Embibe Experts Solutions for Chapter: Area under Curves, Exercise 3: Level 3

The area bounded by the curves $y={\cos }^{-1}x;$ $y={\sin }^{-1}x$ and $y=-\mathrm{\pi }{x}^3,$ where $-1\le x\le 1,$ is:

If $f\left(x\right)=\left\{\begin{array}{lll}\sqrt{\left\{x\right\}}& \mathrm{for}& x\notin Z\\ 1& \mathrm{for}& x\in Z\end{array}\right.$ and $g\left(x\right)={\left\{x\right\}}^2$ where $\left\{·\right\}$ denotes fractional part of $x$, then area bounded by $f\left(x\right)$ and $g\left(x\right)$ for $x\in \left(0, 6\right)$ is:

Let $S$ is the region of points which satisfies $y^2<16x,$ $x<4$ and $\frac{xy\left(x^2-3x+2\right)}{x^2-7x+12}>0$. Its area is

The area bounded by the curves $-x\sqrt{3}+y=2{\log }_e\left(x+y\sqrt{3}\right)-2{\log }_{e}2, y=\sqrt{3}x$ and $y=-\frac{1}{\sqrt{3}}x+2$ is

Area of region bounded by the curves $y=\frac{4-x^2}{4+x^2}$, $25y^2=9x$ and $y=\frac{3}{5}\left|x\right|-\frac{6}{5}$ which contains $(1, 0)$ point in its interior, is

If the area bounded by the curves $2y^2+x=0, 3y^2+x=1$ and $y$-axis in the second quadrant is $A$ sq. units, then the value of $\frac{9A}{3-\sqrt{3}}$ is equal to

Let the area (in sq. units) of the region bounded by the curves $x^2+2y-1=0, y^2+4x-4=0$ and $y^2-4x-4=0$ in the upper half plane be $k$, then the value of $k$ is

Consider a line $\mathrm{\ell } : 2x-\sqrt{3}y=0$ and a parametrised $C : x=\tan t, y=\frac{1}{\cos t}\left(0\le t<\frac{\pi }{2}\right)$. If the area of part bounded by $\mathrm{\ell }, C$ and the $y$-axis is equal to $\frac{1}{4}\ln \left(a+\sqrt{b}\right)$, where $a, b, \in N, b$, is not perfect square then find the value of $\left(a+b\right)$.