Area enclosed by the curve $y = f(x)$ that is defined parametrically as $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$ (where $t \in R$) is equal to

The area of the smaller part between the circle $x^2 + y^2 = 4$ and the line $x = 1$ is

Let $g(x) = \cos(x^2)$,$f(x) = \sqrt{x}$,and $\alpha, \beta$ (where $\alpha < \beta$) be the roots of the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Then,the area (in sq. units) bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha$,$x = \beta$,and $y = 0$ is:

If the area of the region $\{(x, y): -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a > 0\}$ is $\frac{e^2 + 8e + 1}{e}$,then the value of $a$ is:

The area bounded by the curve $4y^{2} = x^{2}(4-x)(x-2)$ is equal to ...... .

Using integration,find the area of the region bounded by the triangle whose vertices are $(-1, 0)$,$(1, 3)$,and $(3, 2)$.