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Q.
The area bounded by the curves $x^{2}+y^{2}=25,4 y=\left|4-x^{2}\right|$ and $x=0$, above $x$ - axis is
Application of Integrals
Solution:
The required area
$=\int\limits_{0}^{4} \sqrt{25-x^{2}} d x-\int\limits_{0}^{2} \frac{4-x^{2}}{4} d x-\int\limits_{2}^{4} \frac{x^{2}-4}{4} d x$
$=\left[\frac{x}{2} \sqrt{25-x^{2}}+\frac{25}{2} \sin ^{-1} \frac{x}{5}\right]_{0}^{4}-\frac{1}{4}\left[4 x-\frac{x^{3}}{3}\right]_{0}^{2}-\frac{1}{4}\left[\frac{x^{3}}{3}-4 x\right]_{2}^{4}$
$\quad=2+\frac{25}{2} \sin ^{-1} \frac{4}{5}$
