The area of the region under the curve $y=|\sin x-\cos x|$,$0 \leq x \leq \frac{\pi}{2}$ and above the $x$-axis,is (in square units)

The area bounded by the curves $y = \cos x$ and $y = \sin x$ and the ordinates $x = 0$ and $x = \frac{\pi}{4}$ is:

The area of the region bounded by the curve $y = \cos x$,the $X$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

Let $f: R \rightarrow R$ be a twice differentiable function such that $f(x + y) = f(x) f(y)$ for all $x, y \in R$. If $f^{\prime}(0) = 4a$ and $f$ satisfies $f^{\prime \prime}(x) - 3a f^{\prime}(x) - f(x) = 0$,$a > 0$,then the area of the region $R = \{(x, y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2\}$ is:

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 (in sq. units) of the region bounded by the curves $y = 2^x$ and $y = |x + 1|$ in the first quadrant is