Application of Integrals Test

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Application of Integrals Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The area enclosed by the curves $$|y + x|\leq 1, |y - x|\leq 1$$ and $$2x^{2} + 2y^{2} = 1$$ is

A
$$\left (2 + \dfrac {\pi}{2}\right )$$ sq. units

B
$$\left (2 - \dfrac {\pi}{2}\right )$$ sq. units

C
$$\left (3 + \dfrac {\pi}{2}\right )$$ sq. units

D
$$\left (3 - \dfrac {\pi}{2}\right )$$ sq.units

Solution

The curves $$y + x\leq 1$$$ and $$|y - x|\leq 1$$ form a square $$ABCD$$ and $$2x^{2} + 2y^{2} = 1$$ is a circle of radius $$\dfrac {1}{\sqrt {2}}$$ and centre $$(0, 0)$$.
Therefore, required area $$=$$ Area of square $$-$$ Area of circle
$$= \left (2 -\dfrac {\pi}{2}\right ) $$ sq. units
Question 2
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Area bounded by $$f(x)=max.\left( \sin { x } ,\cos { x } \right) $$ $$\quad \forall 0\le x\le \cfrac { \pi }{ 2 } $$ and the co-ordinate axis is equal to

A
$$\cfrac { 1 }{ \sqrt { 2 } } $$ sq. units

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

C
$$2$$ sq. units

D
$$1$$ sq. unit

Solution

$$f(x)=max(\sin x,\cos x)$$
$$0<x<\pi/4, \cos x>\sin x$$
$$\pi/4<x<\pi/2, \cos x>\sin x$$
$$f(x)=\cos x, 0<x<\pi/4\\ =\sin x,\pi/4<x<\pi /2$$
Area$$=|\int _{ 0 }^{ \pi /4 }{ \cos { x } } dx|+|\int _{ \pi /4 }^{ \pi /2 }{ \sin { x } } dx|$$
$$|\sin \pi/4-\sin 0|+|\cos\pi/2+\cos \pi/4|$$
$$=\sqrt 2$$ units$$^2$$
Question 3
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What is the area of the rectangle , whose length is $$5\sqrt 3\ cm$$ and breadth is $$5\ cm$$.

A
$$43.3\ cm^2$$

B
$$\sqrt {75}\ cm^2$$

C
$$25\ cm^2$$

D
None of these

Solution

Area of rectangle having length $$l$$ and breadth $$b=l\times b$$
If $$l=5\sqrt3 cm$$ and $$b=5 cm$$, then Area$$=5\sqrt3\times 5cm^2\\ =25\sqrt3=25\times1.73=43.25cm^2$$
Question 4
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The area bounded by the circles $${ x }^{ 2 }+{ y }^{ 2 }=1, { x }^{ 2 }+{ y }^{ 2 }=4$$ in the first Quadrant is

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

B
$$\dfrac {3\pi }{ 4 }$$

C
$$3\pi$$

D
$$\dfrac { \pi }{ 4 }$$

Solution

$$x^2+y^2=1$$ and $$x^2+y^2=4$$ are concentric circles
They form Ring in between them.
Area is given by $$\pi(4-1)=3\pi$$
In first Quadrant it is divided by $$4$$
So, Area is given by $$\dfrac{3\pi}{4}$$
Question 5
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Find the area of a shaded portion.

A
$$270cm^2$$

B
$$360cm^2$$

C
$$180cm^2$$

D
$$150cm^2$$

Solution

Area of shaded region$$=ar(PQRS)-ar(\triangle ORS)$$
$$=20\times 18-\cfrac{1}{2}\times20\times 18\\=360-180=180cm^2$$
Question 6
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If the area bounded by the curve $$y=a{ x }^{ 2 }\quad $$ and $$x=a{ y }^{ 2 },\left( a>0 \right) $$ is $$3sq.units$$, then the value of $$a$$ is

