Application of Integrals Test

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Application of Integrals Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

$$\int { [g(x)-f(x)]dx=5 } $$, then the area between two curves for 0 < x < 2, is

A
5

B
10

C
15

D
20

Solution
Question 2
1 / -0

The area bounded by the curve $$y=x^2$$ and $$y \, = \, \dfrac{2}{1 \, + \, x^2}$$ is $$\lambda\ sq.\ unit$$, then the value of $$[\lambda ]$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Required area = $$\int_{-1}^{1}\left(\dfrac{2}{1+x^2} \, - \, x^2\right)dx$$
$$2\int_{0}^{1}\left(\dfrac{2}{1+x^2} \, - \, x^2\right)dx$$
$$2\left(2tan^{-1}x \, - \, \dfrac{x^3}{3}\right)^1_0 \, = \, 2\left(\dfrac{\pi}{2} \, - \, \dfrac{1}{3}\right) \, = \, \pi \, - \, \dfrac{2}{3}$$
$$\therefore \left[\pi \, - , \dfrac{2}{3}\right] \,= \, 2$$
Question 3
1 / -0

Area of region bounded by $$x=0$$,$$y=0$$ $$x=2$$,$$y=2$$,$$y\le {e}^{x}$$&$$y\ge lnx$$ is

A
$$6-4ln2$$

B
$$4 ln2-2$$

C
$$2ln2-4$$

D
$$6-2ln2$$

Solution

$$Area=\int _{ 0 }^{ 2 }{ \left( x-\ln { x } \right) dx } +\int _{ 0 }^{ 2 }{ \left( x-\ln { y } \right) dy } \\ Area={ \left[ \cfrac { { x }^{ 2 } }{ 2 } -\left( x\ln { x-x } \right) \right] }_{ 0 }^{ 2 }+{ \left[ \cfrac { { y }^{ 2 } }{ 2 } -\left( y\ln { y-y } \right) \right] }_{ 0 }^{ 2 }\\ Area=4-4\ln { 2 } +2\\ Area=6-4\ln { 2 } $$
Question 4
1 / -0

The area bounded by the curves $$y=\left| x \right| -1$$ and $$ y=-\left| x \right| +1$$ is

A
$$1$$

B
$$2$$

C
$$2\sqrt{ 2 }$$

D
$$4$$

Solution

We can see that it is a square of side $$\sqrt{2}$$units
Hence
$$Area=a^{2}$$
$$Area=(\sqrt{2})^{2} =2$$units
Question 5
1 / -0

Area common to the curves $${ y }^{ 2 }$$=ax and $${ x }^{ 2 }$$+$${ y }^{ 2 }$$$$= 4ax$$ is equal to

A
$$(9\sqrt { 3 } +4\pi )\cfrac { { a }^{ 2 } }{ 3 } $$

B
$$(9\sqrt { 3 } +4\pi ){ a }^{ 2 }$$

C
$$(9\sqrt { 3 } -4\pi )\cfrac { { a }^{ 2 } }{ 3 }$$

D
none of the above

Solution

$$x^{2}+y^{2}=4ax$$
$$\Rightarrow$$ $$(x-2a)^{2}+y^{2}=(2a)^{2}$$
$$\int_{0}^{2a}{\sqrt{ax}}dx+\int_{3a}^{4a}{\sqrt{(4ax-x^{2})}dx}$$
$$\Rightarrow$$ $$\left(\dfrac{4\pi}{6}-3^{{3}/{2}}\right)a^{2}+\dfrac{2}{3}\sqrt{a}(3^{{3}/{2}}a^{{3}/{2}})$$
$$\Rightarrow$$ $$a^{2}\left(\dfrac{4\pi}{6}+2\sqrt{3}-\dfrac{3\sqrt{3}}{6}\right)$$
Due to symmetry
$$=\dfrac{a^{2}}{3}[4\pi+9\sqrt{3}]$$
Question 6
1 / -0

Find the area bounded by the curves $$x = a\cos t, y = b\sin t$$ in the first quadrant

A
$$\dfrac{\pi ab}{4}$$

B
$$\dfrac{\pi a^2b}{4}$$

C
None of these

D
$$\dfrac{\pi ab^2}{4}$$

Solution

$$x=a\cos t$$ , $$y=b\sin t$$
$$\rightarrow$$ $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$
It's a hyperbola whose area in any quadrant is given by
$$\dfrac{\pi ab}{4}$$
Question 7
1 / -0

A point P moves in xy-plane In such a way that [|x|] + [|y|] = 1 were [.] denotes the greatest integer function. Area of the region representing all possible positions of the point 'P' is equal to

A
$$4$$ sq. units

B
$$16$$ sq. units

C
$$2 \sqrt{2}$$ sq. units

D
$$8$$ sq. units

Solution

Graph of the given function is in the figure,

From figure , graph is a square with side=$$2$$ units.

