Application of Definite Integral

Question

I been solving this problem for an hour, but I cant get the correct answer. Please help me with this.

Answer 1

If you split the desired violet area into $2$ equals parts by drawing the line $x=\frac32$, you can evaluate it. In fact you only need to solve the system below \begin{cases} \int_0 ^\frac32 \sqrt{4-x^2} = A+S\\ \int_0 ^2 \sqrt{4-x^2} = A+2S\\ \end{cases} letting $S=$ half of the violet area, and $A =$ the sector of the first circle obtained by removing the entire violet area. By solving the system you get \begin{cases} \left[\frac x2\sqrt{4-x^2}+2\arcsin\left(\frac x2 \right)\right]_0 ^\frac32 = A+S\\ \left[\frac x2\sqrt{4-x^2}+2\arcsin\left(\frac x2 \right)\right]_0 ^2 = A+2S\\ \end{cases} \begin{cases} \frac 34\sqrt{4-\frac{9}{16}}+2\arcsin\left(\frac 34 \right)= A+S\\ 2\arcsin\left(1\right)= A+2S\\ \end{cases} \begin{cases} \frac{3\sqrt{55}}{16}+2\arcsin\left(\frac 34 \right)= A+S\\ \pi = A+2S\\ \end{cases} $$\Rightarrow S=A+2S-(A+S)=\pi-\left(\frac{3\sqrt{55}}{16}+2\arcsin\left(\frac 34 \right)\right)\approx 0.0549\Rightarrow \\\text{Violet Coloured Area}\approx 2\cdot 0.0549 \approx 0.109$$