Find the area bounded by the $x$-axis, part of the curve $y=1+\frac{8}{x^{2}}$ and the ordinates at

Question

Find the area bounded by the $x$-axis, part of the curve $y=1+\frac{8}{x^{2}}$ and the ordinates at $x=2$ and $x=4$. If the ordinate $x=a$ divides the area into two equal parts find a.

Professor PC Thomas Classes · Answer

$y=1+\frac{8}{x^{2}} x=2 \Rightarrow y=3 	herefore A=(2,3)$\ $x=4 \Rightarrow y=3 / 2$ and $B=\left(4, \frac{3}{2}ight)$\ Shaded region in the figure is the required area\ Required area $=\int_{2}^{4} y d x=\int_{2}^{4}\left(1+\frac{8}{x^{2}}ight) d x$\ $=\left[x-\frac{8}{x}ight]_{2}^{4}=\left(4-\frac{8}{4}ight)-\left(2-\frac{8}{2}ight)=4$ sq. units.\ If $x=a$, bisects the area, then we have,\ $\int_{2}^{a} y d x=\int_{a}^{4} y d x$\ $\Rightarrow \int_{2}^{a}\left(1+\frac{8}{x^{2}}ight) d x=\int_{a}^{4}\left(1+\frac{8}{x^{2}}ight) d x$\ $\Rightarrow\left[x-\frac{8}{x}ight]_{2}^{a}=\left[x-\frac{8}{x}ight]_{a}^{4}$\ $\Rightarrow\left(a-\frac{8}{a}ight)+2=2-\left(a-\frac{8}{a}ight)$\ $\Rightarrow \quad a^{2}-8=0 \quad 	herefore \quad a= \pm 2 \sqrt{2}$\ $\Rightarrow \quad a=2 \sqrt{2}$\ $(\because a>0)$\