Therefore , Area= $$2\times2=4$$ sq. units
Question 8
1 / -0

The area common to the circles $$r=a\sqrt{2}$$ and $$r=2a\cos{\theta}$$ is:

A
$${a}^{2}\dfrac{\pi}{2}$$

B
$${a}^{2}\pi$$

C
$${a}^{2}\left(\pi+1\right)$$

D
$${a}^{2}\left(\pi-1\right)$$

Solution

The equations of the circles are $$r=a\sqrt{2}$$ ................$$\left(1\right)$$
and $$r=2a\cos{\theta}$$ .................$$\left(2\right)$$
eqn$$\left(1\right)$$ represents a circle with centre at $$\left(0,0\right)$$ and radius $$a\sqrt{2}$$
eqn$$\left(2\right)$$ represents a circle symmetrical about $$OX$$ with centre at $$\left(a,0\right)$$ and radius $$a$$
The circles are shown in fig.At their point of intersection $$P$$,eliminating $$r$$ from $$\left(1\right)$$ and $$\left(2\right)$$,
$$a\sqrt{2}=2a\cos{\theta}$$
$$\Rightarrow \cos{\theta}=\dfrac{1}{\sqrt{2}}$$
or $$\theta=\dfrac{\pi}{4}$$
$$\therefore$$ Required Area$$=2\times$$area$$OAPQ$$ (By symmetry)
$$=2$$(area$$OAP+$$area$$OPQ$$)
$$=2\left[\dfrac{1}{2}\int_{0}^{\frac{\pi}{4}}{{r}^{2}d\theta}+\dfrac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{r}^{2}d\theta\right]$$ for $$\left(1\right)$$ and $$\left(2\right)$$ respectively.
$$=\int_{0}^{\frac{\pi}{4}}{\left(a\sqrt{2}\right)}^{2}d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{{\left(2a\cos{\theta}\right)}^{2}}d\theta$$
$$=2{a}^{2}\left|\theta\right|_{0}^{\frac{\pi}{4}}+4{a}^{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\dfrac{1+\cos{\theta}}{2}}d\theta$$
$$=2{a}^{2}\left(\dfrac{\pi}{4}-0\right)+2{a}^{2}\left|\theta+\dfrac{\sin{2\theta}}{2}\right|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$
$$=\dfrac{\pi{a}^{2}}{2}+2{a}^{2}\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}-\dfrac{1}{2}\right)$$
$${a}^{2}\left(\pi-1\right)$$
Question 9
1 / -0

The area under the curve $$y= 2\sqrt x$$ bounded by the lines $$x=0$$ and $$x= 1$$ is

A
$$\frac{4}{3}$$

B
$$\frac{2}{3}$$

C
$$1$$

D
$$\frac{8}{3}$$

Solution

$$A=(0,0)$$ , $$B=(1,2)$$ , $$C=(1,0)$$
$$=2\int_{0}^{1}{\sqrt{x}}dx$$
$$[\dfrac{2x^{\dfrac{3}{2}}}{\dfrac{3}{2}}]_{0}^{1}$$
$$=\dfrac{4}{3}$$
Question 10
1 / -0

The area common to the cardioids $$r=a\left(1+\cos{\theta}\right)$$ and $$r=a\left(1-\cos{\theta}\right)$$ is:

A
$$\left(\dfrac{3\pi}{2}+4\right){a}^{2}$$

B
$$\left(\dfrac{3\pi}{2}-4\right){a}^{2}$$

C
$$\left(\dfrac{\pi}{2}+4\right){a}^{2}$$

D
$$\left(\dfrac{\pi}{2}-4\right){a}^{2}$$

Solution

The cardioid $$r=a\left(1+\cos{\theta}\right)$$ is $$ABCO{B}^{\prime}A$$ and the cardioid $$r=a\left(1-\cos{\theta}\right)$$ is $$O{C}^{\prime}B{A}^{\prime}{B}^{\prime}O$$
Both the cardioids are symmetrical about the initial line $$OX$$ and intersect at $$B$$ and $${B}^{\prime}$$
$$\therefore$$ Required Area$$=2$$Area $$O{C}^{\prime}BCO$$
$$=2$$[area$$O{C}^{\prime}BO+$$area$$OBCO$$]
$$=2\left[\left(\int_{0}^{\frac{\pi}{2}}{\dfrac{1}{2}{r}^{2}}d\theta\right)_{r=a\left(1-\cos{\theta}\right)}+\int_{\frac{\pi}{2}}^{\pi}\left({\left(1+\cos{\theta}\right)}^{2}d\theta\right)_{r=a\left(1+\cos{\theta}\right)}\right]$$
$$={a}^{2}\left[\int_{0}^{\frac{\pi}{2}}{\left(1-2\cos{\theta}+{\cos}^{2}\theta\right)d\theta}+\int_{\frac{\pi}{2}}^{\pi}{\left(1+2\cos{\theta}+{\cos}^{2}\theta\right)}d\theta\right]$$
$$={a}^{2}\left[\int_{0}^{\pi}{\left(1+{\cos}^{2}\theta\right)d\theta}-2\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}+2\int_{\frac{\pi}{2}}^{\pi}{\cos{\theta}d\theta}\right]$$
$$={a}^{2}\left[\int_{0}^{\pi}{\left(1+\dfrac{1+\cos{2\theta}}{2}\right)d\theta}-2\left|\sin{\theta}\right|_{0}^{\frac{\pi}{2}}+2\left|\sin{\theta}\right|_{\frac{\pi}{2}}^{\pi}\right]$$
$$={a}^{2}\left[\left|\dfrac{3}{2}\theta+\dfrac{\sin{2\theta}}{4}\right|_{0}^{\pi}-2\left(1-0\right)+2\left(0-1\right)\right]$$
$$=\left(\dfrac{3\pi}{2}-4\right){a}^{2}$$