The area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and $x = 0$ is

The curve $y=ax^2+bx$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$. If the area enclosed by this curve,the $X$-axis and the line $x=6$ is $108$ square units,then $2b-a=$

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 of the region bounded by the curve $y=|x-3|$,the $X$-axis,and the lines $x=0$ and $x=2$ is . . . . . . sq. units.

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to

The area under the curve $y = x^2 - 4x$ within the $x$-axis and the line $x = 2$ is: