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The whole area contained between the curve y2(a - x) = x2(a + x) and the line x = a(a > 0) is

Option: 1

2{a^2}\;\left( {1 + \frac{\pi }{4}} \right)$sq. units


Option: 2

{a^2}\;\left( {1 + \frac{\pi }{4}} \right)$sq. units


Option: 3

2{a^2}\;\left( {1 + \frac{\pi }{3}} \right)$sq. units


Option: 4

{a^2}\;\left( {1 + \frac{\pi }{3}} \right)$sq. units


Answers (1)

best_answer

As we learnt 

 

Area between two curves -

 

\\*I\! \! f \: f\left ( x \right )\geqslant g\left ( x \right )\\* in[a,c)\: \: and \: \: g\left ( x \right ) \geqslant f\left ( x \right )\:in(c,b]\\* Then\: area = \\*\\* \int_{a}^{c}\left ( f\left ( x \right )-g\left ( x \right ) \right )dx+\int_{c}^{b}\left ( g\left ( x \right )-f\left ( x \right ) \right )dx

- wherein

 

 The curve is symmetrical about the x-axis and cuts it at (-a, 0) and (a, 0).

Area =

 2\int\limits_0^a {y\,dx} = \int\limits_0^a {x\sqrt {\frac{{a + x}}{{a - x}}} \,} dx = 2\int\limits_0^a x \frac{{a + x}}{{\sqrt {{a^2} - {x^2}} }}\;dx$\\Let\: x = a \sin\theta. \:So \:that\: dx\: = a\: cos\theta d\theta A=2a^{2}\int\limits_0^{\pi /2} {(\sin \theta + {{\sin }^2}\theta )\;d\theta } $

     = 2{a^2}\;\left[ {1 + \frac{1}{2}\; \cdot \;\frac{\pi }{2}} \right] = 2{a^2}\left[ {1 + \frac{\pi }{4}} \right]$

      

Posted by

Deependra Verma

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