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Let $\mathrm{f}(\mathrm{x})=\max \left\{3, \mathrm{x}^{2}, \frac{1}{\mathrm{x}^{2}}\right\}$ for $\frac{1}{2} \leq \mathrm{x} \leq 2 .$ Then the value of the integral $\int_{1 / 2}^{2} f(x) d x$ is
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The correct answer is:
$\frac{14}{3}$
Given Integral can be distributed into
$\int_{1 / 2}^{1 / \sqrt{3}} \frac{1}{x^{2}} \cdot d x+\int_{1 / \sqrt{3}}^{\sqrt{3}} 3 d x+\int_{\sqrt{3}}^{2} x^{2} \cdot d x=\frac{14}{3}$
$\int_{1 / 2}^{1 / \sqrt{3}} \frac{1}{x^{2}} \cdot d x+\int_{1 / \sqrt{3}}^{\sqrt{3}} 3 d x+\int_{\sqrt{3}}^{2} x^{2} \cdot d x=\frac{14}{3}$
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