The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

The area bounded by the straight lines $x = 0, x = 2$ and the curves $y = 2^x, y = 2x - x^2$ is

The area (in sq. units) of the region described by $A = \{ (x,y) | y \ge x^2 - 5x + 4, x + y \ge 1, y \le 0 \}$ is:

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is:

If $A$ is the area bounded by the curves $y = |\cos x|$ and $y = 5 - \frac{4}{\pi} |x - \pi|$ in $x \in [0, 2\pi]$,then the value of $\left( \frac{A}{2} + 2 \right)$ is