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?


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 $\displaystyle \int_{-1/2}^{1/ \sqrt2} \frac{1}{\sqrt{1-x^2}} \ dx$


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!

  • $\begingroup$ @AlbusDumbledore good luck for your exam! By the way I did not understand your comment about integrating both sides of the inequality. $\endgroup$ – Math Lover Jan 17 at 16:48
  • $\begingroup$ @MathLover oh that integrating we get $\int_0^1 f(x)\le \pi/4$ which means the equalty must occur $\endgroup$ – Albus Dumbledore Jan 17 at 16:50
  • $\begingroup$ @AlbusDumbledore Integrating we get $\frac{\pi}{4}$ which is area. There is no inequality here. It is a very standard practice to represent a closed curve with inequality when it is about area otherwise we remain on the curve. Similarly for volume in a closed surface otherwise remain on the surface. If you read $f(x) = y$ in the question, it becomes more clear, may be. $\endgroup$ – Math Lover Jan 17 at 16:56
  • $\begingroup$ @MathLover i integrated both sides of inequality $f(x)\le \sqrt{1-x^2}$ which gave $\int_0^1f(x)\le \pi/4$ but as $\int_0^1 f(x)=\pi/4$ this means $f(x)=\sqrt{1-x^2}$ for all $x $ in domain $\endgroup$ – Albus Dumbledore Jan 17 at 17:02
  • $\begingroup$ That is not true. Take $x = 0.1, f(x) = 0.1$. Is this point in the domain or not? Also not clear what is both sides of inequality. $\endgroup$ – Math Lover Jan 17 at 17:07

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