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# find the area of the region enclosed by the curves y=cos(px÷2) and y=1-x^2where x is variable in interval [-1,1]HINT:If f(x) is even then f(x)=f(-x).

7 years ago

y1=cos(pi*x/2), y2=1-x^2 .

the area is integeral of (y2-y1 )dx between -1 and 1 , let represent integral by I

I(1-x^2-cos(pi*x/2))dx, between -1 and 1 on , calculating it comes out to be .060.

Thanks & Regards
B.Tech.
IIT Delhi

7 years ago

`since, from curve you can see the both the function are even valued , you can integrate function between 0 to 1 and take the twice the value of integral, To calculate area integrate1-x^2-cos (px÷2)between 0 to 1 and take twice of this integral , [x-x^3/3-(2/p)*sin(px÷2)], between 0 and 1, it comes = 1-1/3-(2/p)=.03, and twice this is .06.`
7 years ago
z
mamillapelly bhagath
33 Points
7 years ago
lokesh palingi
44 Points
7 years ago