The area included between the parabola $y=\frac{x^2}{4 a}$ and the curve $y=\frac{8 a^3}{\left(x^2+4 a^2\right)}$ is

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Aicharya · Accepted Answer

Given curves, $\begin{aligned} & y=\frac{x^2}{4 a} \ & y=\frac{8 a^3}{x^2+4 a^2} \end{aligned}$ For point of intersection equation (i) and (ii), $\begin{aligned} & \frac{x^2}{4 a}=\frac{8 a^3}{x^2+4 a^2} \ & x^2\left(x^2+4 a^2ight)=4 a\left(8 a^3ight) \ & x^4+4 a^2 x^2-32 a^2=0 \ & \left(x^2-4 a^2ight)\left(x^2+8 a^2ight)=0 \ & x^2-4 a^2=0 	ext { or } x^2+8 a^2=0 \ & x^2=4 a^2 \quad\left[x^2=-8 a^2 	ext { is not possible }ight] \ & x= \pm 2 a \quad \Rightarrow x= \pm 2 a \end{aligned}$ So, we take limits from 0 to $2 a$. Now, Area enclosed by the two curves $\mathrm{A}=2 	imes \int_0^{2 a}\left[\frac{8 a^3}{\left(x^2+4 a^2ight)}-\frac{x^2}{4 a}ight] d x$ $=2 	imes\left[\int_0^{2 a} \frac{8 a^3}{\left(x^2+4 a^2ight)} d x-\int_0^{2 a} \frac{x^2}{4 a} d xight]$ $\begin{aligned} & =2 	imes\left[8 a^3 	imes \int_0^{2 a} \frac{1}{x^2+\left(2 a^2ight)} d x-\frac{1}{4 a} \int_0^{2 a} x^2 d xight] \ & =2 	imes\left\{8 a^3\left[\frac{1}{2 a} 	an ^{-1}\left(\frac{x}{2 a}ight)ight]_0^{2 a}-\frac{1}{4 a}\left[\frac{x^3}{3}ight]_0^{2 a}ight\} \ & =2 	imes\left\{\frac{8 a^3}{2 a}\left[	an ^{-1}\left(\frac{2 a}{2 a}ight)-	an ^{-1}\left(\frac{0}{2 a}ight)ight]ight. \ & =2 	imes\left\{4 a^2\left[	an ^{-1}(1)-	an ^{-1}(0)ight]-\frac{1}{4 a}\left(\frac{8 a^3}{3}-0ight)ight\} \ & =2 	imes\left\{4 a^2\left(\frac{\pi}{4}-0ight)-\frac{1}{4 a}\left(\frac{8 a^3}{3}ight)ight\} \ & =2 	imes\left\{4 a^2 	imes \frac{\pi}{4}-\frac{1}{4 a} 	imes \frac{8 a^3}{3}ight\} \ & =2\left\{a^2\left(\pi-\frac{2}{3}ight)ight\} \ & 	herefore \quad \mathrm{A}=a^2\left(2 \pi-\frac{4}{3}ight) \mathrm{sq} . 	ext { units } \end{aligned}$

The area included between the parabola $y=\frac{x^2}{4 a}$ and the curve $y=\frac{8 a^3}{\left(x^2+4…

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