That was an obvious answer...

Let's consider the equivalent problem which has $f(x)=x^2$ and $g(x)=x$ . First, they intersect each other in $(0,0)$ and $(1,1)$ , and $f(x)<g(x)$ for $0<x<1$ . Hence we want to find $c$ such that

$\int_0^cx\,dx-\int_0^cx^2\,dx=\int_c^1x\,dx-\int_c^1x^2\,dx$

or, equivalently,

$2\int_0^cx\,dx-2\int_0^cx^2\,dx=\int_0^1x\,dx-\int_0^1x^2\,dx$

Therefore:

$2\frac{c^2}{2}-2\frac{c^3}{3}=\frac{1}{2}-\frac{1}{3}$

$\Rightarrow4c^3-6c^2+1=0$

It's easy to see that $\frac{1}{2}$ is a solution. That was an obvious answer, indeed!

Sorry for too much work but this is just another basic approach towards this problem, while I am aware there are easier methods I would like to present my own

bit

.

In this method, to get the y co-ordinate of the point where area is half we use inverse of the function

finding the area difference of corresponding inverse functions from [0,1]

Inversing functions are similar to flipping x and y axes , so we will end up with a parabola and a straight line with decreasing slope,

[You can see from the picture and ignore the co-ordinate marking]

inverse of $sqrt(x) =x^{2}$ and its obvious that the inverse is lifted by a factor of 1 from center and its facing downwards now so we end up with $-x^{2} +1$

inverse of straight line is same except this time we have lift of +1 and - slope so we end up with $-x+1$

integral of this will be same as the actual function=1/6

we can find the point where integral sum =1/12 which is half of the area as following

$-x^{3}/3 + x - (-x^{2}/2 + x) =1/12$ (This function is the difference of the antiderivatives of corresponding functions ,so it represents the

area since lower limit is 0)

we will get 3 distinct points as solution but only $0.5$ lies in the interval [0,1]