Application of Integrals Test

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Application of Integrals Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the curves $$y=|x-2|$$ and $$y=4-|x| is- $$

A
2

B
4

C
5

D
6

Solution

From the attached fig., $$ABCD$$ is the region bounded by the curves $$y = \left| x - 2 \right|$$ and $$y = 4 - \left| x \right|$$.
Since $$ABCD$$ is a rectangle, therefore
$$ar{\left( ABCD \right)} =$$ area of rectangle $$= AB \times BC$$
Now,
$$AB = \sqrt{{\left( -1 - 0 \right)}^{2} + {\left( 4 - 3 \right)}^{2}} = \sqrt{2}$$
$$BC = \sqrt{{\left( 3 - 0 \right)}^{2} + {\left( 1 - 4 \right)}^{2}} = 3 \sqrt{2}$$
$$\therefore ar{\left( ABCD \right)} = \sqrt{2} \times 3 \sqrt{2} = 6 \text{ sq. units}$$
Hence the area of region bounded by the curves $$y = \left| x - 2 \right|$$ and $$y = 4 - \left| x \right|$$ is $$6$$ units.
Question 2
1 / -0

The area of the region bounded by $$y=(x-4)^2, y=16-x^2$$ and the x axis,is

A
$$16$$

B
$$32$$

C
$$\dfrac{64}{3}$$

D
$$64$$

Solution

$$y_1=x^2+16-8x$$
$$y_2=16-x^2$$
$$16-x^2=x^2=16-8x$$
$$\Rightarrow 2x^2-8x=0$$
$$x^2-4=0$$
$$x=0,4$$
$$\overset { 4 }{ \underset { 0 }{ \int } } (y_1-y_2)dx$$
$$\Rightarrow { }{ \underset { 0 }{ \int } } (2x^2-8x)dx$$
$$\Rightarrow { \left[ \dfrac { { 2x }^{ 3 } }{ 3 } -{ 4x }^{ 2 } \right] }_{ 0 }^{ 4 }$$
$$\left| \dfrac { 128 }{ 3 } -64 \right| =\dfrac { 64 }{ 3 }\\$$
Question 3
1 / -0

Find area bounded by curves $$\left \{ (x,y):y\geq x^{2}andy=\mid x\mid \right \}$$

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

B
$$\dfrac {1}{2}$$

C
$$\dfrac {1}{3}$$

D
$$\dfrac {1}{9}$$

Solution

$$y=\mid x\mid$$$$=\left\{ x\quad ;\quad x\ge 0\\ -x\quad ;x<0 \right\} $$
$$P$$ and $$Q$$ are
$$x^2=x\\x^2-x=0\\x(x-1)=0\\$$
$$x=0,1$$
$$Q=1$$
similarly
$$p=-1$$
$$A=\int _{ 0 }^{ 1 }{ x } -{ x }^{ 2 }dx\\ A={ [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 3 } }{ 3 } ] }_{ 0 }^{ 1 }\\ A=\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 3 } \\ A=\cfrac { 1 }{ 3 } $$
Question 4
1 / -0

The area enclosed between the curves $$y^2 = x$$ and $$y=|x|$$ is

A
$$\dfrac {1}{2}$$

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

C
$$\dfrac {1}{8}$$

D
$$\dfrac {1}{16}$$
Question 5
1 / -0

The area enclosed between the curve $$y = |x|^{3}$$ and $$x = y^{3}$$ is

A
$$\dfrac {1}{2}$$

B
$$\dfrac {1}{4}$$

C
$$\dfrac {1}{8}$$

D
$$\dfrac {1}{16}$$

Solution
Question 6
1 / -0

Find the area bounded by the curve $$y = sin\;x$$ with x-axis between $$x = 0$$ to x = 2$$\pi$$.

A
4 sq. unit

B
8 s q . unit

C
4$$\pi 4$$ sq. unit

D
8$$\pi 4$$ sq. unit

Solution

Consider the problem

We have to draw the graph of $$y=sin\;x$$

Therefore,

Required area $$=Area\; OABO+Area \;BCDB$$

$$\begin{array}{l} =\int _{ 0 }^{ \pi }{ \sin xdx+\left| { \int _{ \pi }^{ 2\pi }{ \sin xdx } } \right| } \\ ={ \left[ { -\cos x } \right] _{ 0 } }^{ \pi }+\left| { { { \left[ { -\cos x } \right] }_{ \pi } }^{ 2\pi } } \right| \\ =\left[ { -\cos \pi +\cos 0 } \right] +\left| { -\cos 2\pi +\cos \pi } \right| \\ =1+1+\left| { \left( { -1-1 } \right) } \right| \\ =2+\left| { -2 } \right| \\ =2+2 \\ =4units \end{array}$$
Question 7
1 / -0

From a piece of cardboard, in the shape of a trapezium ABCD, and AB$$||$$CD and $$\angle$$BCD$$=90^o$$, quarter circle is removed. Given AB$$=$$BC$$=3.5$$cm and DE$$=2$$cm. Calculate the area of the remaining piece of the cardboard.(Take $$\pi$$ to be $$\dfrac{22}{7}$$)

A
$$9.625cm^2$$

B
$$6.125cm^2$$

C
$$2.625cm^2$$

D
None of these

Solution

REF.Image
EC=BC=3.5 cm
DC = DE + EC
= 5.5 cm
Area of rem. piece = Area of trapezoid - Area of quarter circle
Question 8
1 / -0

The area between the curves y=$$x^{2}$$ and $$y=\frac{2}{1+x^{2}}$$ is-

A
$$\pi -\frac{1}{3}$$

B
$$\pi -2$$

C
$$\pi -\frac{2}{3}$$

D
$$\pi +\frac{2}{3}$$

Solution

we have,

$$y={{x}^{2}}\,\,......\,\,\left( 1 \right)$$

$$y=\dfrac{2}{{{x}^{2}}+1}\,\,.......\,\,\left( 2 \right)$$

Taking limit then,

By equation (1) and (2) to, and get,

$${{x}^{2}}=\dfrac{2}{{{x}^{2}}+1}$$

Put x=1 and we get,

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

$$ 1=\dfrac{2}{2} $$

$$ 1=1 $$

Now, taking’

$$x=-1$$ and we get,

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

$$ 1=1 $$

Then, the limit of this function of \[1\,\,to\,\,-1\].

The required area\[=\int_{-1}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}\]

$$ =\int_{-1}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}=2\int_{0}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx} $$

$$ =2{{\left[ 2{{\tan }^{-1}}x-\dfrac{{{x}^{3}}}{3} \right]}_{0}}^{1} $$

$$ =2\left[ 2{{\tan }^{-1}}\left( 1 \right)-2{{\tan }^{-1}}0-\dfrac{1-0}{3} \right] $$

$$ =2\left[ 2\dfrac{\pi }{4}-2\times 0-\dfrac{1}{3} \right] $$

$$ =\pi -\dfrac{2}{3} $$
Hence, this is the answer.
Question 9
1 / -0

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

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

B
$$\pi - \frac{8}{3}$$

C
$$\pi - \frac{{4\sqrt 2 }}{3}$$

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

Solution
Question 10
1 / -0

The area of the region $$[(x, y) : x^{2} + y^{2} \leq 1 \leq x + y|$$ is

A
$$\dfrac {\pi}{5}$$

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

C
$$\dfrac {\pi^{2}}{3}$$

D
$$\dfrac {\pi}{4} - \dfrac {1}{2}$$

Solution