3 Marks Question

Question 13 Marks

Find the area bounded by the curve y = sin x between x = 0 and $x = 2\pi $

Answer

Equation of the curve is y = sin x …(i)
$\therefore y = \sin x \geqslant 0$ for $0 \leqslant x \leqslant \pi $ i.e., graph is in first and second quadrant.
And $y = \sin x \leqslant 0$ for $\pi \leqslant x \leqslant 2\pi $ i.e., graph is in third and fourth quadrant.

If tangent is parallel to x - axis, then $\frac{{dy}}{{dx}} = 0$
$ \Rightarrow \cos x = 0$
$ \Rightarrow x = \frac{\pi }{2},\frac{{3\pi }}{2}$
Table of values for curve y = sin x between x = 0 and $x = 2\pi $

x	0	$\frac{\pi }{2}$	$\pi $	$\frac{{3\pi }}{2}$	$2\pi $
y	0	1	0	- 1	0

Now Required shaded area = Area OAB + Area BCD
$= \left| {\int\limits_0^\pi {ydx} } \right| + \left| {\int\limits_\pi ^{2\pi } {ydx} } \right|$
$= \left| {\int\limits_0^\pi {\sin xdx} } \right| + \left| {\int\limits_\pi ^{2\pi } {\sin xdx} } \right|$
$= \left| { - \left( {\cos x} \right)_0^\pi } \right| + \left| { - \left( {\cos x} \right)_\pi ^{2\pi }} \right|$
$= \left| { - \left( {\cos \pi - \cos 0} \right)} \right| + \left| { - \left( {\cos 2\pi - \cos \pi } \right)} \right|$
$= \left| { - \left( { - 1 - 1} \right)} \right| + \left| { - \left( {1 + 1} \right)} \right|$
= 2 + 2 = 4 sq. units

View full question & answer→

Question 23 Marks

Sketch the graph of y = |x + 3| and evaluate $\int_{ - 6}^0 {\left| {x + 3} \right|dx} $

Answer

y = |x + 3|
$ \Rightarrow y = \left( {x + 3} \right)$, $if\ x\geq-3$
y = -(x + 3), if x < -3
$\int_{ - 6}^0 {\left| {x + 3} \right|dx = ?} $
Area $= \int_{ - 6}^{ - 3} {-\left( {x + 3} \right)dx + \int_{ - 3}^0 {\left( {x + 3} \right)dx} } $
$= \left[ {-\frac{x^2}{2}-3x} \right]_{-6}^{-3}+ \left[ {\frac{x^2}{2}+3x} \right]_{-3}^{0}$
$= \left[ {(-\frac{9}{2}+9)-(-\frac{36}{2}+18)} \right]+ \left[ {(0+0)-(\frac{9}{2}-9)} \right]$
$= \left[ {(\frac{9}{2}+0)+(0+\frac{9}{2})} \right]$
= 9 sq units

View full question & answer→

Question 33 Marks

Find the area enclosed by the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

Answer

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$ = 4\int_0^a {\frac{b}{a}\sqrt {{a^2} - {x^2}} dx} $
$=\frac{4b}{a} \left[ {\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})} \right]_0^a$
$=\frac{4b}{a} \left[ {(\frac{a}{2}×0+\frac{a^2}{2}sin^{-1}1)-0} \right]$
$= \frac{4b}{a}\frac{a^2}{2}\frac{π}{2}$
$ = \pi ab$ sq. units

View full question & answer→

Question 43 Marks

Find the area enclosed by the circle $x^2 + y^2 = a^2$

Answer

The equation of circle is $x^2 + y^2 = a^2$ .... (i)
Its centre is origin and radius a. We know that circle $x^{2 }+ y^2 = a^2$ is symmetrical about both axes.
$\therefore$ required area = $4 \int \limits_{0}^{2} y d x=4 \int \limits_{0}^{2} \sqrt{a^{2}-x^{2}} d x$
$=4\left[\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a}$
$=4\left[\left(0+\frac{\mathrm{a}^{2}}{2} \sin ^{-1} 1\right)-\left(0+\frac{\mathrm{a}^{2}}{2} \sin ^{-1} 0\right)\right]$
$=4\left[\left(0+\frac{a^{2}}{2} \sin ^{-1} 1\right)-\left(0+\frac{a^{2}}{2} \sin ^{-1} 0\right)\right]$
$=4\left[\frac{a^{2}}{2} \times \frac{\pi}{2}\right]$ $\left[\because \sin ^{-1} 0=0\right]$
$=\pi a$

View full question & answer→

Maths 3 Marks Question

Take a timed test