Application of Integrals Test

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Application of Integrals Test - 50

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Weekly Quiz Competition

Question 1
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The area of the region bounded by the circle $$x^{2} + y^{2} = 1$$ is

A
$$2\pi sq\ units$$

B
$$\pi sq\ units$$

C
$$3\pi sq\ units$$

D
$$4\pi sq\ units$$

Solution

We have, $$x^{2} + y^{2} = 1^{2} [\because r = \pm 1]$$

$$\Rightarrow y^{2} = 1 - x^{2} \Rightarrow = \sqrt {1 - x^{2}}$$

$$\therefore$$ Area enclosed by circle $$= 2\int_{-1}^{1} \sqrt {1^{2} - x^{2}} dx = 2.2\int_{0}^{1} \sqrt {1^{2} - x^{2}}dx$$

$$= 4\left [\dfrac {x}{2} \sqrt {1^{2} - x^{2}} + \dfrac {1^{2}}{2} \sin^{-1} \dfrac {x}{1} \right ]_{0}^{1}$$

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

$$= 4. \dfrac {\pi}{4} = \pi sq\ units$$.
Question 2
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The area of the region bounded by the curve $$y = \sin x$$ between the ordinates $$x = 0, x = \dfrac {\pi}{2}$$ and the x-axis is

A
$$2\ sq\ units$$

B
$$4\ sq\ units$$

C
$$3\ sq\ units$$

D
$$1\ sq\ units$$

Solution

Area of the region bounded by the curve $$y = \sin x$$ between the ordinates
$$x = 0, x = \dfrac {\pi}{2}$$ and the X-axis is

$$A = \int_{0}^{\pi/2}\sin x dx$$

$$= -[\cos x]_{0}^{\pi/2} $$

$$= - \left [\cos \dfrac {\pi}{2} - \cos 0\right ]$$

$$= -[0 - 1] = 1\ sq\ unit$$.
Question 3
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The area of the region bounded by the curve $$x = 2y + 3$$ and the $$y$$ lines. $$y = 1$$ and $$y = -1$$ is

A
$$4\ sq\ units$$

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

C
$$6\ sq\ units$$

D
$$8\ sq\ units$$

Solution

Required area, $$A = \int_{-1}^{1} (2y + 3)dy$$
$$= \left [\dfrac {2y^{2}}{2} + 3y\right ]_{-1}^{1}$$
$$= [y^{2} + 3y]_{-1}^{1}$$
$$= [1 + 3 - 1 + 3]$$
$$= 6\ sq\ units$$.
Question 4
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The area of the region bounded by the curve $$x^{2} = 4y$$ and the straight line $$x = 4y - 2$$ is

A
$$\dfrac {3}{8} sq\ units$$

B
$$\dfrac {5}{8} sq\ units$$

C
$$\dfrac {7}{8} sq\ units$$

D
$$\dfrac {9}{8} sq\ units$$

Solution

Given equation of curve is $$x^{2} = 4y$$ and the straight line $$x = 4y - 2$$.
For intersection point, put $$x = 4y - 2$$ in equation of curve, we get
$$(4y - 2)^{2} = 4y$$
$$\Rightarrow 16y^{2} + 4 - 16y = 4y$$
$$\Rightarrow 16y^{2} - 20y + 4 = 0$$
$$\Rightarrow 4y^{2} - 5y + 1 = 0$$
$$\Rightarrow 4y^{2} - 4y - y + 1 = 0$$
$$\Rightarrow 4y(y - 1) - 1(y - 1) = 0$$
$$\Rightarrow (4y - 1)(y - 1) = 0$$
$$\therefore y = 1, \dfrac {1}{4}$$
For $$y = 1, x = \sqrt {4.1} = 2$$ [since, negative value does not satisfy the equation of line]
For $$y = \dfrac {1}{4}, x = \sqrt {4.\dfrac {1}{4}} = -1$$ [positive value does not satisfy the equation of line]
So, the intersection points are $$(2, 1)$$ and $$\left (-1, \dfrac {1}{4}\right )$$
$$\therefore$$ Area of shaded region $$= \int_{-1}^{2}\left (\dfrac {x + 2}{4}\right ) dx - \int_{-1}^{2} \dfrac {x^{2}}{4}dx$$

$$= \dfrac {1}{4} \left [\dfrac {x^{2}}{2} + 2x\right ]_{-1}^{2} - \dfrac {1}{4} \left |\dfrac {x^{3}}{3}\right |_{-1}^{2}$$

$$= \dfrac {1}{4} \left (\dfrac {4}{2} + 4 - \dfrac {1}{2} + 2\right ) - \dfrac {1}{4} \left (\dfrac {8}{3} + \dfrac {1}{3}\right )$$

$$= \dfrac {1}{4} . \dfrac {15}{2} - \dfrac {1}{4} . \dfrac {9}{3} $$

$$= \dfrac {45 - 18}{24}$$

$$= \dfrac {27}{24} = \dfrac {9}{8} sq\ units$$.
Question 5
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The area of the region bounded by parabola $$y^{2} = x$$ and the straight line $$2y = x$$ is

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

B
$$1 sq\ units$$

C
$$\dfrac {2}{3} sq\ units$$

D
$$\dfrac {1}{3} sq\ units$$

Solution

We have to find the area enclosed by parabola $$y^{2} = x$$ and the straight line $$2y = x$$.
$$\therefore \left (\dfrac {x}{2}\right )^{2} = x$$
$$\Rightarrow x^{2} = 4x \Rightarrow x (x - 4) = 0$$
$$\Rightarrow x = 4 \Rightarrow y = 2$$ and $$x = 0 \Rightarrow y = 0$$
So, the intersection points are $$(0, 0)$$ and $$(4, 2)$$.
Area enclosed by shaded region,
$$A = \int_{0}^{4} \left [\sqrt {x} - \dfrac {x}{2}\right ] dx$$

$$= \left [\dfrac {x^{\frac {1}{2} + 1}}{\dfrac {1}{2} +1} - \dfrac {1}{2} . \dfrac {x^{2}}{2} \right ]_{0}^{4} $$

$$= \left [2 . \dfrac {x^{\frac {3}{2}}}{3} - \dfrac {x^{2}}{4}\right ]_{0}^{4}$$

$$= \dfrac {2}{3} 4^{3/2} - \dfrac {16}{4} - \dfrac {2}{3} . 0 + \dfrac {1}{4} . 0$$

$$= \dfrac {16}{3} - \dfrac {16}{4} $$

$$= \dfrac {64 - 48}{12} $$

$$= \dfrac {16}{12} = \dfrac {4}{3} sq\ units$$.
Question 6
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The area of the region bounded by the curve $$y = x + 1$$ and the lines $$x = 2$$ and $$x = 3$$ is

A
$$\dfrac {7}{2} sq\ units$$

B
$$\dfrac {9}{2} sq\ units$$

C
$$\dfrac {11}{2} sq\ units$$

D
$$\dfrac {13}{2} sq\ units$$

Solution

Required area, $$A = \int_{2}^{3} (x + 1)dx $$

$$= \left [\dfrac {x^{2}}{2} + x\right ]_{2}^{3}$$

$$= \left [\dfrac {9}{2} + 3 - \dfrac {4}{2} - 2\right ] $$

$$= \dfrac {7}{2} sq\ units$$.
Question 7
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The area of the region bounded by the curve $$y = \sqrt {16 - x^{2}}$$ and x-axis is

A
$$8\pi \ sq. units$$

B
$$20\pi \ sq. units$$

C
$$16\pi \ sq. units$$

D
$$256\pi \ sq. units$$

Solution

Given equation of curve is $$y = \sqrt {16 - x^{2}}$$ and the equation of line is $$X-axis$$ i.e., $$y = 0$$
$$\therefore \sqrt {16 - x^{2}} = 0 ..... (i)$$
$$\Rightarrow 16 - x^{2} = 0$$
$$\Rightarrow x^{2} = 16$$
$$\Rightarrow x = \pm 4$$
So, the intersection points are $$(4, 0)$$ and $$(-4, 0)$$.

$$\therefore$$ Area of curve, $$A = \int_{-4}^{4} (16 - x^{2})^{1/2} dx$$

$$= \int_{-4}^{4} \sqrt {(4^{2} - x^{2})} dx$$

$$= \left [\dfrac {x}{2} \sqrt {4^{2} - x^{2}} + \dfrac {4^{2}}{2} \sin^{-1} \dfrac {x}{4}\right ]_{-4}^{4}$$

$$= \left [\dfrac {4}{2} \sqrt {4^{2} - 4^{2}} + 8\sin^{-1} \dfrac {4}{4}\right ] - \left [-\dfrac {4}{2} \sqrt {4^{2} - (-4)^{2}} + 8\sin^{-1} \left (-\dfrac {4}{4}\right )\right ]$$

$$= \left [2.0 + 8.\dfrac {\pi}{2} - 0 + 8.\dfrac {\pi}{2}\right ] = 8\pi sq\ units$$.
Question 8
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The area of the region bounded by the ellipse $$\dfrac {x^{2}}{25} + \dfrac {y^{2}}{16} = 1$$ is

A
$$20\pi sq\ units$$

B
$$20\pi^{2} sq\ units$$

C
$$16\pi^{2} sq\ units$$

D
$$25\pi sq\ units$$

Solution

We have, $$\dfrac {x^{2}}{5^{2}} + \dfrac {y^{2}}{4^{2}} = 1$$

Here, $$a = \pm 5$$ and $$b = \pm 4$$

And $$\dfrac {y^{2}}{4^{2}} = 1 - \dfrac {x^{2}}{5^{2}}$$

$$\Rightarrow y^{2} = 16\left (1 - \dfrac {x^{2}}{25}\right )$$

$$\Rightarrow y = \sqrt {\dfrac {16}{25}} (25 - x^{2})$$

$$\Rightarrow y = \dfrac {4}{5} \sqrt {(5^{2} - x^{2})}$$

$$\therefore$$ Area enclosed by ellipse, $$A = 4 . \dfrac {4}{5} \int_{-5}^{5}\sqrt {5^{2} - x^{2}} dx$$

