The area (in sq. units) bounded by the parabola $y=x^2+3$, the tangent to the parabola at $(3,12)$ and the coordinate axes and lying in the first quadrant is

The area bounded by the curve $y=x^{3}$,the $X$-axis and the lines $x=1$ and $x=4$ is

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The area bounded by the curve $y = f(x)$ and the line $y = 3$ is: (in $/3$)

The area of the figure bounded by $y = e^x$,$y = e^{-x}$,and the straight line $x = 1$ is

Using the method of integration,find the area of the triangle $ABC$,whose vertices are $A(2,0)$,$B(4,5)$,and $C(6,3)$.

If the area bounded by the curve $x^2=4y$,the $X$-axis,and the line $x=4$ is divided into two equal areas by the line $x=\alpha$,then the value of $\alpha$ is ...