The area we are interested can be written as
#int_(x_1)^(x_2) sqrt x - x^2 dx#.
Before explaining why it's #sqrt x - x^2#, let's look at #x_1# and #x_2#.
#x_1# and #x_2# are real solutions to the equation #sqrt x = x^2#; this means that they are the #x#-coordinates of the intersections of #y = sqrt x# and #y = x^2#.
These are equal to #0# and #1#, respectively, as #color(red)("proved")# below:
#sqrt x = x^2#
#x^(color(red)(1/2)) = x^color(red)2#
Square both sides.
# x = x^color(red)4#
We can already take #0# off the list, as #0 = 0^k#, for any #k#.
#color(red)(x_1 = 0)#
If we don't account for #x=0#, we can divide both sides by #x#:
#1 = x^color(red)3#
Let's rewrite things to make it easier to handle :
#x^3 - 1^3 = 0#
We can use the difference of cubes formula, which states that
#color(red)a^3 - color(red)b^3 = (color(red)a-color(red)b)(color(red)a^2 + color(red)(ab) + color(red)b^2)#
Plug in #color(red)a = x# and #color(red)b = 1#:
#x^3 - 1^3 = (x-1)(x^2+x+1)#
#(x-1)(x^2+x+1)=0#
The first case is when #x-1 = 0#, so #color(red)(x=1)# is a solution.
The second case is when #x^2+x+1 = 0#. We can use the quadratic formula for this to get :
#x=(-1+-sqrt(color(red)(-3)))/2#
However, we are only interested in the real solutions, so #x=1# is the solution we need. This means
#color(red)(x_2 = 1)#
So our integral is now
#int_(x_1)^(x_2) sqrt x - x^2 dx# = #int_0^1 sqrt x - x^2#
Now, it's #sqrt x - x^2# because for any #x = color(blue)a, 0<=color(blue)a<=1#, #sqrt x# is #color(red)"bigger or equal"# when compared to #x^2#, as #color(green)("proved")# below:
#sqrt x >= x^2#
#color(blue)(a)^color(green)(1/2) >= color(blue)a^color(green)2#
Square both sides.
#color(blue)a >= color(blue)a^color(green)4#
Again, #0# is clearly a possible case, so we won't account for it.
This means we can divide by #color(blue)a# :
#1>=color(blue)a^color(green)3#
Take the #color(green)("third root")# of both sides.
#1>=color(blue)a#
Which is true by definition.
Note: We just called #x# a different name. It doesn't really have an effect, so let's get back into the #x# world.
Back to our integral; it has become
#int_0^1 sqrt x - x^2 dx#
We can rewrite this as
#int_0^1 sqrt xdx - int_0^1 x^2 dx#
#int_0^1 x^color(red)(1/2) dx - int_0^1 x^color(red)(2) dx#
We know that any integral of the form
#int_0^n x^color(red)t dx = n^color(red)(t+1)/color(red)(t+1)#, for any #color(red)t# and #n#.
And finally, we have
#int_0^1 x^(1/2) dx - int_0^1 x^2 dx = #
#= 1^(3/2)/(3/2) - 1^3/3#
#= 2/3 - 1/3 = 1/3#.
