Categories > definite-integral > Area bounded by region > If the ordinate $x=a$ divides the area bounded by the curve $y=\left( 1+\frac{8}{{{x}^{2}}} \right)\,,$ $x-$axis and the ordinates $x=2,$ $x=4$ into two equal parts, then $a=$

definite-integral

Question: If the ordinate $x=a$ divides the area bounded by the curve $y=\left( 1+\frac{8}{{{x}^{2}}} \right)\,,$ $x-$axis and the ordinates $x=2,$ $x=4$ into two equal parts, then $a=$


1) 8
2) $2\sqrt{2}$
3) 2
4) $\sqrt{2}$
Solution: Explanation: No Explanation
Area bounded by region Volume and surface area of solids of revolution

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