The area bounded by the line $y=x$,the $x$-axis,and the ordinates $x=-1$ and $x=2$ is

Area of the region bounded by the curve $y=\cos x$,$x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ is . . . . . . sq. units.

The area (in sq. units) of the region $A = \{(x, y) : (x-1)[x] \leq y \leq 2\sqrt{x}, 0 \leq x \leq 2\}$,where $[t]$ denotes the greatest integer function,is

Find the area of the region bounded by the line $y=3x+2$,the $x$-axis,and the ordinates $x=-1$ and $x=1$.

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = \frac{\pi}{4}$ and $x = \beta > \frac{\pi}{4}$ is given by $\left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Then,find the value of $f\left( \frac{\pi}{2} \right)$.

The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then $A_1 : A_2$ equals (in $: 1$)