Application of Integrals Test

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Application of Integrals Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Area of the region bounded by the curve $$y^{2}=4x, y-$$ axis and the line $$y=3$$ is

A
$$2$$

B
$$\dfrac{9}{4}$$

C
$$\dfrac{9}{3}$$

D
$$\dfrac{9}{2}$$

Solution

(1) Curve $$y^{2}=4x$$ is right handed parabola
(2) line $$y=3\Rightarrow$$ points in line area $$(0,3), (1,3), (2,3), (3,3), (-1,3)(-2,3)$$
(3) Pt where line meets $$y^{2}=4x$$ is solved by. solving the equation
as we know $$y=3$$ so $$x=\dfrac{3^{2}}{4}\Rightarrow \dfrac{9}{4}$$
so pt is $$\left(\dfrac{9}{4},3\right)$$
(If we solve along $$'y'$$ axis)
[Here to solve along $$'x'$$ axis, it is tough as to get equation of line in terms of $$x$$ is tough.
(4) So area is area under parabola from $$y=0$$ to $$y=3$$.
= Area of shade $$=\displaystyle\int_{0}^{3}|x|dy=\int_{0}^{3}\dfrac{y^{2}}{4}dy$$
$$=\left.\dfrac{1}{4}\times \displaystyle\int_{0}^{3}y^{2}dx=\dfrac{1}{4}\times \dfrac{y^{3}}{3}\right]_{0}^{3}=\dfrac{1}{4}\times \left(\dfrac{3^3}{3}-\dfrac{0^3}{3}\right)$$
$$=\left(\dfrac{1}{4}\times \dfrac{27}{3}\right)-0=\dfrac{9}{4}sq.units$$ So Answer is $$(b)\dfrac{9}{4}$$.
Option $$(B)$$
Question 2
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I: The area bounded by the curves $$y=\sin x,\ y=\cos x$$ and $$\mathrm{Y}$$-axis is $$\sqrt{2}-1$$ sq. units.
II: The area bounded by $$y=\cos x,\ y=x+1,\ y=0$$ is 3/2 sq. units.

Which of the above statement is correct?

A
Only I

B
Only II

C
Both I and II

D
Neither I nor II.

Solution

$$I)$$
The required area will be
$$\int_{0} ^{\frac{\pi}{4}} cos(x)-sin(x) .dx$$

$$=[sin(x)+cos(x)]_{0} ^{\frac{\pi}{4}}$$

$$=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-1$$

$$=\sqrt{2}-1$$ sq units.

$$II)$$
Required area will be
$$\int_{-1} ^{0} x+1 .dx+\int_{0} ^{\frac{\pi}{2}} cos(x).dx$$

$$=[\dfrac{x^{2}}{2}+x]_{-1} ^{0} +[sin(x)]_{0} ^{\frac{\pi}{2}}$$

$$=-(\dfrac{1}{2}-1)+1$$

$$=\dfrac{3}{2}$$ sq units.
Question 3
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Arrangement of the following areas between the curves is descending order:
$$\mathrm{A}:y^{2}=4x,\ x^{2}=4y$$
$$ \mathrm{B}.\ y=x,\ y=x^{3}$$
$$\mathrm{C}.\ y^{2}=8x,y=2x$$
$$ \mathrm{D}.\ y=\sqrt{x},\ y=x^{2}$$

A
A,B,C,D

B
A,C,B,D

C
D,B,C,A

D
D,C,B,A

Solution

A $$\frac{16}{3}a^2=16/3\:sq\:units.$$ $$(a=1)$$
B    $$1/2\:sq\:units.$$
C $$\overset {2 }{ \underset { 0 }{ \int } } (2x-\sqrt{8x})dx$$ . $$4-\frac{2}{3}\sqrt{8}   2\sqrt{2}$$
$$\left | 4-\frac{4}{3} \times 4\right |=4/3\:sq\:units.$$
D $$\frac{16}{3}(1/16)=1/3\:sq\:units.$$   $$(a=1/4)$$
   So, descending order is A,C,B,D

Question 4

1 / -0

Match the following:

