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.
