals are one aspect of calculus that I have always struggled with. Even more so now that I am back in school after taking ~3 years off without having to even think about them. Anyway, so I am rather stuck on this problem. I imagine the answer is pretty simple, and maybe I'm just tired and need a fresh perspective.

So here is the question:

So this is what I have for part a. I feel like I am missing something here, or doing something stupid, but I can't figure out what.

1 = ∫f(x)dx
1 = ∫ (a/(2a+x)^2)dx + ∫(Ka/(a+2x)^2)dx

This is incorrect because you don't have limits on the integrals. It should be
\(\displaystyle \int_0^\alpha \dfrac{\alpha}{(2\alpha+ x)^2}dx+ \int_\alpha^\infty \dfrac{K\alpha}{(\alpha+ 2x)^2} dx\)

1 = -a/(2a+x) - Ka/2(a+2x)
1 + a/(2a+x) = - Ka/2(a+2x)
(-2a-4x)(1 + a/(2a+x)) = Ka
((-2a-4x)(1 + a/(2a+x)))/a = K

So this gives me the awfully complicated looking term
K = -2a-4x + -2a-4x/(2a+x))

What am I missing, here?

As for part b of this question, I am utterly stumped.

I know if you want the median of f(x), you do:

f(x) dx = 1/2.
But I don't understand how to solve it out. What do I plug in for f(x), here? I have two different functions depending on the value of a, but I don't know the value of a. I don't even know where to begin, here. Could somebody just do the first step of this one for me? I think once I see the first step I can figure it out.

 

[CocoatheSeaLion]

This is incorrect because you don't have limits on the integrals. It should be
\(\displaystyle \int_0^\alpha \dfrac{\alpha}{(2\alpha+ x)^2}dx+ \int_\alpha^\infty \dfrac{K\alpha}{(\alpha+ 2x)^2} dx\)

I just didn't know how to type integrals into a forum post so I left them out.

So ... I mean, no offense but is that really all you have to say? No comments on the rest of it? Because this isn't really helpful at all ...

 

[DrPhil]

hi all,

for some reason, integrals are one aspect of calculus that i have always struggled with. Even more so now that i am back in school after taking ~3 years off without having to even think about them. Anyway, so i am rather stuck on this problem. I imagine the answer is pretty simple, and maybe i'm just tired and need a fresh perspective.

So here is the question:

so this is what i have for part a. I feel like i am missing something here, or doing something stupid, but i can't figure out what.

1 = ∫f(x)dx
1 = ∫ (a/(2a+x)^2)dx + ∫(ka/(a+2x)^2)dx
1 = -a/(2a+x) - ka/2(a+2x)
1 + a/(2a+x) = - ka/2(a+2x)
(-2a-4x)(1 + a/(2a+x)) = ka
((-2a-4x)(1 + a/(2a+x)))/a = k

so this gives me the awfully complicated looking term
k = -2a-4x + -2a-4x/(2a+x))

what am i missing, here?

As for part b of this question, i am utterly stumped.

I know if you want the median of f(x), you do:

f(x) dx = 1/2.
but i don't understand how to solve it out. What do i plug in for f(x), here? I have two different functions depending on the value of a, but i don't know the value of a. I don't even know where to begin, here. Could somebody just do the first step of this one for me? I think once i see the first step i can figure it out.

this is incorrect because you don't have limits on the integrals. It should be

\(\displaystyle \displaystyle \int_0^\alpha \dfrac{\alpha}{(2\alpha+ x)^2}dx+ \int_\alpha^\infty \dfrac{K\ \alpha}{(\alpha+ 2x)^2} dx \)

Do you remember definite integrals? After you have found the integral, evaluate the result at the upper limit and at the lower limit, and subtract. The result is the area under specified region, and will NOT be a factor of the (dummy) variable of integration. When I saw your result for \(\displaystyle K\) being a function of \(\displaystyle x\), I knew something was wrong.

