The area bounded by the curve y = |x2 - 9| and the line y = 3 is :

Question

Accepted Answer

$y = 3$ and $y = |{x^2} - 9|$

Intersect in first quadrant at $$x = \sqrt 6 $$ and $$x = \sqrt {12} $$

Required area

$$ = 2\left[ {{2 \over 3}\left( {6 \times \sqrt 6 } \right) + \int\limits_{\sqrt 6 }^3 {\left( {3 - \left( {9 - {x^2}} \right)} \right)dx + \int\limits_3^{\sqrt {12} } {\left( {3 - \left( {{x^2} - 9} \right)} \right)dx} } } \right]$$

$$ = 2\left[ {4\sqrt 6  + \left. {\left( {{{{x^3}} \over 3} - 6x} \right)} \right|_{\sqrt 6 }^3 + \left. {\left( {12x - {{{x^3}} \over 3}} \right)} \right|_3^{\sqrt {12} }} \right]$$

$$ = 2\left[ {4\sqrt 6  + \left( {4\sqrt 6  - 9} \right) + \left( {8\sqrt {12}  - 27} \right)} \right]$$

$$ = 2\left[ {8\sqrt 6  + 16\sqrt 3  - 36} \right] = 8\left[ {2\sqrt 6  + 4\sqrt 3  - 9} \right]$$

Solutions · 1