List-I	List-II
1. Area of the region bounded by $$y=\|5\sin x\|$$ from $$\mathrm{x}=0$$ to $$ x=4\pi$$ and x-axis	a. 3/2
2. The area bounded by $$\mathrm{y}=$$ cosx in $$[0,2\pi]$$ and the $$\mathrm{X}$$-axis	b. $$\sqrt{2}-1$$
3. The area bounded by $$y=sinx, y=cosx$$ and the y-axis	c. 4
4. The area bounded by $$y = cos x , y = x +1, y=0$$	d. 40

The correct match is

a,b,c,d

d,b,a,c

d,c,b,a

d,a,b,c

Solution

1. $$y=|5\sin x|$$
   $$\overset {4\pi }{ \underset { 0 }{ \int } } |5\sin x|dx=5\times4\overset {\pi }{ \underset { 0 }{ \int } } \sin x=40\:sq\:units.$$
2. $$y=\cos x$$ $$\leftarrow [0,2\pi]$$
     $$4\overset {\pi/2 }{ \underset { 0 }{ \int } } cos x\:dx=4\:sq\:units.$$
3. $$\overset {\pi /4}{ \underset { 0 }{ \int } } (\cos x-\sin x)dx=(\sqrt{2}-1)sq\:units.$$
4. $$1+1/2=3/2sq\:units.$$
      d,c,b,a

Question 5
1 / -0

The area bounded by the $$y=\left| \sin { x } \right| $$, x-axis and the lines $$\left| x \right| =\pi $$ is

A
$$2$$ square units

B
$$1$$ square units

C
$$4$$ square units

D
None of these

Solution

Required Area $$=A_1+A_2=2A_2$$

$$\displaystyle =2\left[\int^{\pi}_0 y\,dx=\int^{\pi}_0\,|\sin x|dx\right]$$

$$\displaystyle \Rightarrow 2\int^{\pi}_0\sin x\,dx=2[-\cos x]^{\pi}_0$$

$$\Rightarrow 2[-\cos\pi+\cos 0]=2[-(-1)+1]$$

$$\Rightarrow 2\times (1+1)=2\times 2=4$$ Sq. unit

Question 6

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Match the following:

List-I	List-II
1. Area of region bounded by $$y=2x-x^{2}$$ and $$x-$$axis	a. $$\dfrac13$$
2. Area of the region $$\{(x, y):x^{2}\leq y\leq\|x\|\}$$	b. $$\dfrac12$$
3. Area bounded by $$y=x$$ and $$y=x^{3}$$	c. $$\dfrac23$$
4. Area bounded by $$y=x\|x\|$$, $${x}$$-axis and $${x}=-1,\ {x}=1$$	d. $$\dfrac43$$

The correct match for $$1\ 2\ 3\ 4$$ is

$$1-b,2- c.3- d,4- a$$

$$1-c,2- d,3- a,4- b$$

$$1-d,2- a,3- b,4- c$$

$$1-a,2- b,3- c,4- d$$

Solution

$$y=2x-x^2$$

The values of $$x$$ where $$y$$ is $$0$$ are

$$x = 0,2$$

Now, the area of the region bounded by $$y=2x-x^2$$ and $$x-$$axis is

$$\displaystyle \int_0^2(2x-x^2)\ dx = \left[x^2 - \dfrac{x^3}{3}\right]_0^2$$

$$=4-\dfrac83 = \dfrac43\ \ \rightarrow d$$

$$x^2≤y≤|x|$$

We have to find the area of the region bounded by $$y=x^2$$ and $$y = |x|$$ for $$x\in[-1,1]$$

$$\displaystyle\int_{-1}^1-(x^2-|x|)\ dx = -2\int_0^1(x^2-x)\ dx$$

$$ = -2\left[\dfrac{x^3}3 - \dfrac{x^2}2\right]_0^1 = \dfrac13\ \ \rightarrow a$$

We have to find the area of the region bounded by $$y=x$$ and $$y = x^3$$ for $$x\in[-1,1]$$

$$\displaystyle\int_{-1}^1(x-x^3)\ dx = 2\int_0^1(x-x^3)\ dx$$

$$ =2\left[\dfrac{x^2}2 - \dfrac{x^4}4\right]_0^1 = \dfrac14\ \ \rightarrow b$$

We have to find the area of the region bounded by $$y=x|x|$$ and $$y = 0$$ for $$x\in[-1,1]$$

$$\displaystyle\int_{-1}^1(x|x|)\ dx = 2\int_0^1(x^2)\ dx$$

$$ =2\left[\dfrac{x^3}3\right]_0^1 = \dfrac23\ \ \rightarrow c$$

Hence, option C.

