Application of Integrals Test

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Application of Integrals Test - 24

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Weekly Quiz Competition

Question 1
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The line $$2y=3x+12$$ cuts the parabola $$4y=3x^2$$. What is the area enclosed by the parabola and the line?

A
$$27$$ square unit

B
$$36$$ square unit

C
$$48$$ square unit

D
$$54$$ square unit

Solution

The graph is shown in the image.
The intersection point of $$2y=3x+12$$ and $$4y=3x^2$$
$$2\Big(\dfrac{3x^2}{4}\Big)=3x+12$$
$$x^2-2x-8=0$$
$$x=-2,4$$
Hence, intersection point are $$(-2,3)$$ and $$(4,12)$$
Now, $$A=\int_{-2}^{4}\Bigg[\dfrac{3x+12}{2}\Bigg]-\int_{-2}^{4}\Bigg[\dfrac{3x^2}{4}\Bigg]$$
$$A=\Bigg[\dfrac{3x^2}{4}+6x \Bigg]\Bigg|_{-2}^4 -\Bigg[\dfrac{x^3}{4}\Bigg]\Bigg|_{-2}^4 $$
$$A=9+36-18 $$
$$A=27$$ $$sq.unit$$
Question 2
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The area in the first quadrant between $$x^2 + y^2 = \pi^2$$ and $$y = sin x$$ is

A
$$\dfrac{\pi^3 - 8}{4}$$

B
$$\dfrac{\pi^3}{4}$$

C
$$\dfrac{\pi^3 - 16}{4}$$

D
$$\dfrac{\pi^3 - 8}{2}$$

Solution

$$x^2 + y^2 = \pi^2$$ is a circle of radius $$\pi$$ and centre at origin.
Required area
= Area of circle (1st quadrant) $$- \int_0^{\pi} sin x dx$$
$$= \dfrac{\pi. \pi^2}{4} - [-cos x]_{0}^{\pi} = \dfrac{\pi^3}{4} + (cos \pi - cos 0)$$
$$= \dfrac{\pi^3}{4} + (-1 - 1) = \dfrac{\pi^3}{4} - 2 = \dfrac{\pi^3 - 8}{4}$$
Hence, option A is correct.
Question 3
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Consider the curves $$y = \sin x$$ and $$y = \cos x$$.
What is the area of the region bounded by the above two curves and the lines $$x = 0$$ and $$x = \dfrac {\pi}{4}$$?

A
$$\sqrt {2} - 1$$

B
$$\sqrt {2} + 1$$

C
$$\sqrt {2}$$

D
$$2$$

Solution

The area enclosed by $$y=sinx, y=cosx,x=0,x=\frac { \pi }{ 4 } $$ is given by
$$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ (\cos x-\sin x)dx } $$
$$=|\sin x+\cos x|_{0}^{\frac{\pi}{4}}$$
$$=\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 2 } } -0-1$$
$$=\sqrt { 2 } -1$$
Question 4
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The area bounded by the curves $$y = \cos x$$ and $$y = \sin x$$ between the ordinates $$x = 0$$ and $$x = \dfrac {3\pi}{2}$$ is

A
$$(4\sqrt {2} - 2)sq\ units$$

B
$$(4\sqrt {2} + 2)sq\ units$$

C
$$(4\sqrt {2} - 1)sq\ units$$

D
$$(4\sqrt {2} + 1)sq\ units$$

Solution

Required area
$$= \int_{0}^{\pi /4} (\cos x - \sin x) dx + \int_{\pi/4}^{5\pi/4}(\sin x - \cos x) dx + \int_{5\pi/4}^{3\pi/2} (\cos x - \sin x)dx$$
$$= [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{5\pi/4} + [\sin x + \cos x]_{5\pi/4}^{3\pi/2}$$
$$=\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-1 \right)-\left( -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} \right) + \left( -1+0+\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right)$$
$$=\dfrac{8}{\sqrt{2}} - 2$$
$$= (4\sqrt {2} - 2)sq\ units$$
Hence, option A is correct.
Question 5
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Consider the curves $$y = \sin x$$ and $$y = \cos x$$.
What is the area of the region bounded by the above two curves and the lines $$x = \dfrac {\pi}{4}$$ and $$x = \dfrac {\pi}{2}$$?

A
$$\sqrt {2} - 1$$

B
$$\sqrt {2} + 1$$

C
$$2\sqrt {2}$$

D
$$2$$

Solution

The area enclosed by $$y=\sin x, y=\cos x, x=\dfrac { \pi }{ 2}, x=\dfrac { \pi }{ 4 }$$ is given by
$$\displaystyle \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2} }{ (-\cos x+\sin x)dx }$$
$$=|-\sin x-\cos x|_{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }$$
$$=-1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } }$$
$$=\sqrt { 2 } -1$$
Question 6
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The area of the region bounded by the lines $$y = 2x + 1, y = 3x + 1$$ and $$x = 4$$ is

A
$$16\ sq.unit$$

B
$$\dfrac {121}{3}\ sq.unit$$

C
$$\dfrac {121}{6}\ sq.unit$$

D
$$8\ sq.unit$$

Solution

$$A$$(Shaded region) $$= A(\triangle ABD) - A(\triangle ABC) = \dfrac {1}{2} [4\times 12 - 4 \times 8] = \dfrac {1}{2} (48 - 32) = 8\ sq.units$$.
OR
$$A (\triangle ACD) = \dfrac {1}{2}\begin{vmatrix} 0& 1 & 1\\ 4 & 9 & 1\\ 4 & 13 & 1\end{vmatrix}$$
$$= \dfrac {1}{2}\times 16 = 8$$
Question 7
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The area bounded by the curves $$y=\cos x$$ and $$y=\sin x$$ between the ordinates $$x=0$$ and $$x=\displaystyle\frac{3\pi}{2}$$ is?

