Mathematics - Area Under Curves Question with Solution | TestHub

The equation of the circle is $x^2+y^2=4$.

The area of the smaller region bounded by the circle and the line $x=1$ is given by:

$A=2{\int }_1^2\sqrt{4-x^2}dx$

Using the standard integral formula $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)$, we get:

$A=2{\left[\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_1^2$

Substituting the limits:

$A=2\left[\left(0+2{\sin }^{-1}(1)\right)-\left(\frac{1}{2}\sqrt{3}+2{\sin }^{-1}\left(\frac{1}{2}\right)\right)\right]$

$A=2\left[2\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{2}-2\left(\frac{\pi }{6}\right)\right]$

$A=2\left[\pi -\frac{\sqrt{3}}{2}-\frac{\pi }{3}\right]$

$A=2\left[\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right]$

$A=\frac{4\pi }{3}-\sqrt{3}$

Answer: $\frac{4\pi }{3}-\sqrt{3}$