Integration- Area between a line and a curve

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KingRich
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Hi, can someone advice on this question.

The equation of a curve is y=√x

Find the shaded area in terms of P.

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I divided the area between 0 and P into two areas A and B. A (the shaded part under curve) B( beneath the line)

I found the integral of y=x^½ to be 2/3P^3/2, therefore area AB.

I assume correctly the line is y=x, so that the integral B=1/2p^2

If my Maths is correct.

2/3p^3/2 - 1/2 p^2
Last edited by KingRich; 4 weeks ago
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Cupcakes12
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you need to integrate the area under the curve and then find the area of the triangle and then take it away, so what you've done seems correct
Last edited by Cupcakes12; 4 weeks ago
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KingRich
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(Original post by Cupcakes12)
you need to integrate the area under the curve and then find the area of the triangle and then take it away, so what you've done seems correct
I thought so too, however according to the answer in my textbook it’s incorrect. Although, it wouldn’t be the first time the text book was wrong
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Cupcakes12
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(Original post by KingRich)
I thought so too, however according to the answer in my textbook it’s incorrect. Although, it wouldn’t be the first time the text book was wrong
it just occurred to me, do you need to subsititute in the value for P, i.e. where y=x meets y=x^1/2

Also do you know for certain the straight line is y=x
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KingRich
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(Original post by Cupcakes12)
it just occurred to me, do you need to subsititute in the value for P, i.e. where y=x meets y=x^1/2

Also do you know for certain the straight line is y=x
I don’t know for certain the line is y=x, although there’s no indication to what it may be other than that, nor is there any constraints suggesting that it is not.

I’m already told the point at which they meet is p. Therefore, y would be p^1/2
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mqb2766
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(Original post by KingRich)
I don’t know for certain the line is y=x, although there’s no indication to what it may be other than that, nor is there any constraints suggesting that it is not.

I’m already told the point at which they meet is p. Therefore, y would be p^1/2
As the other poster said, the line isn't y=x, but it is a triangle with base p and height p^(1/2) so you can combine the two expressions into a single one.

Note the problem is basically the famous Archimedes - parabola area
https://en.wikipedia.org/wiki/Quadra...f_the_Parabola
So only 2,200 years old :-)
Last edited by mqb2766; 4 weeks ago
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Muttley79
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(Original post by KingRich)
I thought so too, however according to the answer in my textbook it’s incorrect. Although, it wouldn’t be the first time the text book was wrong
Check the straight line .... y = mx meets y = x^0.5 wen x = P.

I don't think m = 1
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KingRich
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(Original post by Cupcakes12)
it just occurred to me, do you need to subsititute in the value for P, i.e. where y=x meets y=x^1/2

Also do you know for certain the straight line is y=x
Okay, so I believe it may be my assumption of the line.

If we propose from the equation given y=√x that would suggest where they meet like you suggested in your question that the coordinate y would be p^1/2 and if that the case the area of the triangle would be p^3/2 /2, therefore, AB- new found B would be 1/6p^3/2.

Which is the answer. Thank you for questioning my assumption
Last edited by KingRich; 4 weeks ago
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KingRich
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(Original post by Muttley79)
Check the straight line .... y = mx meets y = x^0.5 wen x = P.

I don't think m = 1
Thank you. I realised my error after cupcake questioned my assumption, upon realising y is x^1/2 which made me realise where I went wrong
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KingRich
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(Original post by mqb2766)
As the other poster said, the line isn't y=x, but it is a triangle with base p and height p^(1/2) so you can combine the two expressions into a single one.

Note the problem is basically the famous Archimedes - parabola area
https://en.wikipedia.org/wiki/Quadra...f_the_Parabola
So only 2,200 years old :-)
I was wondering where you were. Few minutes late to my assistance lol
Last edited by KingRich; 4 weeks ago
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mqb2766
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(Original post by KingRich)
I was wondering where you were. Few minutes late to my assistance
Ask for a refund.
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KingRich
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(Original post by mqb2766)
Ask for a refund.
Lol.
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KingRich
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(Original post by mqb2766)
Ask for a refund.
Just wondering, if my comment sound arrogant? As that was not my attention. I appreciate your help as you’ve helped me the most through my questions.
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mqb2766
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(Original post by KingRich)
Just wondering, if my comment sound arrogant? As that was not my attention. I appreciate your help as you’ve helped me the most through my questions.
Course not - would have thought my reply would have made that clear.
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KingRich
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(Original post by mqb2766)
Course not - would have thought my reply would have made that clear.
I just wanted to be sure as I know It’s hard to tell if somebody is having a joke.
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