If the ordinate x-a divides the area bounded by the curve y-1-8-x2- x - axis and the ordinates x-2- x-4 into two equal parts- then a-A- 8B- 2 - 2 -2D- -2

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Area between Two Curves

If the ordina...

Question

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

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2 \sqrt{2}

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\sqrt{2}

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Solution

The correct option is B $2 \sqrt{2}$
Let the ordinate at x=a divide the area into two equal parts

Area of AMNB = $\int_{2}^{4} (1 + \frac{8}{x^{2}}) d x = {[x - \frac{8}{x}]}_{2}^{4} = 4$
Area of ACDM = $\int_{2}^{a} (1 + \frac{8}{x^{2}}) d x = 2$
On solving, we get $a = \pm 2 \sqrt{2}$ ; Since $a > 0 \Rightarrow a = 2 \sqrt{2}$ .

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