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(B) Let the ordinate at $x = a$ divide the area into two equal parts.The total area $A$ bounded by the curve $y = 1 + \frac{8}{x^2}$,the $x$-axis,and

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$2\sqrt{2}$

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(B) Let the ordinate at $x = a$ divide the area into two equal parts.The total area $A$ bounded by the curve $y = 1 + \frac{8}{x^2}$,the $x$-axis,and the lines $x = 2$ and $x = 4$ is given by:$A = \int_{2}^{4} \left( 1 + \frac{8}{x^2} \right) dx$$A = \left[ x - \frac{8}{x} \right]_{2}^{4}$$A = \left( 4 - \frac{8}{4} \right) - \left( 2 - \frac{8}{2} \right) = (4 - 2) - (2 - 4) = 2 - (-2) = 4$ square units.Since the ordinate $x = a$ divides this area into two equal parts,the area from $x = 2$ to $x = a$ must be half of the total area,which is $4 / 2 = 2$.$\int_{2}^{a} \left( 1 + \frac{8}{x^2} \right) dx = 2$$\left[ x - \frac{8}{x} \right]_{2}^{a} = 2$$\left( a - \frac{8}{a} \right) - \left( 2 - \frac{8}{2} \right) = 2$$a - \frac{8}{a} - (2 - 4) = 2$$a - \frac{8}{a} + 2 = 2$$a - \frac{8}{a} = 0$$a^2 - 8 = 0$$a^2 = 8$$a = \pm 2\sqrt{2}$.Since the region is in the first quadrant where $x > 0$,we have $a = 2\sqrt{2}$.

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

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