Solutions - 0080-01

(1) Integral formula for arclength:

$$S={\int }_c^d\sqrt{1+{(g^{\prime })}^2}dx$$

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(2) Work out integrand:

$$\begin{array}{c}g(x)=\frac{1}{12}{y}^3+y^{-1}\gg \gg g^{\prime }(x)=\frac{1}{4}{y}^2-y^{-2}\\ \\ {(g^{\prime })}^2\gg \gg \frac{1}{16}{y}^4-\frac{1}{2}+y^{-4}\\ \\ 1+{(g^{\prime })}^2\gg \gg \frac{1}{16}{y}^4+\frac{1}{2}+y^{-4}\gg \gg {(\frac{1}{4}{y}^2+y^{-2})}^2\\ \\ \sqrt{1+{(g^{\prime })}^2}\gg \gg \frac{1}{4}{y}^2+y^{-2}\end{array}$$

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(3) Integrate:

$$\begin{array}{c}S\gg \gg {\int }_1^2\frac{1}{4}{y}^2+y^{-2}dy\\ \\ \gg \gg {\frac{1}{12}{y}^3-y^{-1}|}_1^2\gg \gg \frac{13}{12}\end{array}$$