equality between variable and integral

Question

I received the following question as part of my homework: Let $f(x)$ be a continuous function onto $[0,1]$. $f(x)\le\frac{1} {2\sqrt{x}}$ for every $0<x\le1$. Prove that x=0 is the only solution for the following equation in $[0,1]$ : $$\int_0^{x^2} f(t) \,dt=x$$ My idea was to say that due to the fact that $\frac{1} {2\sqrt{x}}$ is greater or equals to $f(x)$ Then its integral is also greater or equals to the integral of $f(x)$, and then since the integral of $\frac{1} {2\sqrt{x}}$ is an improper integral, then it equals to the limit: $\lim\limits_{ε \to 0} \int_ε^{x^2} g(t) \,dt$ (where g(t)=$\frac{1} {2\sqrt{t}}$). I was hoping that somehow I'd get from here an expression which approaches zero from the right side of the axis, thus x would be less or equal to something that approaches zero from the right side which would indicate that x must be zero, but unfortunately I got stuck after the last step that I've shown here and would like for your help. Thanks in advance!

Answer 1

I think you need to require that $f(x) < \frac1{2\sqrt{x}}$,

$$\frac{d}{dx}\bigg[\int_0^{x^2} f(t) \ dt\bigg] = 2x\cdot f(x^2) = \frac{dx}{dx} = 1$$ $$f(x^2) = \frac{1}{2x}\implies f(x) = \frac{1}{2\sqrt{x}}$$

$$\int_0^{x^2} \frac{1}{2\sqrt{t}} \ dt = \sqrt{x^2} - \sqrt{0} = x$$