Verified answer

The functions are:

f(x)=√x

and: g(x)=x

The intersection points are:

g(x)=x=f(x)=√x

(0,0) and (1,1)

Now there is two areas between the two functions, the area from x=0 to x=1 and the area from x=1 to ∞.

The first one is

A₁=∫(√x-x)dx [from 0 to 1]or:

A₁=∫√xdx-∫xdx [from 0 to 1]

The second is:

A₂=∫(x-√x)dx [from 1 to ∞]

so:

A₁=∫√xdx-∫xdx=[(2/3)∛(x²)-0.5x²]

A₁=[(2/3)-0.5]=1/6

A₂=∫(x-√x)dx=Divergent

The domain for the first function must necessarily be "real positive numbers", for otherwise, F(X)^2 or the co-domain will have imaginary numbers. (And we do not talk about imaginary areas).

The F function could be alternatively represented as-

F(Y) = Y^(1/2), where Y = (X)^2. The F function is a curve in the first and the fourth quadrant.

The G function represent a straight line through the first and the third quadrant.

The two curves intersect at 0 and 1 (in the first quadrant).

Integrating between 0 and 1, you get:

(2/3) - (1/2) = 1/6

We have: f(x) = sqrt(x), g(x)= x

These two curves intersect when sqrt(x) = x, i.e. x = x^2

x(1 - x^2) = 0

==> x = 0, 1 or -1. Reject -1 as then f(x) is not defined.

So our integration limits are from x = 0 to x = 1

The area between the curves is then:

int (sqrt(x) - x) from x = 0 to x = 1

= 1/6 units square

if you mean x^2 then the answer is 1/6

as integrate x- x^2

as then substitute by the intersection points which is 0 and 1 then get the absolute value so you get 1/6

if u mean square root x then the answer is as well 1/6!!!