Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. (in $\pi$)

Let $A$ denote the area bounded by the curve $y=\frac{1}{x}$ and the lines $y=0, x=1, x=10$. Let $B=1+\frac{1}{2}+\ldots+\frac{1}{9}$ and $C=\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{10}$. Then,

The area bounded by the curve $y = f(x)$,the coordinate axes,and the line $x = x_1$ is given by $x_1 \cdot e^{x_1}$. Therefore,$f(x)$ equals:

The area of the region bounded by the curve $y = \sin 2x$,the $x$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

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 ...

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