Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$,and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals

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 $f\left( \frac{\pi}{2} \right)$.

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 bounded by the lines $y = x$,$x = -1$,$x = 2$ and the $x$-axis is

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 $: 1$)

Area enclosed by the ellipse $9x^2 + 4y^2 = 1$ in the first quadrant is . . . . . . .