Question

prove that the curves y^2 = 4x and x^2 = 4y divide the area of the square bounded by x=0 ,x=4 and
y = 4 and y = 0 in three equal parts?

Answer 1

i think this is from ncert application of integration

Answer 2

the same sum is done in the book's solved example

Answer 3

any way area enclosed by the parabola x^2 = 4yand the x-axis is
obtained by integrating (x^2/4)dx from limits 0 to 4
its integral is [x^3/12] putting the limits we
have
[4^3/12] - [0^3/12]=16/3

Answer 4

similarly area enclosed by the parabola y^2 = 4x and the y-axis is obtained by integrating (y^2/4)dy from limits 0 to 4 its integral is [y^3/12] putting the limits we have [4^3/12] - [0^3/12]=16/3

Answer 5

hence the third part =4*4 -(16/3) -(16/3) =16/3

Answer 6

the curves y^2 = 4x and x^2 = 4y divide the area of the square bounded by x=0 ,x=4 and y = 4 and y = 0 in three equal parts and each part equals 16/3

Answer 7

now just draw tow parabolas

Answer 8

by the way x=0 is the eq for y-axis and y=0 is the eq for x-axis