Finding the area under this unusual curve

In summary, the conversation discusses different approaches to solving a given problem, including one that the question setter deemed necessary to include in the mark scheme. The individual shares their own solution and discusses potential concerns about it receiving full marks. They also mention the use of symmetry in their solution and provide a detailed calculation for the area of the region bounded by specific points.
  • #1
chwala
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Homework Statement
Kindly see attached
Relevant Equations
A=##\int_a^b f(y)\,dy##
I was looking at the problem below in detail, attached find the problem and the mark scheme solution.

1626738834683.png


1626738866036.png


Now this was my approach which is just similar to the Mark Scheme method ##2## above where they expressed ##x=f(y)##...

I did it this way;

1626739385041.png


...There was some work involved particularly in expressing ##x=f(y)##...then on the integrating part of

A=8##\int_1^3 (1.5-0.5y)^{0.5}\,dy##
one had to let ##u=1.5-0.5y## to realize ##du=-0.5dy## then proceed from there...##A_{1}=\frac {32} {3}##, and ##A_{2}=4## bingo!
 
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  • #2
OK, that seems valid. So now you need to perform the integral.

What was your question?

I'll also note that you could have just observed that ##A_2## is a square of side 2, therefore its area is 4.
 
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  • #3
RPinPA said:
OK, that seems valid. So now you need to perform the integral.

What was your question?

I'll also note that you could have just observed that ##A_2## is a square of side 2, therefore its area is 4.
I did not have a question but rather looking at the question in detail... or am i not supposed to share this?i already solved the problem on my own the required area is ##A=A_{1}+A_{2}## as indicated...hence my concluding term, bingo! :cool:
 
  • #4
Nothing wrong with sharing your solution, but if you post in a forum and expect people to respond, it's generally customary to indicate what sort of response you're looking for.
 
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  • #5
RPinPA said:
Nothing wrong with sharing your solution, but if you post in a forum and expect people to respond, it's generally customary to indicate what sort of response you're looking for.
Thanks , I've taken note of that...cheers Rpinpa.
 
  • #6
RPinPA said:
OK, that seems valid. So now you need to perform the integral.

What was your question?

I'll also note that you could have just observed that ##A_2## is a square of side 2, therefore its area is 4.
Maybe i should have asked whether the approach i used would realize the full marks...i know that examiners are supposed to stick solely to given mark scheme guides to the latter. Thanks.
 
  • #7
chwala said:
Maybe i should have asked whether the approach i used would realize the full marks...i know that examiners are supposed to stick solely to given mark scheme guides to the latter. Thanks.
I would say you made life hard for yourself by handling the parabolic area in terms of y, dy instead of x, dx. I presume that in your submitted solution you would have shown all your steps, making it quite long.

It is a bit worrying that the question setter deemed it necessary to write out so many variations of the solution for the benefit of the marker. It suggests the markers are not to be trusted to recognise other valid solutions, in which case your radically different solution could indeed cost marks.
 
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  • #8
haruspex said:
I would say you made life hard for yourself by handling the parabolic area in terms of y, dy instead of x, dx.
I agree.

Due to symmetry, it suffices to calculate the area bounded by points B, C, and O, and then multiply that number by 8. In my work below, I don't show the calculation for the point of intersection C, which is (1, 1).

Area of region bounded by B, C, and O:
##\int_0^1 3 - 2x^2 - x ~dx = \left. 3x - \frac 2 3 x^3 - \frac {x^2} 2 \right|_0^1 = 3 - \frac 2 3 - \frac 1 2 = \frac {11} 6##
Multiply by 8 to get ##\frac {88}6 = \frac{44} 3##
 
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Related to Finding the area under this unusual curve

1. What is the purpose of finding the area under a curve?

The area under a curve represents the total value or quantity of a variable over a specific range. It is commonly used in mathematics and science to determine the total amount of a quantity, such as distance, volume, or energy.

2. How do you find the area under a curve?

To find the area under a curve, you can use various mathematical methods such as integration, numerical integration, or approximation techniques. The method used will depend on the shape of the curve and the available data.

3. Why is it important to find the area under a curve?

Finding the area under a curve is important because it allows us to analyze and understand the behavior of a variable over a given range. It can also help in making predictions and drawing conclusions about the data.

4. Can you find the area under any type of curve?

Yes, the area under any type of curve can be found using mathematical methods. However, the complexity of the curve and the available data may affect the accuracy and ease of finding the area.

5. What is an unusual curve and how does it affect finding the area under it?

An unusual curve is a curve that does not follow a typical or expected pattern. This can make it challenging to find the area under it as traditional methods may not be applicable. In such cases, specialized techniques may be needed to accurately determine the area.

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