Draw the region whose area is $$\int_0^\infty \frac 1 {1+x^2} \,dx$$
Show that $\int_0^∞1/(1+x^2)\,dx = \int_0^1\sqrt{(1-y)/y}\, dy$
Do not evaluate or calculate any integral, use a pictoral argument or algebraic argument.
I drew the graph of the equation, and see what the question is basically asking.
Both expressions are the same: one expressed in terms of x, one in terms of y. One of the integrals uses vertical rectangles to sweep across the graph to cover the area (From 0 to infinity) and the other uses horizontal rectangles(from 0 to 1). I do not know how to prove that they are equal without evaluating them.
LHS is $\frac \pi2$ because it is half the well-known integral which becomes $[\arctan x]_{-\infty}^{\infty } $, that is, it's half the area under $\frac1{1+x^2}$...
RHS is also $\frac\pi2$, as it computes the area of half the unit circle, as seen by a change of variables ...
In a word, both are half the area of the unit circle...