Question

The area enclosed by $x^2+y^2=4,y=x^2+x+1,$ $y=\left[{\sin }^2\frac{x}{4}+\cos \frac{x}{4}\right]$ and $x$-axis (where $[.]$ denotes the greatest integer function) is:

• A. $\frac{2\pi }{3}+\sqrt{3}-\frac{1}{6}$
• B. $\frac{2\pi }{3}+2\sqrt{3}-\frac{1}{6}$
• C. $2\sqrt{3}-\frac{1}{3}$
• D. $\frac{\pi }{3}+\sqrt{3}$

Answer 1

The correct option is A $\frac{2\pi }{3}+\sqrt{3}-\frac{1}{6}$

${\sin }^2\frac{x}{4}+\cos \frac{x}{4}=1-{\cos }^2\frac{x}{4}+\cos \frac{x}{4}$
$=1-({\cos }^2\frac{x}{4}-\cos \frac{x}{4})$
$=\frac{5}{4}-{(\cos \frac{x}{4}-\frac{1}{2})}^2$
Now, for $-2\le x\le 2$
$\Rightarrow -\frac{1}{2}\le \frac{x}{4}\le \frac{1}{2}$
$\Rightarrow {\cos }^2\frac{x}{4}\le \cos \frac{x}{4}$ for all $x\in [-2,2]$

$\Rightarrow {\cos }^2\frac{x}{4}+{\sin }^2\frac{x}{4}\le {\sin }^2\frac{x}{4}+\cos \frac{x}{4},\forall x\in [-2,2]$
$\Rightarrow \left[{\sin }^2\frac{x}{4}+\cos \frac{x}{4}\right]=1\forall x\in [-2,2]$

Now from the figure, to find the enclosed area, we have to find point of intersections of $y=1$ and the circle.

Solving both equations we get $x=-\sqrt{3}$ and $x=\sqrt{3}$

Also we have to find the points of intersection of $y=1$ and $y=x^2+x+1$.

Solving both equations we get $x=-1$ and $x=0$
Now dividing the enclosed area into different intervals as per the points of intersections, we get required area

$={\int }_{-2}^{-\sqrt{3}}\sqrt{4-x^2}dx+{\int }_{-\sqrt{3}}^{-1}dx+{\int }_{-1}^0(x^2+x+1)dx+{\int }_0^{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$
$=\frac{2\pi }{3}+\sqrt{3}-\frac{1}{6}$