Solutions - 0080-02

(1) Integral formula for arclength:

$$L={\int }_a^b\sqrt{1+{\left(f^{\prime }(x)\right)}^2}dx$$

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(2) Calculate, Simplify the integrand:

$$\begin{array}{cc}f(x)& =\frac{1}{8}{x}^4+\frac{1}{4x^2}\\ & \\ \gg \gg f^{\prime }(x)& =\frac{1}{2}{x}^3-\frac{1}{2}{x}^{-3}\\ & \\ \gg \gg {(f^{\prime })}^2& =\frac{1}{4}{x}^6-\frac{1}{2}+\frac{1}{4}{x}^{-6}\\ & \\ \gg \gg 1+{(f^{\prime })}^2& =\frac{1}{4}{x}^6+\frac{1}{2}+\frac{1}{4}{x}^{-6}\\ & \\ & ={(\frac{1}{2}{x}^3+\frac{1}{2}{x}^{-3})}^2\end{array}$$

Therefore, the integrand is:

$$\sqrt{1+{(f^{\prime })}^2}\gg \gg \frac{1}{2}{x}^3+\frac{1}{2}{x}^{-3}$$

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(3) Integrate:

$$\begin{array}{c}L\gg \gg \int \frac{1}{2}{x}^3+\frac{1}{2}{x}^{-3}dx\\ \\ \gg \gg {\frac{1}{8}{x}^4-\frac{1}{4}{x}^{-2}|}_1^2\\ \\ \gg \gg (2-\frac{1}{16})-(\frac{1}{8}-\frac{1}{4})\\ \\ \gg \gg \frac{33}{16}\end{array}$$