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 $a =$

• A. 8
• B. $2\sqrt{2}$
• C. 2
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$2\sqrt{2}$

Answer 2

Let the ordinate at $x = a$divide the area into two equal parts

Area of $AMNB$

$= {\int_{2}^{4}{\left( 1 + \frac{8}{x^{2}} \right)\ dx = \left\lbrack x - \frac{8}{x} \right\rbrack}}_{2}^{4} = 4$

Area of $ACDM = \int_{2}^{a}\left( 1 + \frac{8}{x^{2}} \right)dx = 2$

On solving, we get $a = \pm 2\sqrt{2}$;Since $a > 0$⇒ $a = 2\sqrt{2}$.