You have the two integrals, but haven't applied the limits. Your result should look like this (using the easier to type "\(\displaystyle a\)" instead of "\(\displaystyle \alpha\)"):

\(\displaystyle \displaystyle \left. \dfrac{-a}{(2a+x)}\right|_0^a + \left. \dfrac{-K\ a}{2(a+2x)}\right|_a^\infty = 1\)

Can you proceed from there?

For part b, compare the value of the first definite integral to (1/2) to decide which piece of the function you have to look at to find the median.

 

[CocoatheSeaLion]

Do you remember definite integrals? After you have found the integral, evaluate the result at the upper limit and at the lower limit, and subtract. The result is the area under specified region, and will NOT be a factor of the (dummy) variable of integration. When I saw your result for \(\displaystyle K\) being a function of \(\displaystyle x\), I knew something was wrong.

You have the two integrals, but haven't applied the limits. Your result should look like this (using the easier to type "\(\displaystyle a\)" instead of "\(\displaystyle \alpha\)"):

\(\displaystyle \displaystyle \left. \dfrac{-a}{(2a+x)}\right|_0^a + \left. \dfrac{-K\ a}{2(a+2x)}\right|_a^\infty = 1\)

Can you proceed from there?

Ooooh ... I see. Yeah, I don't know why I have such trouble with integrals. It's like my brain turns off whenever I look at them.

But how do I evaluate the result for the right term, from a to infinity? I can evaluate the left hand side no problem, but I don't understand how to evaluate a definite integral when one term is infinity.

For part b, compare the value of the first definite integral to (1/2) to decide which piece of the function you have to look at to find the median.

So I just set the integral of the first part to 1/2 and solve for a?

 

[Deleted member 4993]

I just didn't know how to type integrals into a forum post so I left them out.

So ... I mean, no offense but is that really all you have to say? No comments on the rest of it? Because this isn't really helpful at all ...

That should be all - you did not write the integrations correctly → which probably is the cause of doing the integrations incorrectly → which is the cause of getting wrong answers.

You did not show the steps in between (finding the integration result and applying the limits)

We don't know whether you put the limits at all or not - or you put the wrong limits - or your integration result was wrong → hence that's all we could say!!

 

[CocoatheSeaLion]

That should be all - you did not write the integrations correctly → which probably is the cause of doing the integrations incorrectly → which is the cause of getting wrong answers.

You did not show the steps in between (finding the integration result and applying the limits)

We don't know whether you put the limits at all or not - or you put the wrong limits - or your integration result was wrong → hence that's all we could say!!

Sorry! I didn't mean to sound snappy. I thought I was just being corrected on syntax rather than being helped, which I found irritating, but now that I see my mistake I realize the problem! Thanks, and sorry again!

Anyway, so to continue with my previous post, to solve a definite integral with infinity do I substitute infinity for an arbitrary constant c and solve from there? So the left hand part (a/(2a+x)^2) reduces to 1.5, and the right hand part becomes ...

-Ka/(-2(a+2c)) - -Ka/(-2(a+2a))?

I know that can be further reduced, but is that the right logic I am following? Or am I still missing something? This looks like it still reduces to something quite ugly...

 

[Deleted member 4993]

Ooooh ... I see. Yeah, I don't know why I have such trouble with integrals. It's like my brain turns off whenever I look at them.

But how do I evaluate the result for the right term, from a to infinity? I can evaluate the left hand side no problem, but I don't understand how to evaluate a definite integral when one term is infinity.

