The area of the region bounded by the curves $y=3x+1$,$y=4x+1$ and the line $x=2$ is

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

The area of the region bounded by the line $y = 3 - x$,the $X$-axis,and the lines $x = 2$ and $x = 5$ is . . . . . . .

Find the area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$.

Area of the region bounded by the curve $x^2 = 4y$ and the line $y = 3$ is . . . . . . . (in $\sqrt{3}$)

Area under the curve $y = \sin 2x + \cos 2x$ between $x = 0$ and $x = \frac{\pi}{4}$ is ......... $sq. \text{ } unit$.