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  4. Area bounded by x2 - 4ay and y = (8a3/x2 + 4 a2) is

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Q. Area bounded by $x^2 - 4ay $ and $y = \frac{8a^3}{x^2 + 4 a^2}$ is

Application of Integrals

Solution:

by equating $ \frac{x^{2}}{4a} = \frac{8a^{3}}{x^{2} + 4a^{2}} $
$ \Rightarrow \pm2a$
Area $ = 2 \int^{2a}_{0} \left(\frac{8a^{3}}{x^{2}+4a^{2}} - \frac{x^{2}}{4a}\right)dx $