The area bounded by the curve $y = \log x$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

Area lying in the first quadrant and bounded by the ellipse $4x^2 + 9y^2 = 144$ is . . . . . . . (in $\pi$)

Find the area of the region bounded by the curves $y=x^{2}+2, \,y=x,\, x=0$ and $x=3$.

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 $f\left( \frac{\pi}{2} \right)$.

The area of the region bounded by the curve $x + 2y + 8 = 0$ and the lines $y = -3$ and $y = -1$ is . . . . . . sq. units.

Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$,and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals