On the geometrical view of an integral

Evaluate the integral

$\mathcal{J} = \int_0^1 \sqrt{4-x^2} \; \mathrm{d}x$

using geometric methods.

Solution

We are working on the following figure

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Thus,

$\begin{align*} \int_{0}^{1} \sqrt{4-x^2} \; \mathrm{d}x &= \left ( \overset{\triangle}{\mathrm{OAB}} \right ) + \left ( \mathrm{A\overset{\frown}{O} \Gamma} \right )\\ &= \frac{\sqrt{3}}{2} + \pi \cdot 2 ^2 \cdot \frac{30}{360} \\ &= \frac{\sqrt{3}}{2} + \frac{4\pi}{12} \\ &= \frac{\pi}{3} + \frac{\sqrt{3}}{2} \end{align*}$

since the red angle is $60^\circ$ due to the triangle $\mathrm{OAB}$ since $\tan \hat{\mathrm{O}} = \sqrt{3}$ ( $\mathrm{OB}=1 \; , \; \mathrm{AB}=\sqrt{3}$ ). Therefore , the green angle is $180^\circ - 90^\circ - 60^\circ = 30^\circ$ . Finally, the area of the circular sector is equal to