SOLUTION 1: $ \ \ $ If $ y=3x+2 $ for $ -1 \le x \le 2 $, then $ \displaystyle{ { dy \over dx} = 3 } $
so that
$$ ARC = \displaystyle{ \int_{-1}^{2} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx } $$
$$ = \displaystyle{ \int_{-1}^{2} \sqrt{ 1 + (3)^2 } \ dx } $$
$$ = \displaystyle{ \int_{-1}^{2} \sqrt{ 10 } \ dx } $$
$$ = \displaystyle{ \sqrt{ 10 } x } \ \Big\vert_{-1}^{2} $$
$$ = \sqrt{10} \ (2 - (-1)) $$
$$ = 3\sqrt{10} $$
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