The area of the region bounded by the line $2y + x = 8$,the $x$-axis,and the lines $x = 2$ and $x = 4$ is:

The graphs of sine and cosine functions intersect each other at a number of points,and between two consecutive points of intersection,the two graphs enclose the same area $A$. Then $A^{4}$ is equal to ............

$A$ 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 region bounded by the $y$-axis,$y = \cos x$,and $y = \sin x$ for $0 \leq x \leq \frac{\pi}{2}$ is:

Let $q$ be the maximum integral value of $p$ in $[0, 10]$ for which the roots of the equation $x^2 - px + \frac{5}{4}p = 0$ are rational. Then the area of the region $\{(x, y): 0 \leq y \leq (x - q)^2, 0 \leq x \leq q\}$ is

Find the area of the region lying in the first quadrant and bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$.