Using Integration find the area enclosed by the curve y^2=8x-x^2 above x-axis.

Answer 1

integral of 8x-x^2 = 4x^2 -x^3/3, you can ignore the constant

evaluate it where the parabola intersects the x axis

x(8-x)=0, that's x=0 and x =8, just plug in 8 into the integral

4(8)^2 - (8)^3/3 = 4(64) - (64)(8)/3 = 64(4-8/3) = 64(4/3)= 256/3 = 85 1/3