Average Distance Between Two Points on f(x)=sin(x)
Let x 1 and x 2 be chosen randomly and independently on the interval [ 0 , 2 π ] and let f ( x ) = sin ( x ) .
If a is the average distance between the two points P 1 = ( x 1 , f ( x 1 ) ) and P 2 = ( x 2 , f ( x 2 ) ) , then find a to three significant figures.
The answer is 2.29.
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2 solutions
can you explane it more pleace
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The average value of a function f ( x , y ) on a compact domain D with area A is defined as f a v g = A 1 ∫ D f ( x , y ) d A . In our case, this function is the distance between the points ( x , sin x ) and ( y , sin y ) , namely, f ( x , y ) = ( x − y ) 2 + ( sin x − sin y ) 2 on D = [ 0 , 2 π ] × [ 0 , 2 π ] .
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no im saying how did you get that formula for averege where did come from
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@Nahom Assefa – As I said, it's the definition of "average value." You will find it in any good introductory calculus text. Here is a random link that I googled. You can think of it as a continuous version of the notion of "average" we use in discrete maths; take the limit of a Riemann sum!
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@Otto Bretscher – you cant find this in calculus text book
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@Nahom Assefa – Sure you can! Look in Stewart or Thomas... I have them right in front of me... search the index: average value of a function of two variables
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@Otto Bretscher – oh i see it now i understand
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@Nahom Assefa – Averages are certainly an important application of integrals, both single-variable and multi-variable, and they should be discussed in every course on calculus.
@Otto Bretscher – but why do you have a doubel inegraton.
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@Nahom Assefa – we need to make two independent choices, for the two points
@Mark Hennings @Otto Bretscher @David Vreken Sir, can the Integral be evaluated manually??? Does it have an elementary closed form?? Because, I used a calculator to evaluate the integral.......
I certainly used numerical techniques.
One of my applied maths profs at Harvard used to say that "finding integrals in closed form is not a marketable skill these days"; it is merely a passtime for those who are bored ;)
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Can you pls post the solution
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sure...it's just a straightforward computation of an average, according to the definition.
Hah.......No offense Sir, but I don't know why, but I just can't agree with your prof........Because, well according to me, it is more fun in evaluating the EXACT closed form........even if they are much much complicated...........because, that is what makes it fun........even if it isn't a marketable skill, it IS a skill which sometimes lead to BEAUTIFUL results...........!!!! Personal perspectives tho........:)
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Finding antiderivatives is an (almost) algorithmic process, and, as such, not very creative. I don't think it should be emphasised in maths education. But, if you find it a fun thing to do in your spare time, go for it! It simply isn't my thing; you will rarely find me doing (single-variable) integration problems, on Brilliant or otherwise.
that would take me a long time.
Actually, I wrote a small computer program that simulated finding the distance between two random points on the curve one million times and gave me the average. It wasn't elegant or exact, but close enough to obtain the correct answer.
It's a straightforward computation of an average, ( 2 π ) 2 1 ∫ 0 2 π ∫ 0 2 π ( x − y ) 2 + ( sin x − sin y ) 2 d y d x ≈ 2 . 2 8 8