Question 7
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The area of the region bounded by the curves $$y=ex\log x$$ and $$y=\displaystyle \frac{\log x}{ex}$$ is:

A
$$\displaystyle \frac{e^{2}-5}{4e}$$

B
$$e-\displaystyle \frac{5}{4}$$

C
$$\displaystyle \frac{e}{4}-5$$

D
$$\displaystyle \frac{e}{4}-\frac{1}{4e}$$

Solution

$$\left | \overset {1 }{ \underset { 1/e}{ \int } } ex\log x-\dfrac{\log x}{ex}dx\right |$$

$$\left |e\overset {1 }{ \underset { 1/e}{ \int } } x\log x-\dfrac{1}{e}\overset {1 }{ \underset { 1/e}{ \int } } \frac{\log x}{x}dx\right |$$

$$\left |e \left ( log x \frac{x^2}{2}-\frac{1}{4}x^2\right) \left |_{1/e}^{1}-\dfrac{1}{e}\dfrac{(\log x)^2}{2} \right |_{1/e}^{1}\right|$$

$$\left |e \left (\dfrac{-1}{4}-\left (\dfrac{(-1)e^{-2}}{2} -\dfrac{1}{4}e^{-2} \right ) -1/e\left (0-\left ( \dfrac{1}{2} \right ) \right ) \right )\right|$$

$$\dfrac{-e}{4}+\left ( \dfrac{3}{4}e^{-1} \right )+\dfrac{2}{4e}$$

$$\left |\dfrac{-e}{4} +\dfrac{5}{4e} \right |=\dfrac{e^2-5}{4e}sq\:units.$$
Question 8
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Area of the figure bounded by the lines $$ y=\sqrt{x},x\in [0,1],\ y=x^{2},\ x\in[1,2]$$ and $$y=-x^{2}+2x+4,x\in0,2]$$ is:

A
$$\dfrac{10}{7}$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{4}{3}$$

D
$$\dfrac{19}{3}$$

Solution

As shown in the figure, the area bounded
$$ = \int _0 ^2 (-x^{2}+2x+4) dx - \int _0 ^1 \sqrt{x} dx - \int _1 ^2 x^2 dx $$
$$= \cfrac{19}{3}$$
Question 9
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The area bounded by the parabolas $$\mathrm{y}^{2}=5\mathrm{x}+6$$ and $$\mathrm{x}^{2}=\mathrm{y}$$

A
$$\displaystyle \frac{19}{5}$$

B
$$\displaystyle \frac{21}{5}$$

C
$$\displaystyle \frac{23}{5}$$

D
$$\displaystyle \frac{27}{5}$$

Solution

$$y^{2}=5x+6$$ and $$x^{2}=y$$
Then
$$x^{4}=5x+6$$
Or
$$x^{4}-5x-6=0$$
$$x=-1$$ is an obvious root.
Then
$$x^{4}-5x-6=0$$
$$(x+1)(x-2)(x^{2}+x+3)=0$$
Now $$x^{2}+x+3=0$$ has no real roots since $$D<0$$.
Hence we have to find the area between the curves for x=-1 to x=2.
Therefore
$$Area$$
$$=|\int_{-1} ^{2}x^{2}-\sqrt{5x+6}.dx|$$

$$=|[\dfrac{x^{3}}{3}-\dfrac{2}{15}(5x+6)^{\frac{3}{2}}]_{-1} ^{2}|$$

$$=|\dfrac{8}{3}-\dfrac{128}{15}-[\dfrac{-1}{3}-\dfrac{2}{15}]|$$

$$=|\dfrac{9}{3}-\dfrac{126}{15}|$$

$$=\dfrac{126-45}{15}$$

$$=\dfrac{81}{15}$$

$$=\dfrac{27}{5}$$.
Question 10
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The ratio of the areas into which the circle $$x^{2}+y^{2}=64$$ is divided by the parabola $$y^{2}=12x$$ is:

