Question

Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0,y = 2 and y = 4.

Answer 1

The equation of the curve is y = 4x², i.e. `x^2=y/4`

∴ `x=sqrt(y)/2`

Required area = area of the region ABCDA

`=int_2^4x  dy, "where"  x=sqrt(y)/2`

`=int_2^4sqrt(y/2)dy=1/2int_2^4y^(1/2)dy`

`=1/2[y^(3/2)/(3/2)]_2^4=1/3[y^(3/2)]_2^4`

`=1/3[4^(3/2)-2^(3/2)]`

`=1/3(8-root(2)(2))`sq units