A
$$\cfrac { 2 }{ 3 } $$

B
$$\cfrac { 1 }{ 3 } $$

C
$$1$$

D
$$4$$

Solution

The parabolas intersect at $$(0,0)$$ and $$(\dfrac{1}{a},\dfrac{1}{a})$$

Area bounded $$=\displaystyle \int_{0}^{\frac{1}{a}}(\sqrt{\dfrac{x}{a}}-ax^2)dx$$
$$=\left [ \dfrac{2}{3}\dfrac{x^{\frac{3}{2}}}{\sqrt{a}}-a\dfrac{x^3}{3} \right ]_{0}^{\frac{1}{a}}$$

$$=\left [ \dfrac{2}{3}\dfrac{1}{\sqrt{a}(a^{\frac{3}{2}})} - \dfrac{a}{3a^3} \right ]$$

$$=\left [ \dfrac{2}{3a^2} - \dfrac{1}{3a^2} \right ]$$

$$ = \dfrac{1}{3a^2}$$

It is given that area $$= 3$$ sq.units

Therefore, $$\dfrac{1}{3a^2}=3$$

$$\implies a = \dfrac{1}{3}$$

Hence, answer is option (B).
Question 7
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Find the area of the closed figure bounded by the following curves
$$y =$$ $$\sqrt{x}, y \, = \, \sqrt{4 - 3x}$$, y = 0.

A
$$\dfrac 89$$

B
$$\dfrac 79$$

C
$$\dfrac 59$$

D
$$\dfrac 13$$

Solution

Given curve, $$y = \sqrt{x} , y = \sqrt{4-3x}, y = 0$$

$$y > 0, x \ge 0 , y - 3x \ge 0$$ (basic square root conditions)

$$\Rightarrow x \le 4/3$$ __(1)

Now, $$y = \sqrt{4 - 3x} \Rightarrow y^2 = 4-3x$$

$$x = \dfrac{4 - y^2}{3} = \dfrac{(2-y)(2 + y)}{3}$$

It represents a parabola cutting y-axis at $$(0,2), (0,-2)$$ & x-axis at $$(4/3, 0)$$

Ref. image

Point of intersection

$$\sqrt{4 - 3x} = \sqrt{x}$$

$$\Rightarrow 4-3x = x$$

$$x = 1 \Rightarrow y = \sqrt{1} = 1$$

$$(1, 1)$$ is req. point

Area bounded by curves (shaded)

$$= \displaystyle \int_0^1 \sqrt{x} dx + \int_1^{4/3} \sqrt{4 - 3x} dx$$
$$= \dfrac{2}{3} \left[x^{3/2}\right]_0^1 + \left[\dfrac{2}{3} \times \dfrac{(4 - 3x)}{(-3)}^{3/2} \right]^{4/3}_1$$

$$= \dfrac{2}{3} (1 - 0) + \left(\dfrac{-2}{9} \right) (0-1)$$

$$= \dfrac{2}{3} + \dfrac{2}{9} = \dfrac{8}{9}$$

$$\therefore$$ option A is correct.
Question 8
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The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

A
1:2

B
2:3

C
2:5

D
3:4

E
3:5

Solution

Let side of square $$=a$$
& radii of circle $$=r$$
$$(b=a/2)$$
$${ r }^{ 2 }=\dfrac { { a }^{ 2 } }{ 4 } +{ a }^{ 2 }=\dfrac { { 5a }^{ 2 } }{ 4 } $$
Area of square $$={ a }^{ 2 }=\dfrac { { 4r }^{ 2 } }{ 5 } $$
Semicircle
$${ a }^{ 2 }={ 2r }^{ 2 }$$
$$\dfrac { Ar.\quad of\quad sq }{ Ar.\quad of\quad circle } =\dfrac { { a }^{ 2 } }{ { \pi r }^{ 2 } } =\dfrac { { 2a }^{ 2 } }{ { \pi a }^{ 2 } } =\dfrac { 2 }{ \pi } $$
Ar of $${ \left( square \right) }_{ circle }={ a }^{ 2 }={ 2r }^{ 2 }$$
Ratio $$\rightarrow \dfrac { { 4r }^{ 2 } }{ 5\times { 2 }r^{ 2 } } =\dfrac { 2 }{ 5 } $$
Question 9
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The area enclosed by the curves
$$f(x) = \vert sin x - cos x \vert + \vert cos x + sin x \vert \ \text {and} \ g(x) = 2\vert cos x + sin x \vert , 0 \leq x \leq \pi$$

A
$$2(2 - \sqrt2)$$

B
$$4(2 + \sqrt2)$$

C
$$4(2 - \sqrt2)$$

D
$$2(2 + \sqrt2)$$
Question 10
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The area of figure bounded by the curve $$y=2x-{x}^{2}$$ and the straight line $$y=-x$$ is

A
$$\frac { 9 }{ 2 } $$

B
$$9$$

C
$$\frac { 7 }{ 2 } $$

D
$$7$$

Solution