A
$$\displaystyle \frac{4\pi-\sqrt{3}}{8\pi+\sqrt{3}}$$

B
$$\displaystyle \frac{4\pi+\sqrt{3}}{8\pi-\sqrt{3}}$$

C
$$\displaystyle \frac{4\pi-\sqrt{3}}{8\pi-\sqrt{3}}$$

D
$$\displaystyle \frac{4\pi+\sqrt{3}}{8\pi+\sqrt{3}}$$

Solution

$$x^2+y^2 = 64$$ ... (i)
$$y^2 = 12x$$ ... (ii)
Substituting (ii) in (i), we get
$$x^2 + 12x -64 = 0$$
$$x^2 + 16x -4x -64 = 0$$
$$(x-4)(x+16) = 0$$
$$x =4$$ is the feasible solution.
$$ y^2 = 12x$$
$$ y = \pm \sqrt{48} = \pm 4\sqrt{3}$$
Angle made at the center by the two points $$= 2 tan^{-1}(4\sqrt{3}/4) = \cfrac{2\pi}{3}$$

Area of Region inside Parabola $$=$$ Area of sector $$+$$ area b/w parabola and Straight line from origin (as shown in figure)
$$= _0 ^4\int (\sqrt{12x} - \sqrt{3}x)dx $$
$$= 8 \sqrt{3} $$
So Total area inside parabola $$= \cfrac{2 \pi/3}{2 \pi} \pi 64 + \cfrac{16 \sqrt{3}}{3}$$
$$= \cfrac{16}{3}(4 \pi + \sqrt{3})$$
Area of remaining part of circle $$= 64 \pi - \cfrac{16}{3}(4 \pi + \sqrt{3}) $$
Ratio $$= \displaystyle \dfrac{4\pi+\sqrt{3}}{8\pi-\sqrt{3}}$$

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Application of Integrals Test - 51

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Application of Integrals Test - 51

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SHARING IS CARING

Question 1

$$2$$

$$\dfrac{9}{4}$$

$$\dfrac{9}{3}$$

$$\dfrac{9}{2}$$

Solution

Question 2

Only I

Only II

Both I and II

Neither I nor II.

Solution

Question 3

A,B,C,D

A,C,B,D

D,B,C,A

D,C,B,A

Solution

Question 4

a,b,c,d

d,b,a,c

d,c,b,a

d,a,b,c

Solution

Question 5

$$2$$ square units

$$1$$ square units

$$4$$ square units

None of these

Solution

Question 6

$$1-b,2- c.3- d,4- a$$

$$1-c,2- d,3- a,4- b$$

$$1-d,2- a,3- b,4- c$$

$$1-a,2- b,3- c,4- d$$

Solution

Question 7

$$\displaystyle \frac{e^{2}-5}{4e}$$

$$e-\displaystyle \frac{5}{4}$$

$$\displaystyle \frac{e}{4}-5$$

$$\displaystyle \frac{e}{4}-\frac{1}{4e}$$

Solution

Question 8

$$\dfrac{10}{7}$$

$$\dfrac{3}{5}$$

$$\dfrac{4}{3}$$

$$\dfrac{19}{3}$$

Solution

Question 9

$$\displaystyle \frac{19}{5}$$

$$\displaystyle \frac{21}{5}$$

$$\displaystyle \frac{23}{5}$$

$$\displaystyle \frac{27}{5}$$

Solution

Question 10

$$\displaystyle \frac{4\pi-\sqrt{3}}{8\pi+\sqrt{3}}$$

$$\displaystyle \frac{4\pi+\sqrt{3}}{8\pi-\sqrt{3}}$$

$$\displaystyle \frac{4\pi-\sqrt{3}}{8\pi-\sqrt{3}}$$

$$\displaystyle \frac{4\pi+\sqrt{3}}{8\pi+\sqrt{3}}$$

Solution

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