Application of Integrals Test

Result

Application of Integrals Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

Area bounded between the curves $$y=\sqrt{4-x^2}$$ and $$y^2=3|x|$$ is/are?

A
$$\dfrac{\pi -1}{\sqrt{3}}$$

B
$$\dfrac{2\pi -1}{3\sqrt{3}}$$

C
$$\dfrac{2\pi -\sqrt{3}}{3}$$

D
$$\dfrac{2\pi -\sqrt{3}}{3\sqrt{3}}$$

Solution
Question 2
1 / -0

Let $$A_{n}$$be the area bounded by the curve $$y=(\tan x)^{n}$$ and the lines $$x=0,y=0$$ and $$4x-\pi=0$$, where

A
$$A_{n+2}+A_{n}=\dfrac{1}{(n+1)}$$

B
$$A_{1}=\dfrac{1}{2}\ln { 2 } $$

C
$$A_{n}

D
$$A_{2}=1=\dfrac{\pi}{4}$$

Solution
Question 3
1 / -0

The area of a region bounded by $$X$$ -axis and the curves defined by $$y = \tan x$$ $$0 \leq x \leq \frac { \pi } { 4 }$$ and $$y = \cot x , \frac { \pi } { 4 } \leq x \leq \frac { \pi } { 2 }$$ is

A
$$\log 3$$ sq. units

B
$$\log 5$$ sq. units

C
$$\log 1$$ sq. unit

D
$$\log 2$$ sq. unit

Solution
Question 4
1 / -0

The area bounded by the curve $$ y = \dfrac { \sin { x } }{ { x } } , x-$$ axis and the ordinates $$ x=0,x=\dfrac { \pi }{ { 4 } }$$ is:

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

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

C
$$>\dfrac { \pi }{ 4 }$$

D
None of these
Question 5
1 / -0

The parabolas $$y^2=4x, x^2=4y$$ divide the square region bounded by the lines $$x=4$$, $$y=4$$ and the coordinate axes. If $$S_1, S_2, S_3$$ are respectively the area of these parts numbered from top to bottom then $$S_1 : S_2 : S_3$$ is?

A
$$2:1:1$$

B
$$1:1:1$$

C
$$1:2:1$$

D
$$1:2:3$$

Solution
Question 6
1 / -0

Area of the figure bounded by $$x$$ -axis, $$y = \sin ^ { - 1 } x , y = \cos ^ { - 1 } x$$ and the first point intersection from the origin is

A
$$2$$ $$\sqrt { 2 }$$

B
$$2 \sqrt { 2 } + 1$$

C
$$\sqrt { 2 } - 1$$

D
$$\sqrt { 2 } + 1$$

Solution
Question 7
1 / -0

The area (in sq.units) of the region $$\left\{ ( x , y ) :{ y} ^ { 2 } \ge 2 x\right.$$ and $${x} ^ { 2 } +{ y} ^ { 2 } \le 4 x , x \ge 0 , y \ge 0$$ is

A
$$\pi - \dfrac { 8 } { 3 }$$

B
$$\pi - \dfrac { 4 \sqrt { 2 } } { 3 }$$

C
$$\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }$$

D
$$\pi - \dfrac { 4 } { 3 }$$

Solution

Consider $$y^2=2x$$ and $$x^2+y^2=4x$$, on solving,

we get $$x^2+2x=4x$$

$$x^2=2x\Rightarrow x=0, 2$$

The integral lies between $$0$$ and $$2$$

$$y^2=2x$$ $$y^2+x^2=4x$$

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

Area $$=\displaystyle\int^2_0\sqrt{4x-x^2}-\sqrt{2x}dx$$

$$=\displaystyle\int^2_0\sqrt{2^2-(2-x)^2}-\displaystyle\int^2_0\sqrt{2x}dx$$

$$=\left[\dfrac{x-2}{2}\sqrt{4x-x^2}-\dfrac{4}{2}\sin^{-1}\left(\dfrac{x-2}{2}\right)\right]^2_0-\left[\dfrac{\sqrt{2}}{3/2}x^{3/2}\right]^2_0$$

$$=\left|2\sin^{-1}(-1)\right|-\dfrac{2\sqrt{2}}{3}\cdot 2\sqrt{2}$$

$$=\pi -\dfrac{8}{3}$$.
Question 8
1 / -0

Area bounded by $$y=-x^{2}+6x-5,y=-x^{2}+4x-3$$ and $$y=3x-15$$ for $$x > 1$$, is (in $$sq.\ units$$)

A
$$73$$

B
$$\dfrac {13}{6}$$

C
$$\dfrac {73}{6}$$

D
$$13$$

Solution
Question 9
1 / -0

Find the area of shaded portion

A
$$90\ cm^{2}$$

B
$$80\ cm^{2}$$

C
$$110\ cm^{2}$$

D
$$70\ cm^{2}$$

Solution

Area of rectangle $$=18\times 10$$
$$=180sq\;cm$$
$$\Rightarrow$$ Triangle area $$DEF=\dfrac{1}{2}\times 10\times 6$$
$$=30\;sq\;cm$$
$$\Rightarrow$$ Triangle area $$BCE=\dfrac{1}{2}\times 10\times 8$$
$$=40\;sq\;cm$$
$$\Rightarrow$$ Area shaded region $$=180-30-40$$
$$=110\;sq\;cm$$
Hence, the answer is $$110\;sq\;cm.$$
Question 10
1 / -0

The area bounded by the curves $$\sqrt{x}+\sqrt{y}=1$$ and $${x}+{y}=1$$ is ?

A
$$\dfrac{1}{3}$$

B
$$\dfrac{1}{6}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{5}{6}$$

E
$$\dfrac{1}{4}$$

Solution

$$\textbf{Step -1: Finding the point of intersection.}$$
$$\text{Find the points of intersection of the curves}$$
$$\implies \sqrt x=1-\sqrt y$$
$$\implies x=(1-\sqrt y)^2=1+y-2\sqrt y$$
$$\implies 1+y-2\sqrt y +y=1$$
$$\implies 2y=2\sqrt y$$
$$\implies y=0,1$$
$$\implies \text{Point of intersections are }(1,0)\text{ and }(0,1)$$
$$\textbf{Step -2: Calculating the area .}$$
$$\sqrt y=1-\sqrt x$$
$$\implies y=1+x-2\sqrt x$$
$$\implies f_1(x)=x-2\sqrt x+1$$
$$\implies f_2(x)=1-x$$
$$\implies \text{Area}=\int_0^1(f_1(x)-f_2(x))dx$$
$$\implies \int_0^1(2x-2\sqrt x)dx$$
$$\implies x^2-2\dfrac{x^{\frac32}}{\frac32}|_0^1$$
$$\implies |1-\dfrac43|=|-\dfrac13|=\dfrac13$$
$$\textbf{Hence, The required area is }\mathbf{\dfrac13}\textbf{ and correct answer is option A .}$$