The area bounded by the curve $y = \ln(x)$ and the lines $y = 0$,$y = \ln(3)$,and $x = 0$ is equal to

If $x^{2}+y^{2}=a^{2}$,then $\int_{0}^{a} \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx=$

The area of the region bounded by the line $y=3x$ and the curve $y=x^2$ in square units is

The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

The area (in square units) of the region bounded by the curve $y = |\sin 2x|$ and the $X$-axis in the interval $[0, 2\pi]$ is:

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = \frac{\pi}{4}$ and $x = \beta > \frac{\pi}{4}$ is given by $\left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Then,find the value of $f\left( \frac{\pi}{2} \right)$.