Explanation
$$x=y^{2}$$ and the line $$y=x-6$$ area in $$QI$$+ area in QIV $$|\overset {3 }{ \underset { -2}{ \int } } [(y+6)-y^2]dy|$$ $$=|\left | \dfrac{y^2}{2} +6y-y^3/3\right |_{-2}^{3}$$ $$=\left |\dfrac{9}{2} -\dfrac{4}{2}+6(5)-\left [ \dfrac{27}{3} +8/3\right ] \right |$$ $$\left | \dfrac{5}{2}+30-\left [ \dfrac{35}{3} \right ] \right | = \dfrac{15+180-70}{6} = \dfrac{125}{6}sq\:units.$$
Given curves are $$y^2=8x$$ and $$y=2x$$ Let's find out their intersection point
$$\Rightarrow 4x^{2}=8x$$
$$\Rightarrow x^{2}-2x=0$$
$$\Rightarrow x(x-2)=0$$
$$\Rightarrow x=0,2$$
The required area is $$A=\displaystyle \overset{2}{\underset{0}{\int}}(\sqrt{8x}-2x)dx$$ $$=\displaystyle \overset{2}{\underset{0}{\int}}\sqrt{8x}\:dx-\overset{2}{\underset{0}{\int}}2x\:dx$$ $$=\sqrt{8}\times\dfrac{2}{3}\times [x^{3/2}]_{0}^{2}-[x^2]_{0}^{2}$$ $$=\dfrac{2\sqrt{8}}{3}\sqrt{8}-4$$ $$=\dfrac{16}{3}-4=\dfrac{4}{3}\:sq\:units$$
So, area of the shaded region $$-\overset{2\sqrt{2}a}{\underset{0}{\int}}x^2/4a\:dx+\overset{2\sqrt{2}a}{\underset{0}{\int}}(2a)\:dx$$ $$2a(2\sqrt{2a})-\dfrac{}{4a}\dfrac{x^3}{3}|_{0}^{2\sqrt{2a}}$$ $$=4\sqrt{2}a^2-\dfrac{(2^{3/2}a)^3}{4a\times 3}$$ $$4\sqrt{2}a^2-\dfrac{2\times2\sqrt{2}a^2}{3}$$ $$=\dfrac{16\sqrt{2}a^2}{3}sq\:units.$$
$$I$$: $$\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{3} \right )^2=1$$ $$Area=\pi \times 2 \times3$$ $$=6\pi sq\:units.$$ $$I$$ $$\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{2} \right )^2=1$$ $$\Rightarrow x^2+y^2=4$$ $$\pi (2)^2$$ $$=4\pi\:sq\:units.$$
$$\mathrm{y}=\sqrt{x}$$ ; $$y=\sqrt{4-3x}$$ $$\overset{1}{\underset{0}{\int}}\sqrt{x}dx+\overset{4/3}{\underset{1}{\int}}\sqrt{4-3x}dx$$ $$\dfrac{2}{3}1+\dfrac{2}{9}(4-3x)^{3/2}|_{1}^{4/3}$$ $$\dfrac{2}{3}-\dfrac{2}{9}(0-1)$$ $$=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}sq\:units$$
The area bounded by the curves $${y}=4{x}^{2},\ y=\displaystyle \dfrac{x^{2}}{9}$$ and the line $$y=2$$ is $$A=2\times \displaystyle \overset{2}{\underset{0}{\int}}\left (3\sqrt{y}-\dfrac{\sqrt{y}}{2} \right )dy$$ $$=2\left [\left[\dfrac{2}{3} 3 y^{3/2}\right]_{0}^{2} -\left[\dfrac{2}{3}\dfrac{y^{3/2}}{2} \right ]_{0}^{2} \right ]$$ $$=2\left [ 2(\sqrt{8})-\dfrac{\sqrt{8}}{3} \right ]$$ $$=2\left [ \sqrt{8}\left ( \dfrac{5}{3} \right ) \right ]=\dfrac{20\sqrt{2}}{3}sq\:units.$$
$${\textbf{Step -1: Sketching the rough figure in the Coordinate plane according to the given equation.}}$$
$${\text{Firstly from the first curve}}$$
$${\text{Given function is }} = \left| {\left. {{\text{x}} - {\text{1}}} \right|} \right.$$
$$ \Rightarrow {\text{y}} = {\text{x}} - {\text{1 if x > 1}}$$
$$ = - {\text{x}} + 1{\text{ if x < 1}}$$
$${\text{For 2nd function }}\left| {\left. {\text{x}} \right| = 2} \right.$$
$$ \Rightarrow {\text{x = + 2 or x}} = - 2$$
$${\textbf{Step -2: Finding the required coordinates}}$$
$${\text{So the point will be for A }} = \left( {1,0} \right)$$
$${\text{As EF is x}} = 2{\text{ line so E }} = \left( {2,0} \right)$$
$${\text{F}} = \left( {2,1} \right)$$
$${\text{D}} = \left( { - 2,0} \right)$$
$${\text{Put }} - 2{\text{ in }}\left( { - {\text{x}} + 1} \right){\text{ so y}} = 3$$
$${\text{C}} = \left( {-2,3} \right)$$
$${\textbf{Step -3: Finding the total Area.}}$$
$${\text{Total area}} = \Delta CDA + \Delta AEF$$
$${\text{Area of those }}\Delta {\text{ will be }}\dfrac{1}{2} \times {\text{base}} \times {\text{height}}$$
$$ = \ \dfrac{1}{2} \times {\text{DA}} \times {\text{CD + }}\dfrac{1}{2} \times {\text{AF}} \times {\text{EF}}$$
$$ = \dfrac{1}{2} \times \left( {1 - \left( { - 2} \right)} \right) \times 3 + \dfrac{1}{2} \times \left( {2 - 1} \right) \times 1$$
$$ = \dfrac{1}{2} \times 3 \times 3 + \dfrac{1}{2} \times 1 \times 1$$
$$ = \dfrac{9}{2} + \dfrac{1}{2}$$
$$ = \dfrac{{10}}{2}$$
$$ = 5{\text{ square unit}}$$
$${\textbf{ Hence, the area bounced by given curves is 5 square units. Thus, option B is the}}$$
$${\textbf{correct answer.}}$$
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