The area included between the parabola y= x^2 4 a and the curve y= 8 a^3 (x^2+4 a^2 ) is

Step-by-step solution

Given curves, aligned & y= x^2 4 a & y= 8 a^3 x^2+4 a^2 aligned For point of intersection equation (i) and (ii), aligned & x^2 4 a = 8 a^3 x^2+4 a^2 & x^2 (x^2+4 a^2 )=4 a (8 a^3 ) & x^4+4 a^2 x^2-32 a^2=0 & (x^2-4 a^2 ) (x^2+8 a^2 )=0 & x^2-4 a^2=0 or x^2+8 a^2=0 & x^2=4 a^2 [x^2=-8 a^2 is not possible ] & x= 2 a x= 2 a aligned So, we take limits from 0 to 2 a . Now, Area enclosed by the two curves A =2 ₀^ 2 a [ 8 a^3 (x^2+4 a^2 ) - x^2 4 a ] d x=2 [ ₀^ 2 a 8 a^3 (x^2+4 a^2 ) d x- ₀^ 2 a x^2 4 a d x ] aligned & =2