[jacks]

The number of continuous function \( \displaystyle g:\bigg[0,\frac{3}{2}\bigg]\rightarrow \bigg(0,\infty\bigg) \)

satisfying the equation \( \displaystyle 4\int^{\frac{3}{2}}_{0}g(x)dx+125\int^{\frac{3}{2}}_{0}\frac{1}{\sqrt{g(x)+x^2}}dx=108\;, \) is

[Ignacio Larrosa]

The number of continuous function \( \displaystyle g:\bigg[0,\frac{3}{2}\bigg]\rightarrow \bigg(0,\infty\bigg) \)

satisfying the equation \( \displaystyle 4\int^{\frac{3}{2}}_{0}g(x)dx+125\int^{\frac{3}{2}}_{0}\frac{1}{\sqrt{g(x)+x^2}}dx=108\;, \) is

Possibly zero. What follows is not a demonstration, but it could be a guide. Consider \( g(x)=a \) (constant) and

\( F(a)=\displaystyle \int^{\frac{3}{2}}_{0}4a+125\frac{1}{\sqrt{a+x^2}}\ dx=\displaystyle\frac{125}{2}\ln\left(\displaystyle\frac{\left(\sqrt[ ]{4a+9}+3\right)^2}{4a} \right)+6a \)
\( F'(a)=\displaystyle\frac{12a\sqrt[ ]{4a+9}-375}{2a\sqrt[ ]{4a+9}} \)
\( F'(a)=0\Rightarrow{}4a\sqrt[ ]{4a+9}=125\Rightarrow{}64a^3+144a^2-15625=0 \)
The unique real solution is
\( a=\displaystyle\frac{\sqrt[ 3]{62284-500\sqrt[ ]{15517}}+\sqrt[ 3]{62284+500\sqrt[ ]{15517}}-6}{4}\approx{}5.582889 \)
It is easy to see that it corresponds to a minimum of \( F(a) \), whose exact expression is very cumbersome, but whose value is \( \approx{}108.3056 \)

It remains to be seen what happens if \( g(x) \) is not constant, although positive...

Regards,

[geómetracat]

Probably there is a smart (but ad hoc) way of doing this problem, but you can solve it quite easily by using standard techniques from variational calculus.

In order to find the function \( g \) minimizing the expression
\( \displaystyle \int^{\frac{3}{2}}_{0}\left(4g(x)+125\frac{1}{\sqrt{g(x)+x^2}}\right)dx \),
where, in the variational calculus terminology, the Lagrangian functional of the minimization problem is
\( L(x,g(x)):= 4g(x)+125\frac{1}{\sqrt{g(x)+x^2}} \),
variational calculus tells us we must find the function \( g \) satisfying the Euler-Lagrange equation, which in our case is just:
\( 0=\frac{\partial L}{\partial g} = 4-\frac{125}{2(g(x)+x^2)^{3/2}} \)
So, solving for \( g \), we find:
\( g(x)=\left(\frac{125}{8}\right)^{2/3}-x^2 \)

For this \( g \) the value of the integral is easily computed and gives exactly \( 108 \). Since this function satisfies the requirements (it is positive in \( [0,3/2] \)), there is at least one solution.

By analyzing the second variation of \( L \) we can show that the function \( g \) we have found is indeed a minimum and moreover it's the only one. This shows that there's only one solution to the given equation.

[jacks]

Thanks moderators.