The value of a (a 0) for which the area bou | vedclass

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(A) The area $A$ is given by the integral:$A = \int_{a}^{2a} \left( \frac{x}{6} + \frac{1}{x^2} \right) dx$Evaluating the integral:$A = \left[ \frac{x

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(A) The area $A$ is given by the integral:$A = \int_{a}^{2a} \left( \frac{x}{6} + \frac{1}{x^2} \right) dx$Evaluating the integral:$A = \left[ \frac{x^2}{12} - \frac{1}{x} \right]_{a}^{2a}$$A = \left( \frac{(2a)^2}{12} - \frac{1}{2a} \right) - \left( \frac{a^2}{12} - \frac{1}{a} \right)$$A = \left( \frac{4a^2}{12} - \frac{1}{2a} \right) - \left( \frac{a^2}{12} - \frac{1}{a} \right)$$A = \frac{3a^2}{12} - \frac{1}{2a} + \frac{1}{a}$$A = \frac{a^2}{4} + \frac{1}{2a}$Let $f(a) = \frac{a^2}{4} + \frac{1}{2a}$. To find the minimum area,we find the derivative $f'(a)$ and set it to $0$:$f'(a) = \frac{2a}{4} - \frac{1}{2a^2} = \frac{a}{2} - \frac{1}{2a^2}$Setting $f'(a) = 0$:$\frac{a}{2} = \frac{1}{2a^2}$$a^3 = 1$$a = 1$Since $f''(a) = \frac{1}{2} + \frac{1}{a^3} > 0$ for $a > 0$,the function has a minimum at $a = 1$.

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$,and $x = 2a$ has the least value,is

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