G Tewani Solutions for Chapter: Area, Exercise 1: DPP

Let a function $f\left(x\right)$ be defined in $[-2, 2]$ as $f\left(x\right)=\left\{\begin{array}{ll}\left\{x\right\},& -2\le x<-1\\ |sgnx|,& -1\le x\le 1\\ \{-x\},& 1<x\le 2\end{array}\right.$, where $\left\{x\right\}$ and $sgnx$ denotes fractional part and signum function, respectively. Then the area bounded by the graph of $f\left(x\right)$ and $x$-axis is

The area (in sq. units) bounded by $y=x^2+2$ and $y=2\left|x\right|-\mathrm{cos\pi }x$ in sq. units is equal to

Area bounded by $f\left(x\right)=\frac{x^2-1}{x^2+1}$ and the line $y=1$ is

The area (in sq. units) bounded by the curve $y=x{e}^{-x}$, $y=0$ and $x=c$, where $c$ is the $x$-coordinate of the curve's inflection point, is:

Area (in sq. units) of the region bounded by the curve $y=\frac{16-x^2}{4}$ and $y={\sec }^{-1}\left[-{\sin }^{2}x\right]$, (where $[.]$ denotes the greatest integer function) is

If the area bounded between $x$-axis and the graph of $y=6x-3x^2$ between the ordinates $x=1$ and $x=a$ is $19$ square units, then $a$ can take 

Which of the following is the possible value/values of $c$ for which the area of the figure bounded by the curves $y=\sin 2x$, the straight lines $x=\frac{\pi }{6}, x=c$ and the $x$-axis is equal to $1/2$ sq. units ?

Area (in sq. units) of the region bounded by the curve $y=\tan x$ and lines $y=0$ and $x=1$ is