\(\displaystyle \displaystyle \left. \dfrac{-a}{(2a+x)}\right|_0^a + \left. \dfrac{-K\ a}{2(a+2x)}\right|_a^\infty = 1\)

\(\displaystyle \left. \dfrac{-K\ a}{2(a+2x)}\right|_a^\infty \)

\(\displaystyle = \ \dfrac{-K\ a}{2(a+2*\infty)} \ - \ \dfrac{-K\ a}{2(a+2*a)}\)...................................Corrected

\(\displaystyle = \ 0 \ + \ \dfrac{K\ a}{3(a)}\)..................................Corrected

\(\displaystyle = \ \dfrac{K}{3}\)..................................Corrected

Next

\(\displaystyle \displaystyle \left. \dfrac{-a}{(2a+x)}\right|_0^a\)

\(\displaystyle = \ \dfrac{-a}{(2a+a)} \ - \ \dfrac{-a}{(2a+0)}\)

\(\displaystyle = \ -\dfrac{a}{(3a)} \ + \ \dfrac{a}{(2a)}\)

= 1/2 - 1/3 = 1/6

so

\(\displaystyle \displaystyle \left. \dfrac{-a}{(2a+x)}\right|_0^a + \left. \dfrac{-K\ a}{2(a+2x)}\right|_a^\infty = 1\)

\(\displaystyle \displaystyle \frac{1}{6} + \frac{K}{3} = 1\)

 

[CocoatheSeaLion]

Hm. Okay, thanks for doing it out for me!

So the first term becomes 0 because infinity is in the denominator? As it becomes arbitrarily large the term converges on 0?

(Again, sorry if I am missing a lot, here. As I said, integrals just trip me up everywhere)

So the whole solution should be

1 = (3/2) + K/2
Or
K = -1?

 

[DrPhil]

Ooooh ... I see. Yeah, I don't know why I have such trouble with integrals. It's like my brain turns off whenever I look at them.

But how do I evaluate the result for the right term, from a to infinity? I can evaluate the left hand side no problem, but I don't understand how to evaluate a definite integral when one term is infinity.



So I just set the integral of the first part to 1/2 and solve for a?

Since \(\displaystyle x\) is in the denominator, the fraction goes to zero at the upper limit.

Show your work - especially, what is the result of the first integral? If it is bigger than 1/2, then the median is in that domain. But if the area of the first part of the function is less than 1/2, then you have to use the second piece of the function. Remember that if half of the area is below the median value, then it is also true that half of the area is above the median.

 

[CocoatheSeaLion]

For part b, compare the value of the first definite integral to (1/2) to decide which piece of the function you have to look at to find the median.

So I am in a hurry and can't type out every step, but just let me know whether I am on the right track.

I set the integral of the a/(2a+x)^2 term (the first definite integral) to 1/2. So I solve this out and get a = 1/6.

Since 1/6 < 1/2, I use the other term, the second definite integral, as the piece of the function to find the median, correct?

 

[DrPhil]

So I am in a hurry and can't type out every step, but just let me know whether I am on the right track.

I set the integral of the a/(2a+x)^2 term (the first definite integral) to 1/2. So I solve this out and get a = 1/6.

Since 1/6 < 1/2, I use the other term, the second definite integral, as the piece of the function to find the median, correct?

You can't "solve for a" because that is a given parameter of the function. What you CAN say is that the integral from 0 to a is 1/6, so there is no value of x in the domain 0<x<a that will integrate to have an area of 1/2.

Might want to assign a variable name to the median - shouldn't use "x" because that is being used to define the function f(x). How about:
Let M = median value of x.
Call the two pieces of f(x) --> f1(x) and f2(x)

If you integrate from 0 to M: int(0,M)[f(x) dx] = int(0,a)[f1(x) dx] + int(a,M)[f2(x) dx]
........................................= (1/6) + int(a,M)[f2(x) dx] = 1/2

OR integrate from M to infinity: int(M,inf)[f2(x) dx] = 1/2
This second approach may be easier, since you already know the upper limit is going to evaluate to zero.

BTW, I have purposely use "inline" typing to show how you can use parentheses to define the limits of an integral.

 

[CocoatheSeaLion]

Okay, thank you very much for your help! I need to go to my sister's engagement party now so I won't be able to sit down and look at this again until Sunday, but I think I got a grasp on things! At least, now I understand HOW I have to do it instead of blindly stumbling around in the dark, heh.

Thanks for your help!