Sher Mohammad
Last Activity: 11 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 integrate
1-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.
