Application of Integrals Test

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Application of Integrals Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the curve $$y = x^{2}$$ and $$y = 4x - x^{2}$$ is

A
$$\dfrac {16}{3}sq. units$$

B
$$\dfrac {8}{3}sq. units$$

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

D
$$\dfrac {2}{3}sq. units$$

Solution

Here area,
$$\int _{ 0 }^{ 2 }{ (4x-x^{ 2 }-x^{ 2 })dx }$$

$$ =\int _{ 0 }^{ 2 }{ (4x-2x^{ 2 })dx }$$

$$ ={ \left( 2x^{ 2 }-\dfrac { 2 }{ 3 } x^{ 3 } \right) }_{ 0 }^{ 2 }$$

$$ =\dfrac { 8 }{ 3 } $$
Question 2
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The area of the figure bounded by the parabolas $$x = -2y^{2}$$ and $$x = 1 - 3y^{2}$$ is

A
$$\dfrac {4}{3}$$ square units

B
$$\dfrac {2}{3}$$ square units

C
$$\dfrac {3}{7}$$ square units

D
$$\dfrac {6}{7}$$ square units

Solution

Given curves are $$x=-2y^{2}$$ and $$x=1-3y^{2}$$
$$-2y^{2}=1-3y^{2}$$
$$\implies y^{2}=1 \implies y=\pm 1$$
$$\therefore$$ Curves intersect at $$(-2, \pm 1)$$
Area $$ =\int_{-1}^{1} (1-3y^{2}-(-2y^{2}))\ dy$$

$$ =\int_{-1}^{1} (1-y^{2})\ dy$$

$$ =\int_{-1}^{0} (1-y^{2})\ dy +\int_{0}^{1} (1-y^{2})\ dy$$
Substituting $$y=-y$$ in first integral we get,
Area $$ =\int_{0}^{1} (1-y^{2})\ dy +\int_{0}^{1} (1-y^{2})\ dy$$

$$= 2\int_{0}^{1} (1 - y^{2}) dy$$

$$=2\left|y-\dfrac{y^{3}}{3}\right|_{0}^{1}$$

$$= 2\left (1 - \dfrac {1}{3}\right ) = \dfrac {4}{3}$$
Hence, area bounded by the given curves is $$\dfrac{4}{3}$$ square units.
Question 3
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The area formed by triangular shaped region bounded by the curves $$y=\sin { x } ,y=\cos { x } $$ and $$x=0$$ is

A
$$\left( \sqrt { 2 } -1 \right) $$ sq unit

B
$$1$$ sq unit

C
$$\sqrt { 2 } $$ sq units

D
$$\left( \sqrt { 2 } +1 \right) $$ sq units

Solution

$$\left( \because y=\sin { x } \text{ and } y=\cos { x } \text{ meet at } x={ \pi }/{ 4 } \right) $$
Then, the upper limit will be $$\dfrac{\pi}{4}$$.
Required Area $$=\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \left( \cos { x } -\sin { x } \right) dx }$$
$$=\left[\sin x+\cos x\right]_{0}^{\pi/4}$$
$$ =\left( \sqrt { 2 } -1 \right) $$ sq unit
Question 4
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The area of the figure bounded by the curves $$y = |x - 1|$$ and $$y = 3 - |x|$$ is

A
$$2\ sq. units$$

B
$$3\ sq. units$$

C
$$4\ sq. units$$

D
$$1\ sq. units$$

Solution

Area of the curve figure bounded by curves $$y=\left| x-1 \right| $$ and $$y=3-\left| x \right| $$
$$y=x-1\longrightarrow (1)\\ y=1-x\longrightarrow (2)\\ y=3-x\longrightarrow (3)\\ y=3+x\longrightarrow (4)$$
For : (B)
$$y=1-x\\ y=3-x\\ x=-1\\ y=2$$
Area required = Area of rectangle ABCD
$$=(AB)(AD)$$
$$=\sqrt { 8 } \times \sqrt { 2 } \\ =4\quad sq.\quad units$$
Question 5
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The area of the region of the plane bounded by $$max(|x|, |y|) \leq 1$$ and $$xy\leq \dfrac {1}{2}$$ is

A
$$\dfrac{1}{2} + ln \ 2$$ sq. units

B
$$3 + ln\ 2$$ sq. units

C
$$\dfrac{31}{4}$$ sq. units

D
$$1 + 2\ ln \ 2$$ sq. units
Question 6
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The area bounded between the parabolas $$x^2 = \dfrac{y}{4} $$ and $$x^2 = 9y$$, and the straight line $$y = 2$$ is:

