Area Question 52
Question 52
- 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 at $x=a$ divides the area into two equal parts, then find $a$.
(1983, 3M)
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Solution:
- Here, $\int_{2}^{a} 1+\frac{8}{x^{2}} d x=\int_{a}^{4} 1+\frac{8}{x^{2}} d x$
$$ \begin{array}{lc} \Rightarrow & x-\frac{8}{x}^{a}=x-\frac{8}{x}_{a}^{4} \ \Rightarrow \quad a-\frac{8}{a}-(2-4)=(4-2)-a-\frac{8}{a} \end{array} $$
$$ \begin{aligned} & \Rightarrow \quad a-\frac{8}{a}+2=2-a+\frac{8}{a} \Rightarrow 2 a-\frac{16}{a}=0 \ & \Rightarrow \quad 2\left(a^{2}-8\right)=0 \ & \Rightarrow \quad a= \pm 2 \sqrt{2} \quad \text { [neglecting }- \text { ve sign] } \ & \therefore \quad a=2 \sqrt{2} \end{aligned} $$
