The area of the region bounded by the curve $y = |x^3 - 4x^2 + 3x|$ and the $X$-axis,for $0 \leq x \leq 3$,is

The area (in sq. units) of the smaller part of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$ 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 $: 1$)

The area of the region bounded by the curve $y^2=9x$ and the line $y=3x$ is

The area of the region bounded by the curve $x+2y=8$,the $X$-axis,and the lines $x=1$ and $x=5$ using integration is . . . . . . sq. units.

If $(a, 0); a > 0$ is the point where the curve $y = \sin 2x - \sqrt{3} \sin x$ cuts the $x$-axis first,and $A$ is the area bounded by this part of the curve,the origin,and the positive $x$-axis,then: