Missing 4 in solution to Example 14.1.3

[0-andy-0]

Typo: 1+4t^2 on line 214 became 1+t^2 on line 218 of 14_Line_Integral_Intro.tex

214: {We parametrize our curve $C$ as $\vrt = \langle t,t^2\rangle$ for $-1\leq t\leq 1$; we find $\norm{\vrp(t)} = \sqrt{1+4t^2}$, so $ds = \sqrt{1+4t^2}\ dt$.

218: \int_C f(s)\ ds &= \int_{-1}^1 \Big(1-\cos(t)\sin\big(t^2\big)\Big)\sqrt{1+t^2}\ dt.\\

[APEXCalculus]

Yes - thanks.

[Gomesz785]

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