Find the area of the region bounded by the parabola $y=x^{2}$ and $y=|x|$.

The area of the region (in sq. units),in the first quadrant bounded by the parabola $y = 9x^2$ and the lines $x = 0, y = 1$ and $y = 4$,is

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 curves $y = \sin x$, $y = \cos x$ and the lines $x = 0$, $x = \frac{\pi}{4}$ is

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:

The straight line through the origin which divides the area formed by the curves $y=2x-x^2$, $y=0$, and $x=1$ into two equal halves is