Area Ques 64
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 at $x=a$ divides the area into two equal parts, then find $a$.
(1983, 3M)
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Answer:
Correct Answer: 64.$(2 \sqrt{2})$
Solution:
Formula:
- Here, $\int _{2}^{a} (1+\frac{8}{x^{2}}) d x=\int _{a}^{4} (1+\frac{8}{x^{2}}) d x$
$\Rightarrow \quad [x-\frac{8}{x}] _{2}^{a}=x-[\frac{8}{x}] _{a}^{4} $
$\Rightarrow \quad ( a-\frac{8}{a})-(2-4)=(4-2)-(a-\frac{8}{a})$
$\Rightarrow \quad a-\frac{8}{a}+2=2-a+\frac{8}{a} $
$\Rightarrow \quad 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] } $
$\therefore \quad a=2 \sqrt{2}$
BETA
