Question

Fill in the blanks.
(i) Area of the region enclosed by the curve $y = tan\, x$, the $x$-axis and the line $x=\frac{\pi}{3}$ is P.
(ii) The area under the curve $y=\sqrt{x} $ from $x=0$ to $x=4$ is Q.
(iii) Area enclosed between the $y$-axis, graph of $x=\sqrt{y} $ and the line $y = 4$ is R.
(iv) The area bounded by the axes and the line $y=x+1$ is S.

	P	Q	R	S
(a)	$log\, 2$	$\frac{32}{3}$	$\frac{32}{3}$	$\frac{1}{2}$
(b)	$2$	$16$	$\frac{32}{3}$	$2$
(c)	$\frac{1}{2}$	$\frac{4}{3}$	$\frac{4}{3}$	$2$
(d)	$log\,2$	$\frac{32}{3}$	$\frac{16}{3}$	$\frac{1}{2}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (A)

$\left(i\right)$ We have $y = tan \,x$

$\therefore $ Required area $=\int\limits_{0}^{\pi /3} tan\,x\,dx$
$=-\left|log\right|\left|cos\,x\right|_{0}^{\pi /3}$
$=-\left|log\left(cos\frac{\pi}{3}\right)-log\left(cos\,0\right)\right|$
$=-log \frac{1}{2}=log \left(\frac{1}{2}\right)^{-1}=log\,2$
$\left(ii\right)$ Required area $=2\int\limits_{0}^{4}\left|y\right|dx$
$=2\int\limits_{0}^{4}\sqrt{x}\,dx=\frac{4}{3}\left[x^{3 /2}\right]_{0}^{4}$
$=\frac{4}{3}\left[4^{3/ 2}-0\right]$
$=\frac{32}{3}$ sq. units
$\left(iii\right)$ The equation
$x=\sqrt{y}$ or $x^{2}=y$ represents a parabola.
$\therefore $ Required area

$=2\int\limits_{0}^{4}\left|x\right|dy$
$=2\int\limits_{0}^{4}\sqrt{y}\,dy$
$=\frac{4}{3}\left(y^{3/ 2}\right)_{0}^{4}$
$=\frac{4}{3}\left(4^{3/ 2}-0\right)$
$=\frac{4}{3}.8$
$\frac{32}{3}$ sq. units
$\left(iv\right)$ We have given a line $y = x+ 1$

$\therefore $ Required area
$=\int\limits_{-1}^{0}\left(x+1\right)dx$
$=\frac{x^{2}}{2}\left|_{-1}^{0}+x\right|_{-1}^{0}$
$=\left(0-\frac{\left(-1\right)^{2}}{2}\right)+\left(0+1\right)$
$=\frac{-1}{2}+1$
$=\frac{1}{2}$ sq. units