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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    Area enclosed between the curve \[{{y}^{2}}(2a-x)={{x}^{3}}\] and line \[x=2a\] above x-axis is [MP PET 2001]

    A) \[\pi \,{{a}^{2}}\]               

    B) \[\frac{3\pi \,{{a}^{2}}}{2}\]

    C) \[2\pi \,{{a}^{2}}\]             

    D) \[3\pi \,{{a}^{2}}\]

    Correct Answer: B

    Solution :

    • Curve \[{{y}^{2}}(2a-x)={{x}^{3}}\] is symmetrical about x-axis and passes through origin.  Also \[\frac{{{x}^{3}}}{2a-x}<0\]for \[x>2a\]or \[x<0\]. So curve does not lie in \[x>2a\]and \[x<0,\] curve lies wholly on \[0\le x\le 2a\].
    • \[\therefore \]Area \[=\int_{0}^{2a}{\frac{{{x}^{3/2}}}{\sqrt{2a-x}}dx}\]\[=\int_{0}^{\pi /2}{8\,{{a}^{2}}{{\sin }^{4}}\theta \,d\theta }\],   (Put \[x=2a\,{{\sin }^{2}}\theta )\]                  
    • \[=8{{a}^{2}}\left[ \frac{3}{4}.\frac{1}{2}.\frac{\pi }{2} \right]\]\[=\frac{3\pi {{a}^{2}}}{2}\], (Applying Gamma function).


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