Area between the curve and x axis

In summary, the question is asking for the area between the curve y=2/(x-1)dx and the x-axis over the interval [2,4]. However, the answer is not the definite integral of the function between [2,4], (ln(9)-2) as was originally thought, but the area is instead the signed area between the two curves.
  • #1
TheFallen018
52
0
Hey guys,

I've got this question in my book, and I think that I may be misunderstanding the concept. The book is somewhat lacking on this particular question, and has left me in the dark to some degree.

The question is

Find the area between the curve y=2/(x-1)dx and the x-axis over the interval [2,4].

I was thinking that this should just be the definite integral of the function between [2,4], (ln(9)-2) That being but apparently that is the wrong answer.

Therefore I am left to conclude that I have not understood the question correctly.

If anyone would be able to help clear that up, I would be very grateful.

Kind regards,

TheFallen018
 
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  • #2
$$2\int_2^4\frac{1}{x-1}\,dx=2\ln(3)-2\ln(1)=2\ln(3)$$
 
  • #3
Oops, I just had a look, and I posted this incorrectly. My function was missing something. What I should have posted was this

Find the area between the curve of y=(2/(x-1))-1, and the x-axis over the interval [2,4]

As far as I can figure, I did have the answer correct, but apparently I was missing a little trick to it.

I'll post a screenshot of the problem, just to clear things up.

View attachment 7618

So, the answer to part a should be ln(9)-2, however, part b seems to be asking just about the same thing as part a, so I'm probably missing something there.

If you'd be able to take another look, that would be really appreciated.

Sorry for the mistake in my previous post.
 

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  • #4
Hint: Note (by making a sketch or plot) that the curve given by $y = \frac{2}{x - 1} - 1$ is partly above and partly below the $x$-axis on $[2,4]$. The definite integral gives you the signed area. Where does the curve cross the $x$-axis?

Side remarks: $\ln{9} = \ln{3^2} = 2\ln{3}$. Moreover, it is a bit odd to write the differential $dx$ in the description of the curve (the integrand) itself. It only has meaning when the integrand appears already in the integral.
 
  • #5
Krylov said:
Hint: Note (by making a sketch or plot) that the curve given by $y = \frac{2}{x - 1} - 1$ is partly above and partly below the $x$-axis on $[2,4]$. The definite integral gives you the signed area. Where does the curve cross the $x$-axis?

Side remarks: $\ln{9} = \ln{3^2} = 2\ln{3}$. Moreover, it is a bit odd to write the differential $dx$ in the description of the curve (the integrand) itself. It only has meaning when the integrand appears already in the integral.

I'm not sure if I know what you're getting at. It's clear that part of the curve is above the x axis, and part below. That point of y=0 is obviously at x=3. Perhaps your meaning has escaped me.

As for side remarks, I'm not aware of what your first statement refers to. As for the second, I agree. I also thought it strange that there is a dx after the description of the curve. I assumed that a mistake had been made on the part of whoever wrote up the question sheet.
 
  • #6
TheFallen018 said:
I'm not sure if I know what you're getting at. It's clear that part of the curve is above the x axis, and part below. That point of y=0 is obviously at x=3. Perhaps your meaning has escaped me.

As for side remarks, I'm not aware of what your first statement refers to. As for the second, I agree. I also thought it strange that there is a dx after the description of the curve. I assumed that a mistake had been made on the part of whoever wrote up the question sheet.

For part b), you want the area which is a bit different from the definite integral. The definite integral $I$ of the difference between two functions is:

\(\displaystyle I=\int_a^b f(x)-g(x)\,dx=-\int_a^b g(x)-f(x)\,dx\)

The area $A$ between those curves on the same interval is:

\(\displaystyle A=\int_a^b \left|f(x)-g(x)\right|\,dx=\int_a^b \left|g(x)-f(x)\right|\,dx\)
 
  • #7
TheFallen018 said:
I'm not sure if I know what you're getting at. It's clear that part of the curve is above the x axis, and part below. That point of y=0 is obviously at x=3. Perhaps your meaning has escaped me.

As for side remarks, I'm not aware of what your first statement refers to. As for the second, I agree. I also thought it strange that there is a dx after the description of the curve. I assumed that a mistake had been made on the part of whoever wrote up the question sheet.

MarkFL already commented on the first part (Yes) Indeed, by "signed area" it is meant that the area corresponding to part of the curve under the $x$-axis is measured with a minus-sign in front.

My first statement was to make sure that it is understood that $\ln{9} = 2\ln{3}$, so the equivalence between post #2 and your own computation of that integral would be clear as well. I'm sorry if this was already clear, but I preferred to err on the side clarification.
 
  • #8
MarkFL said:
For part b), you want the area which is a bit different from the definite integral. The definite integral $I$ of the difference between two functions is:

\(\displaystyle I=\int_a^b f(x)-g(x)\,dx=-\int_a^b g(x)-f(x)\,dx\)

The area $A$ between those curves on the same interval is:

\(\displaystyle A=\int_a^b \left|f(x)-g(x)\right|\,dx=\int_a^b \left|g(x)-f(x)\right|\,dx\)

Thanks Mark, I think that's really cleared some things up for me.

If you'd also be so kind, I believe that using the formula you have pointed out, my answer should be

$2\ln{4} -\ln{9}$

Does that sound correct to you?

Thanks again
 
  • #9
TheFallen018 said:
Thanks Mark, I think that's really cleared some things up for me.

If you'd also be so kind, I believe that using the formula you have pointed out, my answer should be

$2\ln{4} -\ln{9}$

Does that sound correct to you?

Thanks again

Yes that's correct:

W|A - area between x-axis and y=2/(x-1)-1 on [2,4]

Note: W|A returns a result in a different form, but is recognizable as equivalent to your result.

I would further simplify:

\(\displaystyle A=2\ln(4)-\ln(9)=2\left(\ln(4)-\ln(3)\right)=2\ln\left(\frac{4}{3}\right)\)
 

FAQ: Area between the curve and x axis

1. What is the area between a curve and the x-axis?

The area between a curve and the x-axis is the region or space that is bounded by the curve and the x-axis. It can be visualized as the space that is under the curve and above the x-axis on a graph.

2. How do you calculate the area between a curve and the x-axis?

The area between a curve and the x-axis can be calculated by using integration. This involves finding the antiderivative of the function that represents the curve and then evaluating it at the upper and lower limits of the region.

3. What is the significance of finding the area between a curve and the x-axis?

The area between a curve and the x-axis is important in calculus as it allows us to find the total amount of a quantity represented by the function. It also helps in understanding the behavior of the function and its relationship with the x-axis.

4. Can negative area exist between a curve and the x-axis?

Yes, negative area can exist between a curve and the x-axis. This happens when the curve dips below the x-axis and the area between them is considered to be negative. It is important to pay attention to the direction of the curve when calculating the area.

5. Is it possible for the area between a curve and the x-axis to be infinite?

Yes, the area between a curve and the x-axis can be infinite. This can happen when the curve is unbounded and extends infinitely in either the positive or negative direction. In such cases, the area cannot be calculated using integration and other methods must be used to approximate it.

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