As you can see on the graph the area in question is enclosed between the positive branch of `y^2 = x^3` (`y = x^(3/2)` ) and y = x
These two graphs intersect at (0,0) and (1,1)
As a double integral this can be written as
`int_0^1 int_(x^(3/2))^ xdydx`
If you take the inner intergral first (`int_(x^(3/2)) ^ x dy` ) this will reduce to
`int_0^1 (x - x^(3/2))dx`
which is what you would write if you were required to express the given area as a difference between the areas under the two curves.
This integrals equals to
`(x^2 / 2 - 2/5 * x^(5/2)) |_0 ^1 = 1/2 - 2/5 - (0 - 0) = 1/10`
The area is 1/10.