The area common to the curves 5x^2 - y = 0 an | vedclass

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(A) Given curves are $y = 5x^2$ and $y = 2x^2 + 9$.To find the intersection points,set $5x^2 = 2x^2 + 9$,which gives $3x^2 = 9$,so $x^2 = 3$,which mea

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$12\sqrt{3}$

Vedclass · Answer

(A) Given curves are $y = 5x^2$ and $y = 2x^2 + 9$.To find the intersection points,set $5x^2 = 2x^2 + 9$,which gives $3x^2 = 9$,so $x^2 = 3$,which means $x = \pm\sqrt{3}$.By symmetry,the area is $2 \int_{0}^{\sqrt{3}} \{(2x^2 + 9) - 5x^2\} dx$.$= 2 \int_{0}^{\sqrt{3}} (9 - 3x^2) dx$.$= 2 [9x - x^3]_{0}^{\sqrt{3}}$.$= 2 [9\sqrt{3} - (\sqrt{3})^3] = 2 [9\sqrt{3} - 3\sqrt{3}] = 2 [6\sqrt{3}] = 12\sqrt{3}$ sq. units.

The area common to the curves $5x^2 - y = 0$ and $2x^2 - y + 9 = 0$ is equal to

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