A
$$4\sqrt{2}-1$$

B
$$4\sqrt{2}+1$$

C
$$4\sqrt{2}-2$$

D
$$4\sqrt{2}+2$$

Solution

We have,
$$y=\cos x$$ and $$y=\sin x$$ between $$x=0$$ and $$x=\displaystyle\frac{3\pi}{2}$$

Required area A is given by
$$A=\displaystyle\int^{3\pi /2}_0|\cos x-\sin x|dx$$

$$\Rightarrow A=\displaystyle\int^{\pi /4}_0|\cos x-\sin x|dx+\int^{5\pi/4}_{\pi /4}|\cos x -\sin x|dx+\displaystyle \int^{3\pi /2}_{5\pi/4}|\cos x - \sin x|dx$$

$$\Rightarrow \displaystyle A=\int^{\pi /4}_0(\cos x-\sin x)dx+\displaystyle \int^{5\pi/4}_{\pi /4}(\sin x -\cos x)dx+ \displaystyle \int^{3\pi /2}_{5\pi /4}(\cos x -\sin x)dx$$

$$\Rightarrow A=\displaystyle [\sin x+\cos x]^{\pi /4}_0+[-\cos x -\sin x]^{5\pi /4}_{\pi/4}+[\sin x+ \cos x]^{3\pi /2}_{\pi /4}$$

$$\Rightarrow A=(\sqrt{2}-1)+(\sqrt{2}+\sqrt{2})+(-1+\sqrt{2})$$

$$\Rightarrow A=4\sqrt{2}-2$$.
Question 8
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The line $$x=\dfrac{\pi}{4}$$ divide the area of the region bounded by $$y=\sin x, y = \cos x$$ and X-axis $$\left(0 \le x \le \frac{\pi}{2}\right)$$ into two regions of areas $$A_1$$ and $$A_2$$. Then, $$A_1:A_2$$ equals

A
$$4:1$$

B
$$3:1$$

C
$$2:1$$

D
$$1:1$$

Solution

Area $$A_1=\displaystyle \int_0^{\frac{\pi}{4}}\sin x\,dx$$
$$=-[\cos x]_0^{\frac{\pi}{4}}$$
$$=1-\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}-1}{\sqrt{2}}$$

and Area $$A_2=\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x\,dx$$
$$=\displaystyle [\sin x]^{\frac{\pi}{2}}_{\frac{\pi}{4}}=\left[1-\dfrac{1}{\sqrt{2}}\right]=\dfrac{\sqrt{2}-1}{\sqrt{2}}$$
$$\therefore A_1:A_2 = \dfrac{\sqrt{2}-1}{\sqrt{2}}: \dfrac{\sqrt{2}-1}{\sqrt{2}}=1:1$$
Question 9
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The area bounded by the parabola $${ y }^{ 2 }=4a(x+a)$$ and $${ y }^{ 2 }=-4a(x-a)$$ is

A
$$\cfrac { 16 }{ 3 } { a }^{ 2 }$$

B
$$\cfrac { 8 }{ 3 } { a }^{ 2 }$$

C
$$\cfrac { 4 }{ 3 } { a }^{ 2 }$$

D
None of these

Solution

Required area $$=4\int _{ 0 }^{ 4 }{ \sqrt { 4a(a-x) } } dx$$
$$=4\times2\sqrt { a } { \left[ \cfrac { -2{ (a-x) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ 4 }=\cfrac { 16 }{ 3 } { a }^{ 2 }$$
Question 10
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Consider an ellipse $$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$$ What is the area included between the ellipse and the greatest rectangle inscribed in the ellipse?

A
$$ab(\pi -1)$$

B
$$2ab(\pi -1)$$

C
$$ab(\pi -2)$$

D
None of the above

Solution

As the rectangle lies inside a standard ellipse, the vertices of the rectangle can be assumed to be $$(h,k),(h,-k),(-h,k)$$ and $$(-h,-k)$$ without any loss of generality.
$$\therefore \dfrac{h^2}{a^2}+\dfrac{k^2}{b^2}=1\implies k=b\sqrt{1-\dfrac{h^2}{a^2}}$$
Clearly, the sides of the rectangle based on the coordinates chosen can be determined as $$2h$$ and $$2k$$.
$$\therefore$$ area of rectangle $$=A(h)=2h\times 2k=4h b\sqrt{1-\dfrac{h^2}{a^2}}\implies 4b\sqrt{h^2-\dfrac{h^4}{a^2}}$$
Now, let $$P(h)=h^2-\dfrac{h^4}{a^2}$$. Thus, the maximization of $$A(h)$$ boils down just to the maximization of $$P(h)$$.
Differentiating $$P(h)$$ wrt $$h$$, we have
$$2h-\dfrac{4h^3}{a^2}=0\implies h=\dfrac{2h^3}{a^2}\implies 2h^2=a^2\implies h=\dfrac{a}{\sqrt{2}}\implies k=\dfrac{b}{\sqrt{2}}$$
Hence, $$A_{max}=2ab$$
$$\therefore$$ area included between this ellipse and rectangle $$=\pi ab-2ab=ab(\pi-2)$$.
this is the required answer.