If $$x=a$$ divides the area bounded by x $$-$$ axis, part of the curve $$y=1+8x2$$ and the ordinates $$x=2$$,$$x=4$$ into equal parts then $$a=$$

∫$$241+8x2dx=2$$∫$$2a1+8x2$$ ⇒$$x-8x24=2x-8x2a$$ ⇒$$a=22=2.82$$

Area of bounded Regions

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Question

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

Moderate

Solution

$\int_{2}^{4} (1 + \frac{8}{x^{2}}) d x = 2 \int_{2}^{a} (1 + \frac{8}{x^{2}}) \Rightarrow {(x - \frac{8}{x})}_{2}^{4} = 2 {(x - \frac{8}{x})}_{2}^{a} \Rightarrow a = 2 \sqrt{2} = 2.82$

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