JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 8)
The area bounded by the curve y = |x2 $$-$$ 9| and the line y = 3 is :
$$4(2\sqrt 3 + \sqrt 6 - 4)$$
$$4(4\sqrt 3 + \sqrt 6 - 4)$$
$$8(4\sqrt 3 + 3\sqrt 6 - 9)$$
$$8(4\sqrt 3 + 2\sqrt 6 - 9)$$
Explanation
$$y = 3$$ and $$y = |{x^2} - 9|$$
Intersect in first quadrant at $$x = \sqrt 6 $$ and $$x = \sqrt {12} $$
_26th_June_Morning_Shift_en_8_1.png)
Required area
$$ = 2\left[ {{2 \over 3}\left( {6 \times \sqrt 6 } \right) + \int\limits_{\sqrt 6 }^3 {\left( {3 - \left( {9 - {x^2}} \right)} \right)dx + \int\limits_3^{\sqrt {12} } {\left( {3 - \left( {{x^2} - 9} \right)} \right)dx} } } \right]$$
$$ = 2\left[ {4\sqrt 6 + \left. {\left( {{{{x^3}} \over 3} - 6x} \right)} \right|_{\sqrt 6 }^3 + \left. {\left( {12x - {{{x^3}} \over 3}} \right)} \right|_3^{\sqrt {12} }} \right]$$
$$ = 2\left[ {4\sqrt 6 + \left( {4\sqrt 6 - 9} \right) + \left( {8\sqrt {12} - 27} \right)} \right]$$
$$ = 2\left[ {8\sqrt 6 + 16\sqrt 3 - 36} \right] = 8\left[ {2\sqrt 6 + 4\sqrt 3 - 9} \right]$$
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