The area bounded by the parabola $y^2=x$,the straight line $y=4$ and the $Y$-axis is

Find the area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2$ and the $x-$ axis.

The area bounded by the curve $y=x|x|$,the $x$-axis,and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x <  0$].

The area of the region bounded by $y=||x-3|-4|-5$ and the $X$-axis is

If $A$ is the area of the region bounded by the curve $y = \sqrt{3x + 4}$,the $x$-axis,and the lines $x = -1$ and $x = 4$,and $B$ is the area bounded by the curve $y^2 = 3x + 4$,the $x$-axis,and the lines $x = -1$ and $x = 4$,then $A:B$ is equal to:

The slope of the tangent to a curve $y = f(x)$ at $(x, f(x))$ is $2x + 1$. If the curve passes through the point $(1, 2)$,then the area of the region bounded by the curve,the $x$-axis,and the line $x = 1$ is: