Application of Integrals Test

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Application of Integrals Test - 49

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Weekly Quiz Competition

Question 1
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The area of the region bounded by $$x^2+y^2-2x-3=0$$ and $$y=|x|+1$$ is

A
$$ \dfrac{\pi}{2} -1 \, sq.units$$

B
$$ 2 \pi \, sq.units $$

C
$$ 4 \pi \, sq.units $$

D
$$ \dfrac{\pi}{2} \, sq.units$$

Solution

$$x^2+y^2-2x-3=0$$

$$ \Rightarrow (x-1)^2+y^2=4 $$

A= $$ \displaystyle \int_{1-\sqrt{2}}^{0} ( \sqrt{4-(x-1)^2} - (-x+1)dx + \displaystyle \int_{0}^{1} ( \sqrt{4-(x-1)^2} - (x+1))dx $$

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

$$ -\left ( -1- \dfrac{\pi}{2}+\dfrac{3}{2}- \sqrt{2}-1+\sqrt{2} \right )- \dfrac{3}{2}=\dfrac{\pi}{2}-1 \, sq.units $$
Question 2
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The value of the parameter $$a$$ such that the area bounded by $$y=a^{2}x^{2}+ax+1,$$ coordinate axes and the line $$x=1$$ attains its least value, is equal to

A
$$ \dfrac {-1}{4} $$

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

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

D
$$ -1 $$

Solution

$$ a^{2}x^{2} + ax+1$$ is clearly positive for all real values of $$x$$. Area under consideration.

$$ A= \displaystyle \int_{0}^{1} (a^{2}x^{2} + ax+1) dx $$

$$ = \dfrac{a^{2}}{3} +\dfrac{a}{2} +1 $$

$$ = \dfrac{1}{6} (2a^{2}+3a+6) $$

$$ = \dfrac{1}{6}\left ( 2\left ( a^{2}+\dfrac{3}{2}a+ \dfrac{9}{16} \right )+6-\dfrac{18}{16} \right ) $$

$$ = \dfrac{1}{6}\left ( 2\left ( a+\dfrac{3}{4} \right )^2+6-\dfrac{18}{16} \right ) $$

Which is clearly minimum for $$ a= -\dfrac{3}{4} $$
Question 3
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The area bounded by $$ y=\sec^{-1}x\, y=cosec^{-1}x \,$$ and line $$x-1=0$$ is

A
$$ log(3+2 \sqrt{2}) \, - \dfrac{\pi}{2} \, sq.units $$

B
$$ \dfrac{\pi}{2} - log(3+2 \sqrt{2}) \,sq.units $$

C
$$ \pi- log_{e}3\, sq.units $$

D
None of these

Solution

Integrating along $$x-axis$$, we get

$$A= \displaystyle \int_{1}^{\sqrt{2}} (\ cosec^{-1}x\,- \ sec^{-1}x)dx $$

Integrating along $$ y-axis$$, we get

$$A= 2\displaystyle \int_{0}^{\dfrac{\pi}{4}} (\ sec\,y-1)dy $$

$$ = 2[log|\ sec\,y+\ tan\,y|-y]_{0}^{\frac{\pi}{4}} $$

$$ = 2\left [ log|\sqrt{2}+1|-\dfrac{\pi}{4} \right ]=log(3+2\sqrt{2})- \dfrac{\pi}{2} $$
Question 4
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The area of the closed figure bounded by $$ x=-1, \, x=2$$ and $$ y= -x^2+2, \, x\leq 1$$ and $$y= 2x-1 ,\, x>1$$ and the abscissa axis is

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

B
$$ \dfrac{10}{3} \, sq.units $$

C
$$ \dfrac{13}{3} \, sq.units $$

D
$$ \dfrac{7}{3} \, sq.units $$

Solution

$$ A= \displaystyle \int_{-1}^{1} (-x^{2}+2)dx+ \displaystyle \int_{1}^{2}(2x-1)dx $$

$$ =\left ( -\dfrac{x^{3}}{3}+2x \right )_{-1}^{1}+(x^{2}-x)_{1}^{2} $$

$$ = \dfrac{16}{3} \, sq.units $$
Question 5
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The area of the region whose boundaries are defined by the curves $$y=2 \ cos\,x, \, y=3 \ tan\,x$$ and the $$y-axis$$ is

A
$$ 1+3ln\left ( \dfrac{2}{\sqrt{3}} \right )\, sq.units$$

B
$$ 1+\dfrac{3}{2}ln3-3ln2\,sq.units $$

C
$$ 1+\dfrac{3}{2}ln3-ln2\,sq.units $$

D
$$ ln3-ln2\, sq.units$$

Solution

Solving $$ 2 \cos \,x = 3 \tan \,x$$, we get

$$ 2-2 \ sin^{2}x=3 \sin \,x \,\,\,\ \Rightarrow \ sin \,x=\dfrac{1}{2}\,\,\,\, \Rightarrow x= \dfrac{\pi}{6} $$

Required Area= $$ \displaystyle \int_{0}^{\dfrac{\pi}{6}} (2\ cos\,x-3 \tan\,x) dx $$

$$ = 2 \ sin\,x - 3 log \ sec\,x|_{0}^{\frac{\pi}{6}} = 1-3ln2 + \dfrac{3}{2}ln3\, sq.units. $$
Question 6
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A tangent having slope of $$-\dfrac{4}{3} $$ to the ellipse $$\dfrac{x^{2}}{18}+ \dfrac{y^{2}}{32}=1 $$ intersects the major and minor axes at points A and B respectively. If C is the center of the ellipse , then area of the triangle ABC is

A
12 sq. units

B
24 sq. units

C
36 sq. units

D
48 sq. units

Solution

One of the tangents of slope m to the given ellipse is
$$y=mx + \sqrt{18m^{2}+ 32}$$
For $$m=-\dfrac{4}{3}$$,
we have $$y=-\dfrac{4}{3}x+ 8.$$
Then points on the axis where tangents meet are A(6,0) and B(0,8).
Then area of triangle ABC is $$\dfrac{1}{2}(6)(8)= 24 units.$$
Question 7
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The area enclosed by the curves $$xy^2 =a^2(a-x)$$ and $$(a - x) y^2 = a^2x$$ is

A
$$(\pi - 2)a^2$$ sq. units

B
$$(4 -\pi)a^2$$ sq. units

C
$$\dfrac{\pi a^2}{3}$$ sq. units

D
None of these
Question 8
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Directions For Questions

Consider the areas $$S_0,S_1,S_2$$..... bounded by the x-axis and half-waves of the curve $$y=e^{-x}\sin\,x$$ where $$x\ge 0$$.

...view full instructions

The sequence $$S_0,S_1,S_2$$.... forms a G.P with common ratio

A
$$\dfrac{e^\pi}{2}$$

B
$$e^{-\pi}$$

C
$$e^{\pi}$$

D
$$\dfrac{e^{-\pi}}{2}$$

Solution
Question 9
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The area of the loop of the curve, $$ay^2 =x^2 (a - x)$$ is

A
$$4a^2 $$ sq. units

B
$$\dfrac{8a^2}{15}$$ sq. units

C
$$\dfrac{16 a^2}{9}$$ sq. units

D
none of these

Solution

$$ay^2 = x^2 (a -x) \Rightarrow y = \pm \ x \sqrt{\dfrac{a-x}{a}}$$

Curve tracing : $$y = x \sqrt{\dfrac{a-x}{a}}$$

We must have $$x \le a$$
For $$0 < x \le a, \ y > 0$$ and for $$x < 0, y < 0$$
Also $$y = 0 \Rightarrow x = 0, a$$
Curve is symmetrical about x-axis.
When $$x \to -\infty, y \to - \infty$$
Also, it can be verified that $$y$$ has only one point of maxima for $$0 < x < a$$.

Area $$=\displaystyle 2\int_0^a x \sqrt{\dfrac{a-x}{a}}dx$$

$$\sqrt{\dfrac{a-x}{a}} = t \Rightarrow - \dfrac{x}{a}=t^2 \Rightarrow x=a(1 - t^2)$$

$$\Rightarrow A = 2\displaystyle \int_1^0 a(1 - t^2)t(-2at)dt$$

$$=\displaystyle 4a^2 \int_0^1 (t^2 - t^4)dt$$

$$= 4a^2 \left[\dfrac{t^3}{3} - \dfrac{t^5}{5}\right]^1_0$$

$$=4a^2\left[ \dfrac{1}{3} - \dfrac{1}{5}\right] = \dfrac{8a^2}{15} $$ sq. units
Question 10
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The area enclosed by the circle $$x^{2} + y^{2} = 2$$ is equal to

A
$$4\pi \ sq\ units$$

B
$$2\sqrt {2 \pi}$$ sq\ units$$

C
$$4\pi^{2} sq\ units$$

D
$$2\pi \,sq\ units$$

Solution

Correct answer is (D); since Area $$= 4\int_{0}^{\sqrt {2}} \sqrt {2 - x^{2}}$$
$$= 4\left (\dfrac {x}{2} \sqrt {2 - x^{2}} + \sin^{-1} \dfrac {x}{\sqrt {2}}\right )_{0}^{\sqrt {2}} = 2\pi \ sq\ units$$.