The area of the region bounded by curves \(x^{2}+y^{2}=4, x=1\) \(\& x=\sqrt{3}\) is..... (a) \((\pi / 3)\) sq. unit (b) \((2 \pi / 3)\) sq. unit (c) \((5 \pi / 6)\) sq. unit (d) \((4 \pi / 3)\) sq. unit

We are given two curves: \(x^2 + y^2 = 4\) (which is a circle of radius 2 centered at the origin) and the vertical lines \(x = 1\) and \(x = \sqrt{3}\). These vertical lines will intersect the circle at different points since the circle's equation is symmetric with respect to both axes. For each vertical line, substitute the given x-value into the circle's equation to find the corresponding y-values: For \(x = 1\), we have: \(1^2 + y^2 = 4\) \(y^2 = 3\) \(y = \pm\sqrt{3}\) For \(x = \sqrt{3}\), we have: \((\sqrt{3})^2 + y^2 = 4\) \(y^2 = 1\) \(y = \pm 1\) So, the intersection points are \((1, \sqrt{3})\), \((1, -\sqrt{3})\), \((\sqrt{3}, 1)\), and \((\sqrt{3}, -1)\).

Let's determine the area enclosed by the two curves and the lines \(x=1\) and \(x=\sqrt{3}\) using integration. Since we know the intersections points, we can set up two separate integrals, one for each side of the circle (above and below the x-axis), and then multiply the result by 2: The integrand will be the function for the upper half of the circle minus the lower half of the circle. The equation of the circle is \(x^2 + y^2 = 4\), so \(y^2 = 4 - x^2\), and \(y = \pm\sqrt{4 - x^2}\). So, the area enclosed by the curve between \(x = 1\) and \(x = \sqrt{3}\) is: \(A = 2 \int_{1}^{\sqrt{3}} (\sqrt{4 - x^2} - (-\sqrt{4 - x^2}))dx\)

Now, we will evaluate the integral: \(A = 2 \int_{1}^{\sqrt{3}} (2\sqrt{4 - x^2})dx\) We can use substitution to solve this integral. Let \(x = 2\sin{u}\), then \(dx = 2\cos{u}du\), and the integral becomes: \(A = 2 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} (2\sqrt{4 - 4\sin^2{u}})(2\cos{u})du\) Simplifying and using the identity \(\sqrt{1 - \sin^2{u}} = \cos{u}\): \(A = 8 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \cos^2{u} du\) Now we can apply the double-angle formula for \(\cos^2{u}\): \(\cos^2{u} = \frac{1 + \cos{2u}}{2}\) Now the integral becomes: \(A = 8 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \frac{1 + \cos{2u}}{2} du\) \(A = 8 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \frac{1}{2}du + 4 \int_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}} \cos{2u} du\) Evaluating those integrals: \(A = 4(\arcsin{\sqrt{3}/2} - \arcsin{1/2}) + 4\left[\frac{1}{2}\sin{2u}\right]_{\arcsin{1/2}}^{\arcsin{\sqrt{3}/2}}\) Now plug in the values and simplify the expression: \(A = 4\left(\frac{\pi}{3} - \frac{\pi}{6}\right) + 4\left[\frac{1}{2}\sin{\left(4\cdot\frac{\pi}{3}\right)} - \frac{1}{2}\sin{\left(4\cdot\frac{\pi}{6}\right)}\right]\) \(A = 4\cdot\frac{\pi}{6} + 4\left[\frac{1}{2}(-\sqrt{3}/2) - \frac{1}{2}(1)\right]\) \(A = \frac{4\pi}{3}\) sq. unit So, the area of the region bounded by the curves is \((d) \frac{4\pi}{3}\) sq. unit.