Question

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

• A. $\frac{11}{3}$
• B. $\frac{13}{3}$
• C. $\frac{14}{3}$
• D. $\frac{16}{3}$

Answer 1

The correct option is C $\frac{14}{3}$

${\int }_{\frac{1}{2}}^{2}f\left(x\right)dx={\int }_{\frac{1}{2}}^{\frac{1}{\sqrt{3}}}\frac{1}{x^2}dx+{\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}3dx+{\int }_{\sqrt{3}}^2{x}^{2}dx={\int }_{\frac{1}{2}}^{\frac{1}{\sqrt{3}}}\frac{-1}{x}+{\int }_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}3x+{\int }_{\sqrt{3}}^2\frac{x^3}{3}=-(\sqrt{3}-2)+3\sqrt{3}-\sqrt{3}+\frac{8}{3}-\sqrt{3}=2+\frac{8}{3}=\frac{14}{3}$