A
$$10\sqrt2$$

B
$$20\sqrt2$$

C
$$\dfrac{10\sqrt2}{3}$$

D
$$\dfrac{20\sqrt2}{3}$$

Solution

The given two parabolas are $$x^2=\dfrac{y}{4}$$ and $$x^2=9y$$
We have to find the area bounded between the parabolas and the straight line $$y=2$$
The required area is equal to the shaded region in the drawn figure.
The area of the shaded region is twice the area shaded in first quadrant.
$$\Rightarrow$$ Required area $$=2\displaystyle\int_0^2\left(3\sqrt{y}-\dfrac{\sqrt{y}}{2}\right)dy$$

$$=2\displaystyle\int_0^2\left(\dfrac{6\sqrt{y}-\sqrt{y}}{2}\right)dy$$

$$=2\displaystyle\int_0^2\left(\dfrac{5\sqrt{y}}{2}\right)dy$$

$$=5\left(\dfrac{y^{\dfrac{3}{2}}}{3/2}\right)^{2}_{0}$$

$$=\dfrac{10}{3}\left(2^{\dfrac{3}{2}}-0\right)$$

$$=\dfrac{20\sqrt{2}}{3}$$
Question 7
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The parabola $${ y }^{ 2 }=4x$$ and $${ x }^{ 2 }=4y$$ divide the square region bounded by the lines $$x=4, y=4$$ and the coordinate axes. If $${ S }_{ 1 }, { S }_{ 2 }$$ and $${ S }_{ 3 }$$ are respectively the areas of these parts numbered from top-to-bottom, then $${ S }_{ 1 } : { S }_{ 2 } : { S }_{ 3 }$$ is

A
$$1 : 1 : 1$$

B
$$2 : 1 : 2$$

C
$$1 : 2 : 3$$

D
$$1 : 2 : 1$$

Solution

$${ S }_{ 1 }={ S }_{ 3 }=\displaystyle\int _{ 0 }^{ 4 }{ \dfrac { { x }^{ 2 } }{ 4 } dx } $$
$$={ \left[ \dfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }=\left[ \dfrac { 64 }{ 12 } -0 \right] =\dfrac { 16 }{ 3 } $$
Now, $${ S }_{ 2 }=\displaystyle\int _{ 0 }^{ 4 }{ \left( \sqrt { 4x } -\dfrac { { x }^{ 2 } }{ 4 } \right) dx } $$
$$={ \left[ 2{ x }^{ { 3 }/{ 2 } }\cdot \dfrac { 2 }{ 3 } -\dfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }=\dfrac { 16 }{ 3 } $$
$$\therefore { S }_{ 1 } : { S }_{ 2 } : { S }_{ 3 }=\dfrac { 16 }{ 3 } : \dfrac { 16 }{ 3 } : \dfrac { 16 }{ 3 } = 1 : 1 : 1$$
Question 8
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The area enclosed between the parabolas $$y^{2} = 16x$$ and $$x^{2} = 16y$$ is

A
$$\dfrac {64}{3} sq. units$$

B
$$\dfrac {256}{3} sq. units$$

C
$$\dfrac {16}{3} sq. units$$

D
None of these

Solution

Both parabolas cut each other at $$(0,0)$$ and $$(16,16)$$
Area enclosed by these parabolas
$$=\displaystyle\int_0^{16}4\sqrt x dx-\int_0^{16} \dfrac{x^2}{16}dx$$

$$=\Bigl[\dfrac{2\times 4\times x^{3/2}}{3}\Bigr]_0^{16}-\Big[\dfrac{x^3}{16\times 3}\Bigr]_0^{16}$$

$$=\dfrac{2\times 4^4}{3}-\dfrac{4^4}{3}=\dfrac{256}{3}$$ sq. units
Question 9
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The area of the region described by $$ \begin{Bmatrix} (x,,y)/x^2 +y^2 \leq 1 and\ y^2\leq1-x\end{Bmatrix}$$ is

A
$$\dfrac{\pi}{2}-\dfrac {2}{3}$$

B
$$\dfrac {\pi}{2} +\dfrac {2}{3}$$

C
$$\dfrac {\pi}{2} + \dfrac {4}{3}$$

D
$$\dfrac {\pi}{2}- \dfrac {4}{3}$$
Question 10
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Area of the region bounded by the curves $$y={ 2 }^{ x },y=2x-{ x }^{ 2 },x=0$$ and $$x=2$$ is given by

A
$$\cfrac { 3 }{ \log { 2 } } -\cfrac { 4 }{ 3 } $$

B
$$\cfrac { 3 }{ \log { 2 } } +\cfrac { 4 }{ 3 } $$

C
$$3\log { 2 } -\cfrac { 4 }{ 3 } $$

D
$$3\log { 2 } +\cfrac { 4 }{ 3 } $$

Solution

Let the required area be $$A$$ sq. units. Then,
$$A=\int _{ 0 }^{ 2 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx } $$
$$=\int _{ 0 }^{ 2 }{ \left( { 2 }^{ x }-(2x-{ x }^{ 2 }) \right) dx } $$
$$={ \left[ \cfrac { { 2 }^{ x } }{ \log { 2 } } -{ x }^{ 2 }+\cfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }$$
$$=\cfrac { 4 }{ \log { 2 } } -4+\cfrac { 8 }{ 3 } \cfrac { 1 }{ \log { 2 } } $$
$$=\cfrac { 3 }{ \log { 2 } } -\cfrac { 4 }{ 3 } sq.units$$