# calc

find a positive value of k such that the area of the region enclosed between y= kcosx and y=kx^2 is 2

i graphed cosx and x^2 without the k, so you know to find the integral of -a to a (its symmetric about the origin) would be the integral of kcosx - kx^2 times dx = 2
i don't know how to solve it after this though!!

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1. Well, you have called your integral from -a to +a (I might do twice the integral from 0 to a but that is minor details).
The real question is: What is a?
Your a is where the intersection of k cos x and k x^2 occurs, in other words:
cos a = a^2
Now one way to find a is to make a table of a , cos a and a^2 and interpolate.
Another way is to define a function that is the difference between cos a and a^2 and look for where that function is 0 using calculus.
Let's say
g(a) = a^2 - cos a
then dg/da = 2a + sin a
Look for where g(a) = 0
As a first guess try a = pi/4 or .785 rad
then
g = -.09
and dg/da = 2.27
sketch a graph of that point (.785 , -.09) and the slope at that point is 2.27
well, that tangent would hit the a axis at a point:
0 = -.09 + 2.27 da
or
da = .04
so for our next try take a = .785+.04 and try again with a = .825
here with a = .825
g(a) = .0012 (notice we are very close to zero)
dg/da = 2.4
sketch that point (.825 , .0012) and slope 2.4 on your graph and look for where the tangent hits the axis
this time our 0 = .0012 + 2.4 da
da = - 5*10^-4
so our new a = .825 - 5*10^-4
a = .8245
We are so close wemight as well have stopped.
Call a = .825 and do your integral from -.825 to +.825 or twice the integral from 0 to .825.
What I am doing is called Newton's method and there are programs for it, but it is really just putting numbers on graphs.

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posted by Damon
2. By the way, if you look up Newton's method in wikipedia, the example they use to demonstrate it is cos x = x^3 so it is very close to what you are doing.

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posted by Damon

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