Application of Integrals Test

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Application of Integrals Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

The area bounded by the circles $$x^{2} + y^{2} = r^{2}, r = 1, 2$$ and the rays given by $$2x^{2} - 3xy - 2y^{2} = 0, y > 0$$ is

A
$$\dfrac {\pi}{4} sq. units$$

B
$$\dfrac {\pi}{2} sq. units$$

C
$$\dfrac {3\pi}{4} sq. units$$

D
$$\pi\ sq. units$$

Solution
Question 2
1 / -0

The area bounding by $$y = 2 - |2 - x|$$ and y = $$\frac { 3 }{ |x| } $$ is :

A
$$\frac { 4+3\ell n3 }{ 2 } $$

B
$$\frac { 4-3\ell n3 }{ 2 } $$

C
$$\frac { 3 }{ 2 } +\ell n3$$

D
$$\frac { 1 }{ 2 } +\ell n3$$

Solution
Question 3
1 / -0

The area enclosed between the curves $$y=ax^2$$ and $$x=ay^2(a>0)$$ is $$1$$ sq.unit, then the value of $$a$$ is

A
$$1/\sqrt{3}$$

B
$$1/2$$

C
$$1$$

D
$$1/3$$

Solution
Question 4
1 / -0

The area inside the parabola $$5x^2-y=0$$ but outside the parabola $$2x^2-y+9=0$$, is

A
$$12\sqrt 3$$

B
$$6\sqrt 3$$

C
$$8\sqrt 3$$

D
$$4\sqrt 3$$

Solution

From given, we have,

$$5x^2-y=0$$

$$2x^2-y+9=0$$

we have boundaries,

$$5x^2=2x^2+9$$

$$3x^2=9$$

$$x=\pm \sqrt 3$$

$$A=\int _{-\sqrt{3}}^{\sqrt{3}}5x^2-\left(2x^2+9\:\right)$$

$$\int _{-\sqrt{3}}^{\sqrt{3}}5x^2-\left(2x^2+9\:\right)$$

$$=\int _{-\sqrt{3}}^{\sqrt{3}}3x^2dx-\int _{-\sqrt{3}}^{\sqrt{3}}9dx$$

$$=6\sqrt{3}-18\sqrt{3}$$

$$=-12\sqrt{3}$$

$$\therefore Area =|-12\sqrt{3}|=12\sqrt{3}$$
Question 5
1 / -0

Area bounded by the curve $$y= sin^{-1}x, y-axis$$ and $$y = cos^{-1}x$$ is equal to

A
$$(2+\sqrt{2})$$

B
$$(2-\sqrt{2})$$

C
$$(1+\sqrt{2})$$

D
$$(\sqrt{2}-1)$$

Solution

Given,

$$y=\sin ^{-1}x,y=\cos ^{-1}x,y-axis$$

$$\Rightarrow \sin y=x, \cos y=x$$

$$\Rightarrow \sin y=\cos y \Rightarrow \tan y=1$$

$$\therefore y=\tan ^{-1}1=\dfrac{\pi }{4}$$

$$Area=\int _0^{\frac{\pi }{4}}\cos \left(y\right)-\sin \left(y\right)dy$$

$$=\int _0^{\frac{\pi }{4}}\cos \left(y\right)dy-\int _0^{\frac{\pi }{4}}\sin \left(y\right)dy$$

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

$$=\sqrt{2}-1$$
Question 6
1 / -0

The area is bounded by $$x+x_1, y=y_1$$ and $$y=-(x+1)^2$$. Where $$x_1, y_1$$ are the values of $$x, y$$ satisfying the equation $$sin^{-1} x +sin^{-1} y = -\pi$$ will be (nearer to origin)

A
$$1/3$$

B
$$3/2$$

C
$$1$$

D
$$2/3$$

Solution

( Wrong question, $$x=x$$, instead of $$x+x_{1}$$ )

Given that:
$$x =x_{1} \\$$
$$y =y_{1} \\$$
$$y =-(x+1)^{2} \\$$
$$\text { & } x_{1}, y_{1} \text { satisfy } \sin ^{-1} x+\sin ^{-1} y=z$$

We know:
$$-\frac{z}{2} \leq \sin ^{-1} x \leq \frac{z}{2}$$
$$\frac{-z}{2} \leq \sin ^{-1} y \leq \frac{z}{2}$$

$$\therefore \quad-z \leq \sin ^{-1} x+\sin ^{-1} y \leq z \\$$
$$\text { So, for } \\$$
$$\qquad \sin ^{-1} x+\sin ^{-1} y=-z \\$$
$$\Rightarrow \sin ^{-1} x=-\frac{\pi}{2} \text { & } \sin ^{-1} y=-z / 2 \\$$
$$x =-1 \text { and } y=-1 \\$$

$$\text { area required } =1-|\int_{1}^{0}-(x+1)^{2} d x \mid \\$$
$$=1-|\int_{1}^{0}\left(x^{2}+1+2 x\right) d x|$$

$$=1- |\frac{-\left.\left(x^{3}+x+x^{2}\right)\right|_{-1} ^{0}}{3}\\$$
$$= 1-|\left(\frac{-1}{3}-1+1\right)| \\$$
$$=1-\frac{1}{3} \\$$
$$=2 / 3 sq \text { . Units }$$
Question 7
1 / -0

The area of the region bounded by x = 0, y = 0, x = 2, y = 2, $$y\le { e }^{ x }$$ and $$y\ge \ell n$$ x, is

A
6 - 4 $$\ell n$$ 2

B
4 $$\ell n$$ 2 - 2

C
2 $$\ell n$$ 2 - 4

D
6 - 2 $$\ell n$$ 2

Solution
Question 8
1 / -0

The area of the figure bounded by the curves $$y=\ln nx $$ & $${(\ln nx)}^{2}$$ is

A
$$e+1$$

B
$$e-1$$

C
$$3-e$$

D
$$1$$

Solution
Question 9
1 / -0

The area of the triangle formed by the lines joining the vertex of the parabola $$x^2 = 12y$$ to the ends of its latus rectum is-

A
$$16$$ sq. units

B
$$12$$ sq. units

C
$$18$$ sq. units

D
$$24$$ sq. units

Solution

Given,

$$x^2=12y$$

$$\Rightarrow 4a=12$$

$$\therefore a=3$$

at $$y=3,x^2=12(3)=36$$

$$\therefore x=\pm 6$$

Therefore the coordinates are,

$$A(-6,3),O(0,0),B(6,3)$$

Area of $$\triangle OAB=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

$$=\dfrac{1}{2}|0(3-3)+(-6)(3-0)+6(0-3)|$$

$$=\dfrac{1}{2}|-36|$$

$$=18sq.units$$
Question 10
1 / -0

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

A
1

B
2

C
3

D
4

Solution

First we form a table for different values of $$x$$ and $$y$$ for both equations as shown .
Here length of $$AC=2$$ units and distance(height) between vertices $$D$$ and line $$AC=1$$units and between vertices $$B$$ and line $$AC=1$$unit
$$\because$$Area of triangle$$=\dfrac{1}{2}\times b\times h$$
$$\therefore\,$$Area of $$ABCD=2\times\,Area\,of\,\triangle{ABC}$$
$$=2\times\dfrac{1}{2}\times 2\times 1=2$$sq.units.