Area under the curves, can you make a right choice?

Let us define the regions $R = \int_{0}^{x_0} f(x) - \sqrt{x} dx$ and $S = \int_{0}^{x_0} f(x_0) - f(x) dx$ about the point $x = x_0.$ If $R = S$ area-wise, then we obtain:

$\int_{0}^{x_0} f(x) - \sqrt{x} dx = \int_{0}^{x_0} f(x_0) - f(x) dx \Rightarrow \boxed{ 2\int_{0}^{x_0} f(x) dx = x_0 \cdot f(x_0) + \int_{0}^{x_0} \sqrt{x} dx}$ (i)

If we now differentiate both sides of (i) with respect to $x_0$ , we next obtain a first-order ODE:

$2f(x_0) = f(x_0) + x_0 f'(x_0) + \sqrt{x_0} \Rightarrow \boxed{ f'(x_0) - \frac{1}{x_0} \cdot f(x_0) = -\frac{1}{\sqrt{x_0}}}$ (ii)

Multiplying (ii) through by the integrating factor $\frac{1}{x_0}$ now yields:

$(\frac{f(x_0)}{x_0})' = -\frac{1}{x_0^{3/2}} \Rightarrow \frac{f(x_0)}{x_0} = \frac{2}{\sqrt{x_0}} + C \Rightarrow \boxed{ f(x_0) = 2\sqrt{x_0} + C \cdot x_0}$ (iii)

Hence, there is an infinite number of functions $f(x)$ that satisfy the above area condition.

Partial solution:

Note that $f(x)=2\sqrt{x}+kx$ , where $k\ge 0$ , are solutions.

From the figure one can immediately deduce: $x_A=x_B$ and $y_A=y_C$ .

Express $R$ and $S$ via integrals:

$R = \int_0^{x_A} (f(x) - \sqrt{x}) dx$

$S = x_A y_A - \int_0^{x_A} f(x) dx$

$R = S \quad \Leftrightarrow \quad \int_0^{x_A} f(x) dx - \underbrace{\int_0^{x_A} \sqrt{x} dx}_{\frac{2}{3} x_A^{\frac{3}{2}}} = x_A y_A - \int_0^{x_A} f(x) dx$

$\Leftrightarrow \quad \boxed{\int_0^{x_A} f(x) dx = \frac{1}{3} x_A^{\frac{3}{2}} + \frac{1}{2} x_A y_A}$ , i.e. $F(0)=0$

Find a singular $f$ to solve the equation:

e.g. $f(x) = \frac{1}{2} \sqrt{x} + \frac{1}{2} y_A$

$\Rightarrow F(x) = \frac{1}{3} x^{\frac{3}{2}} + \frac{1}{2} x y_A$ , which satisfies $F(0)=0$ .

To obtain infinitely many solutions, add a paramterized term to $F(x)$ such that it doesn't change the value of $\int_0^{x_A} f(x) dx$ :

$F_n(x) = \frac{1}{3} x^{\frac{3}{2}} + \frac{1}{2} x y_A + \sin\left(\frac{2 \pi}{x_A} x \pm n \pi \right)$

$\Rightarrow \boxed{f_n(x) = \frac{1}{2} \sqrt{x} + \frac{1}{2} y_A + \frac{2 \pi}{x_A} \cos\left(\frac{2 \pi}{x_A} x \pm n \pi \right)}$ .

It can easily be seen that:

$\int_0^{x_A} f_n(x) dx = \left[\frac{1}{3} x^{\frac{3}{2}} + \frac{1}{2} x y_A + \sin\left(\frac{2 \pi}{x_A} x \pm n \pi \right)\right]_0^{x_A} \\ = \frac{1}{3} x_A^{\frac{3}{2}} + \frac{1}{2} x_A y_A + \underbrace{\sin\left((2+n) \pi \right)}_{0} - \underbrace{\sin\left( n \pi \right)}_{0} \\ = \frac{1}{3} x_A^{\frac{3}{2}} + \frac{1}{2} x_A y_A$

which proves that infinitely many $f(x)$ exist for this problem set.

First, $\forall x\geq 0, \int_0^x f(x)-f(t) dt = \int_0^x f(t)-\sqrt{t}dt$ thus $\forall x>0, f'(x)=f(x)/x + g(x)$ where $g$ is partially contiguous. Using Cauchy linear theorem there is a real vector family of such f functions. Moreover, $f\neq 0$ because $g(1)\neq 0$ . There are infinitely many such f functions.

When the two Areas R and S are evaluated and equated it's found that the expression doesn't contain f (x) which implies that out can be anything.

Alternatively you can compute the integral $\int_{0}^{\sqrt{B}} (f-\sqrt{x})dx$ and set this equal to $BC-\int_{0}^{B} (f.dx)$ where B is the extreme x co-ordinate (on right) now $R=S$ gives $2.\int_{0}^{B} (f.dx)=BC+\int_{0}^{B} (\sqrt{x}dx)=BC+\frac{2}{3}B^{\frac{3}{2}}$ Also note that this implies $\int_{0}^{B} (f-x).dx=\frac{1}{3}B^{\frac{3}{2}}$ the later implies, $\int_{0}^{B} (f.dx)=\frac{1}{3}B^{\frac{3}{2}}+\frac{B^2}{2}$ this implies $B=C$

Although it was not necessary to show that to have consistency $B=C$ is necessary and sufficient yet this scales down the degree of freedom in x-y directions to 1; i.e in the infinite multiset $S(n \times n)$ we are only left with $S(n)$ subset; Thus only $\frac{1}{n}$ of the $R \times R$ contributes to our choice;