The area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and $x = 0$ is

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

The area enclosed between the curve ${y^2}(2a - x) = {x^3}$ and the line $x = 2a$ above the $x$-axis is

Let $q$ be the maximum integral value of $p$ in $[0, 10]$ for which the roots of the equation $x^2 - px + \frac{5}{4}p = 0$ are rational. Then the area of the region $\{(x, y): 0 \leq y \leq (x - q)^2, 0 \leq x \leq q\}$ is

Using the method of integration,find the area bounded by the curve $|x|+|y|=1$.

The area (in square units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is