Integrals with undefined bounds

[Cake]

Homework Statement

Find the area enclosed by the equations:
y=1/x
and
y=1/x^2
and
x=2

Homework Equations

N/A

The Attempt at a Solution

So I solved this analytically after looking at a graph of the two functions. Using integrals I got the following:
ln(2)-1/2
Which is the correct answer. I assumed the bounds of the integral were 0 as well as 2, due to the fact that I thought I'd get infinite area if I used the whole negative x-axis for a bound. My issue is that when I was solving analytically I got:
-ln(0)-1/0
as the lower bound value I was subtracting from the upper bound of the integral. And while I just wrote it off as something I can omit from my answer (and got the right answer anyway), I can't help but feel I'm missing something conceptually from this problem. For instance, It seemed to me at first like y=1/x^2 should have been the top function and I should subtract the area of y=1/x. But that is apparently not the case. And I'm also not sure how I can reasonably pick 0 as the lower bound of the integral when it's undefined in the answer I got.

So where is my understanding failing me?

 

[Ray Vickson]

Homework Statement

Find the area enclosed by the equations:
y=1/x
and
y=1/x^2
and
x=2

Homework Equations

N/A

The Attempt at a Solution

So I solved this analytically after looking at a graph of the two functions. Using integrals I got the following:
ln(2)+1/2
Which is the correct answer. I assumed the bounds of the integral were 0 as well as 2, due to the fact that I thought I'd get infinite area if I used the whole negative x-axis for a bound. My issue is that when I was solving analytically I got:
-ln(0)-1/0
as the lower bound value I was subtracting from the upper bound of the integral. And while I just wrote it off as something I can omit from my answer (and got the right answer anyway), I can't help but feel I'm missing something conceptually from this problem. For instance, It seemed to me at first like y=1/x^2 should have been the top function and I should subtract the area of y=1/x. But that is apparently not the case. And I'm also not sure how I can reasonably pick 0 as the lower bound of the integral when it's undefined in the answer I got.

So where is my understanding failing me?

(1) The area between $y = 1/x$ and $y = 1/x^2$ between their crossing point $x = 1$ and the given upper limit $x = 2$ is NOT $\ln(2) + 1/2$, although it certainly has a $\ln(2)$ and a $1/2$ in it. Check your work. (BTW: if somebody tells you that answer is correct, they are mistaken.)
(2) The area between the two curves from $x \to 0+$ and $x = 1$ is infinite, either $-\infty$ or $+\infty$, depending on how you define the sign of "between". The absolute area = $+\infty$, so if you wanted to paint the region between the two curves, from $x = 0$ to $x = 2$ you would need an infinite amount of paint.

It the problem was stated exactly as you wrote it, there are some important missing items of information---the desired lower bound on $x$.

 

[PWiz]

Are you sure that the lower limit is not defined? Try sketching the graphs. Shouldn't $\frac{1}{x}$ and $\frac{1}{x^2}$ intersect somewhere, hence forming a finite area that must be evaluated? I'm sure you can carry on from here.

 

[Cake]

(1) The area between $y = 1/x$ and $y = 1/x^2$ between their crossing point $x = 1$ and the given upper limit $x = 2$ is NOT $\ln(2) + 1/2$, although it certainly has a $\ln(2)$ and a $1/2$ in it. Check your work. (BTW: if somebody tells you that answer is correct, they are mistaken.)

Got it. I think I see where I went wrong. The lower bound should be 1. So I shouldn't have assumed it was 0. But in response to your second part, the question is written exactly as I put it. I double checked that too. So there's no indication that I should have used x=1 instead of zero. So I'm not sure how I should direct my intuition. Should I do like PWiz says and look for the intersection and just assume that's the lower or upper bound other than the x value named in the problem. Or is there something more subtle I'm missing?

 

[Ray Vickson]

Got it. I think I see where I went wrong. The lower bound should be 1. So I shouldn't have assumed it was 0. But in response to your second part, the question is written exactly as I put it. I double checked that too. So there's no indication that I should have used x=1 instead of zero. So I'm not sure how I should direct my intuition. Should I do like PWiz says and look for the intersection and just assume that's the lower or upper bound other than the x value named in the problem. Or is there something more subtle I'm missing?

I think you should just use x = 1 as the lower limit, with an explanation of why you are doing it. That way, the person making the problem knows that you understand the issue but have opted to make an assumption about the missing information.

 

[Mark44]

Got it. I think I see where I went wrong. The lower bound should be 1. So I shouldn't have assumed it was 0. But in response to your second part, the question is written exactly as I put it. I double checked that too. So there's no indication that I should have used x=1 instead of zero. So I'm not sure how I should direct my intuition. Should I do like PWiz says and look for the intersection and just assume that's the lower or upper bound other than the x value named in the problem. Or is there something more subtle I'm missing?

Sketch a graph of the two functions, and note that they intersect at x = 1. The region whose area you are to find is completely bounded, so is finite.

 

[Cake]

I think you should just use x = 1 as the lower limit, with an explanation of why you are doing it. That way, the person making the problem knows that you understand the issue but have opted to make an assumption about the missing information.

It's not to turn in, so the only one I'm justifying this to is myself. But thanks a lot for the help.

Sketch a graph of the two functions, and note that they intersect at x = 1. The region whose area you are to find is completely bounded, so is finite.

That sums up what I thought better than how I described it. Thank you.

Thread done :D