The area of the region R=\{(x, y): 5 x² ≤ y ≤ 2 x²+9\} is [24 Feb 2021 Shift 2]

Correct Answer is : 12312 {3}123 square units

Correct Answer is : 12312 {3}123 square units

Test Index

Area Under The Curves Part 1

Section: Mathematics

Question : 53 of 100

Marks: +1, -0

The area of the region

R=\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}+9\}

Solution:

Given,

R=\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}+9\}

Here, we have two curves

y=5 x^{2}

and

y=2 x^{2}+9

, point of intersection of both curves is find by solving both equations i.e.

5 x^{2} \;\; =2 x^{2}+9

\Rightarrow \;\; x^{2} \;\; =3 \Rightarrow x=\pm \sqrt{3}

\therefore \;\; \text{ Area } \;\; =\int\limits_{-\sqrt{3}}^{\sqrt{3}}(2 x^{2}+9-5 x^{2}) d x

=2 \int\limits_{0}^{\sqrt{3}}(9-3 x^{2}) d x

=2[9 x-x^{3}]_{0}^{\sqrt{3}}

=2[9 \sqrt{3}-3 \sqrt{3}]

=12 \sqrt{3} \;\text{ sq units }

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