If $\int\limits_0^1 {(4x^3 - f(x))f(x)dx = \frac{4}{7}}$,then the area of the region bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

The area of the region lying in the first quadrant and bounded by the curve $y = 4x^2$, the $Y$-axis, and the lines $y = 2$ and $y = 4$ is:

Area under the curve $y = \sin 2x + \cos 2x$ between $x = 0$ and $x = \frac{\pi}{4}$ is ......... $sq. \text{ } unit$.

If the tangent to the curve $y = 1 - x^2$ at $x = \alpha,$ where $0 < \alpha < 1,$ meets the axes at $P$ and $Q.$ Also $\alpha$ varies,the minimum value of the area of the triangle $OPQ$ is $k$ times the area bounded by the axes and the part of the curve for which $0 < x < 1,$ then $k$ is equal to

The area enclosed between the curve ${y^2}(2a - x) = {x^3}$ and the line $x = 2a$ above the $x$-axis is

The area bounded by the $x-$ axis,the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $\sqrt{b^2 + 1} - \sqrt{2}$ for all $b > 1$. Then $f(x)$ is: