Evaluate the integral using the given data 
Let $f:[0,1]\to [0,1]$ be a continuous function such that $x^2 +(f(x))^2\le 1$ for all $x\in [0,1]$ and $\int_0^1 f(x).dx=\frac{\pi}{4}$, then find $\int_{-1/2}^{1/\sqrt 2} \frac{f(x)}{1-x^2}.dx$

The part that’s throwing me off the inequality function. I have no idea how to solve such that problems. Can I get a hint?
 A: It is a badly written question. It says for all $x \in (0,1)$ and does not define $f(x)$ for $x \lt 0$ and then asks to integrate for negative values of $x$. Anyway I assume the given equation is true for all $x \in (-1, 1)$.
$x^2 + (f(x))^2 \leq 1$ for all $x \ $ and $\displaystyle \int_0^1 f(x) \ dx = \frac{\pi}{4}$.
And we need to find $\displaystyle \int_{-1/2}^{1/ \sqrt2} \frac{f(x)}{1-x^2} \ dx$
We know from the equation that $x^2 + (f(x))^2 \leq 1$ is a circle. The inequality ensures that you cover all the points inside the circle, in addition to covering points on the circle because the question is about area.
The next point which show the integral $I = \frac{\pi}{4} \ $ for $0 \leq x \leq 1$ only reiterates that the function is never negative otherwise the area would be $\frac{\pi}{2}$ for $0 \leq x \leq 1$.
So the integral that we need to find simply translates to $\displaystyle \int_{-1/2}^{1/ \sqrt2} \frac{1}{\sqrt{1-x^2}} \ dx$
A: Hint :
$$f(x) \leq \sqrt{1-x^2}$$
It means graph of $f(x)$ always lies on or below the unit circle in the first quadrant.
Now what can you conclude from following? $$\int_{0}^{1} f(x) \, dx = \frac{\pi}{4}$$
Think area!
