Find the area bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the ordinates $x=0$ and $x=ae$,where $b^{2}=a^{2}(1-e^{2})$ and $e < 1$.

Area of the region (in sq. units) bounded by the curve $y=x^2-5x+4$,$x=0$,$x=2$ and the $X$-axis 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 bounded by the curve $y = x^3$,the $x$-axis,and the two ordinates $x = 1$ and $x = 2$ is equal to:

If the area above the $x$-axis,bounded by the curves $y = 2^{kx}$,$x = 0$,and $x = 2$ is $\frac{3}{\ln 2}$,then the value of $k$ is

The area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is