$$= \dfrac {16}{5} \int_{0}^{5} \sqrt {5^{2} - x^{2}} dx$$

$$= \dfrac {16}{5} \left [\dfrac {x}{2} \sqrt {5^{2} - x^{2}} + \dfrac {5^{2}}{2} \sin^{-1} \dfrac {x}{5} \right ]_{0}^{5}$$

$$= \dfrac {16}{5}\left [\dfrac {5}{2} \sqrt {5^{2} - 5^{2}} + \dfrac {5^{2}}{2} \sin^{-1} \dfrac {5}{5} - 0 - \dfrac {25}{5} . 0\right ]$$

$$= \dfrac {16}{5} \left [\dfrac {25}{2} . \dfrac {\pi}{2}\right ]$$
$$= \dfrac {16}{5} . \dfrac {25\pi}{4}$$

$$= 20\pi sq\ units$$.
Question 9
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Area of the region in the first quadrant enclosed by the x-axis, the line $$y = x$$ and the circle $$x^{2} + y^{2} = 32$$ is

A
$$16\pi \ sq units$$

B
$$4\pi \ sq units$$

C
$$32\pi \ sq units$$

D
$$24\pi \ sq units$$

Solution

We have enclosed by $$X-axis$$ i.e., $$y = 0, y = x$$ and the circle $$x^{2} + y^{2} = 32$$ in first quadrant.
Since, $$x^{2} + (x)^{2} = 32 [\because y = x]$$
$$\Rightarrow 2x^{2} = 32$$
$$\Rightarrow x = \pm 4$$
So, the intersection point of circle $$x^{2} + (x)^{2} = 32$$ and line $$y = x$$ are $$(4, 4)$$ and $$(-4, -4)$$.
And $$x^{2} + y^{2} = (4\sqrt {2})^{2}$$
Since, $$y = 0$$
$$\therefore x^{2} + (0)^{2} = 32$$
$$\Rightarrow x = \pm 4\sqrt {2}$$
So, the circle intersects the $$X-axis$$ at $$(\pm 4\sqrt {2}, 0)$$.

Area of shaded region $$= \int_{0}^{4} xdx + \int_{4}^{4\sqrt {2}} \sqrt {(4\sqrt {2})^{2} - x^{2}} dx$$

$$= \left |\dfrac {x^{2}}{2}\right |_{0}^{4} + \left |\dfrac {x}{2} \sqrt {(4\sqrt {2})^{2} - x^{2}} + \dfrac {(4\sqrt {2})^{2}}{2} \sin^{-1} \dfrac {x}{4\sqrt {2}}\right |_{4}^{4\sqrt {2}}$$

$$= \dfrac {16}{2} + \left [\dfrac {4\sqrt {2}}{2} . 0 + 16\sin^{-1} \dfrac {(4\sqrt {2})}{(4\sqrt {2})} - \dfrac {4}{2} \sqrt {(4\sqrt {2})^{2}} - 16) - 16\sin^{-1} \dfrac {4}{4\sqrt {2}}\right ]$$

$$= 8 + \left (16\dfrac {\pi}{2} - 2\sqrt {16} - 16\dfrac {\pi}{4}\right )$$
$$= 8 + [8\pi - 8 - 4\pi] = 4\pi sq\ units$$.
Question 10
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Area lying in the first quadrant and bounded by the circle $$x^{2}+y^{2}=4$$ and the lines $$x=0$$ and $$x=2$$ is

A
$$\pi$$

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

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

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

Solution

Circle is $$x^{2}+y^{2}=4$$ with centre $$(0,0)$$ and radius $$2$$.
se required area is shaded area of circle from
$$(x=0\ to\ x=2)$$
$$=\displaystyle\int_{0}^{2}\sqrt{4-x^{2}}dx=\displaystyle\int_{0}^{2}\sqrt{2^{2}-x^{2}}dx$$
$$=\left.\left(\dfrac{x}{2}\sqrt{2^{2}-x^{2}}+\dfrac{4}{2}\sin^{-1}\dfrac{x}{2}\right)\right]_{0}^{2}$$ $$\displaystyle\int \sqrt{a^{2}-x^{2}}dx=\dfrac{x}{a}\sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a}$$
$$=\dfrac{2}{2}\sqrt{2^{2}-2^{2}}+\dfrac{4}{2}\sin^{-1}\dfrac{2}{2}-\dfrac{0}{2}\sqrt{2^{2}-0^{2}}-\dfrac{4}{2}\sin^{-1}\dfrac{0}{2}$$
$$=1\sqrt{0}+2\sin^{-1}1-0-2\sin^{-1}0$$
$$=2\times \dfrac{\pi}{2}-2\times 0=\pi\ sq.units$$
so answer is $$(A)\pi $$ unit .....

OR
As circle of radius $$2$$. so area of circle $$=\pi r^{2}$$
$$\pi \times 2^{2}=4\pi$$
As we need area of only $$1$$ quadrant
$$\Rightarrow \dfrac{1}{4}\times$$ area of circle $$=\dfrac{1}{4}\times 4\pi =\pi\ sq.units$$.
Option $$(A)$$