JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 11)
The area of the region : $$R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\} $$ is :
$$6\sqrt 3 $$ square units
$$12\sqrt 3 $$ square units
$$11\sqrt 3 $$ square units
$$9\sqrt 3 $$ square units
Explanation
_24th_February_Evening_Shift_en_11_1.png)
Required area
$$ = 2\int\limits_0^{\sqrt 3 } {\left( {2{x^2} + 9 - 5{x^2}} \right)} dx$$
$$ = 2\int\limits_0^{\sqrt 3 } {\left( {9 - 3{x^2}} \right)dx} $$
$$ = 2|\,9x - {x^3}\,|_0^{\sqrt 3 } = 12\sqrt 3 $$
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