The area included between the parabolas $y^{2} = 5x$ and $x^{2} = 5y$ is

Consider the functions $f, g: R \rightarrow R$ defined by
$f(x)=x^2+\frac{5}{12}$ and $g(x)=\begin{cases} 2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4} \end{cases}$
If $\alpha$ is the area of the region
$\{( x , y ) \in R \times R :| x | \leq \frac{3}{4}, 0 \leq y \leq \min \{f( x ), g( x )\}\}$,
then the value of $9 \alpha$ is.

The area of the region above the $X$-axis included between the parabola $y^{2}=x$ and the circle $x^{2}+y^{2}=2x$ in square units is

The area bounded by the curve $y=\cos x$,the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ is

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is:

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is