Question

Fill in the blanks.
(i) Area of the region enclosed by the curve $y=tanx$, 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)	$log2$	$\frac{32}{3}$	$\frac{32}{3}$	$\frac{1}{2}$
(b)	$2$	$16$	$\frac{32}{3}$	$2$
(c)	$\frac{1}{2}$	$\frac{4}{3}$	$\frac{4}{3}$	$2$
(d)	$log2$	$\frac{32}{3}$	$\frac{16}{3}$	$\frac{1}{2}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (A)

$\left(i\right)$ We have $y=tanx$

$\therefore$ Required area $=\underset{0}{\overset{\pi /3}{\int }}tanxdx$
$=-\left|log\right|{\left|cosx\right|}_0^{\pi /3}$
$=-\left|log\left(cos\frac{\pi }{3}\right)-log\left(cos0\right)\right|$
$=-log\frac{1}{2}=log{\left(\frac{1}{2}\right)}^{-1}=log2$
$\left(ii\right)$ Required area $=2\underset{0}{\overset{4}{\int }}\left|y\right|dx$
$=2\underset{0}{\overset{4}{\int }}\sqrt{x}dx=\frac{4}{3}{\left[x^{3/2}\right]}_0^4$
$=\frac{4}{3}\left[4^{3/2}-0\right]$
$=\frac{32}{3}$ sq. units
$\left(iii\right)$ The equation
$x=\sqrt{y}$ or $x^2=y$ represents a parabola.
$\therefore$ Required area

$=2\underset{0}{\overset{4}{\int }}\left|x\right|dy$
$=2\underset{0}{\overset{4}{\int }}\sqrt{y}dy$
$=\frac{4}{3}{\left(y^{3/2}\right)}_0^4$
$=\frac{4}{3}\left(4^{3/2}-0\right)$
$=\frac{4}{3}.8$
$\frac{32}{3}$ sq. units
$\left(iv\right)$ We have given a line $y=x+1$

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