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  • Can someone help me with this, - Integral Calculus - JEE Main-8

Area bounded by the curves 4y = |{x^2} - 4| and y + |x|\, = 7, is equal to

  • Option 1)

    8 sq. units

  • Option 2)

    16 sq. units

  • Option 3)

    4 sq. units

  • Option 4)

    32 sq. units

 

Answers (1)

best_answer

As we learnt 

 

Area between two curves -

 

If we have two functions intersection each other.First find the point of intersection.  Then integrate to find area

\int_{o}^{a}\left [ f\left ( x \right )-9\left ( x \right ) \right ]dx

- wherein

 

 Required area = 2\left( {\int\limits_0^2 {\left( {7 - x - \left( {\frac{{4 - {x^2}}}{4}} \right)} \right)\,dx + \int\limits_2^4 {\left( {7 - x - \left( {\frac{{{x^2} - 4}}{4}} \right)} \right)\,dx} } } \right)$

                    =2\left( {\int\limits_0^2 {\left( {6 - x - \frac{{{x^2}}}{4}} \right)\,dx + \int\limits_2^4 {\left( {8 - x - \frac{{{x^2}}}{4}} \right)\,dx} } } \right)$
                    =2\left( {6x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{{12}}} \right)_0^2 + 2\left( {8x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{{12}}} \right)_2^4 = 32\: sq. units


Option 1)

8 sq. units

Option 2)

16 sq. units

Option 3)

4 sq. units

Option 4)

32 sq. units

Posted by

gaurav

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