Let f(x) = max left{3, x^2, frac{1}{x^2}right | vedclass

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(C) We are given $f(x) = \max \left\{3, x^2, \frac{1}{x^2}\right\}$ for $x \in \left[\frac{1}{2}, 2\right]$.To evaluate the integral,we determine the

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$\frac{14}{3}$

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(C) We are given $f(x) = \max \left\{3, x^2, \frac{1}{x^2}\right\}$ for $x \in \left[\frac{1}{2}, 2\right]$.To evaluate the integral,we determine the intervals where each function is the maximum:$1$. For $x \in \left[\frac{1}{2}, \frac{1}{\sqrt{3}}\right]$,$\frac{1}{x^2} \geq 3$ and $\frac{1}{x^2} \geq x^2$,so $f(x) = \frac{1}{x^2}$.$2$. For $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$,$3 \geq x^2$ and $3 \geq \frac{1}{x^2}$,so $f(x) = 3$.$3$. For $x \in \left[\sqrt{3}, 2\right]$,$x^2 \geq 3$ and $x^2 \geq \frac{1}{x^2}$,so $f(x) = x^2$.Thus,the integral is:$\int_{1/2}^2 f(x) dx = \int_{1/2}^{1/\sqrt{3}} \frac{1}{x^2} dx + \int_{1/\sqrt{3}}^{\sqrt{3}} 3 dx + \int_{\sqrt{3}}^2 x^2 dx$$= \left[-\frac{1}{x}\right]_{1/2}^{1/\sqrt{3}} + [3x]_{1/\sqrt{3}}^{\sqrt{3}} + \left[\frac{x^3}{3}\right]_{\sqrt{3}}^2$$= (-\sqrt{3} - (-2)) + (3\sqrt{3} - \sqrt{3}) + \left(\frac{8}{3} - \frac{3\sqrt{3}}{3}\right)$$= 2 - \sqrt{3} + 2\sqrt{3} + \frac{8}{3} - \sqrt{3} = 2 + \frac{8}{3} = \frac{14}{3}$.

Let $f(x) = \max \left\{3, x^2, \frac{1}{x^2}\right\}$ for $\frac{1}{2} \leq x \leq 2$. Then,the value of the integral $\int_{1/2}^2 f(x) dx$ is

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