Find the area of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, 0 \leq x \leq 2\}$.

The area of the region $R = \{(x, y) : 0 \le y \le \frac{27}{x}, 1 \le x \le 9\}$ is equal to:

The slope of the tangent to a curve $y = f(x)$ at $(x, f(x))$ is $2x + 1$. If the curve passes through the point $(1, 2)$,then the area of the region bounded by the curve,the $x$-axis,and the line $x = 1$ is:

The area of the region bounded by the parabola $y^2 = 8x$ and its latus rectum is . . . . . . sq. units.

Let $n \geq 2$ be a natural number and $f:[0,1] \rightarrow \mathbb{R}$ be the function defined by
$f(x)= \begin{cases} n(1-2nx) & \text{if } 0 \leq x \leq \frac{1}{2n} \\ 2n(2nx-1) & \text{if } \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1-nx) & \text{if } \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(nx-1) & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$,then the maximum value of the function $f$ is

The area (in square units) bounded by the line $y = x$, the $X$-axis and the lines $x = -2$ and $x = 4$ is...