Area under the curve of an relation.

Question

The area enclosed by the curve, $(y-\tan^{-1} x)^2 =x-x^2$

Solving this we get : (y- $\tan^{-1}x )= \sqrt{x-x^2} \;\;\; and \;\;\;(y-\tan^{-1}x)= -\sqrt{x-x^2} $ .

I plotted these functions on graphing calculators and got to know that they are symmetrical (I mean the areas enclosed seems to be equal about this line). I saw that the functions are equal at x=0 and x=1 and hence the line of symmetry must be y=$\frac{\pi}4 $x . Here is the screenshot.

If there was no graphing calculator with me. How would I determine that the relation is symmetrical about the line y=$\frac{\pi}4 $x . And how would I even know that the relation is symmetrical or not. In general, how do we determine whether a given relation is symmetrical and draw its graph with hand.

Going on with the problem, Required area= 2( $ \int_0^1(\sqrt{x-x^2} +tan^{-1}x -\frac{\pi}4x)dx = \frac{\pi}2 - ln2 , $ which is not the correct answer. By how much I understood of this problem, I guess I am doing it correctly. But why is my answer coming out to be wrong.

Answer 1

Since you have already split the function

Lets define

$f(x)$ = $\arctan$ + $\sqrt{x-x^2}$ ; $g(x)$ = $\arctan$ - $\sqrt{x-x^2}$

Now if the functions intersect at x=a & x=b (a<b), then the area between them is given by

Area = $\displaystyle \vert {\int_a^bf(x)dx - \int_a^bg(x)dx}\vert$

Substituting the equations you get

Area = $\displaystyle \vert \int_0^1\arctan + \sqrt{x-x^2}dx - \displaystyle \int_0^1\arctan - \sqrt{x-x^2}dx \vert $

Simplifying to get

Area = $\displaystyle 2 \int_0^1 \sqrt{x-x^2}dx $

Which I think you can easily solve further

Answer 2

The enclosed area is $$\int_0^1\left((\tan^{-1}x+\sqrt{x-x^2})-(\tan^{-1}x-\sqrt{x-x^2})\right)\,dx$$ $$=2\int_0^1\sqrt{x-x^2}\,dx=\frac\pi4.$$ It is strictly smaller that $2\int_0^1(\sqrt{x-x^2} +\tan^{-1}x -\frac{\pi}4x)\,dx =\pi-\ln2$ (btw, not "$\frac{\pi}2 -\ln2$"), which proves that the two areas on each side of the line $y=\frac\pi4x$ are not equal. More precisely, as can also be viewed from your picture, the area above the line is larger than the area below it.