SOLUTION 2: $ \ \ $ If $ y = x^{3/2} $ for $ 0 \le x \le 4 $, then $ \displaystyle{ { dy \over dx} = (3/2)x^{1/2} } $
so that
$$ ARC = \displaystyle{ \int_{0}^{4} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx } $$
$$ = \displaystyle{ \int_{0}^{4} \sqrt{ 1 + ((3/2)x^{1/2})^2 } \ dx } $$
$$ = \displaystyle{ \int_{0}^{4} \sqrt{ 1 + ((3/2)^2(x^{1/2})^2 } \ dx } $$
$$ = \displaystyle{ \int_{0}^{4} \sqrt{ 1 + (9/4)x } \ dx } $$
(Now integrate using the method of u-substitution. Let $ u= 1 + (9/4)x \ \longrightarrow \ du= (9/4)dx \ \longrightarrow \ (4/9)du = dx \. $)
$$ = \displaystyle{ (4/9) \int_{x=0}^{x=4} \sqrt{ u } \ du } $$
$$ = \displaystyle{ (4/9)(2/3) u^{3/2} \ \Big\vert_{x=0}^{x=4} } $$
$$ = \displaystyle{ (8/27) (1 + (9/4)x)^{3/2} \ \Big\vert_{x=0}^{x=4} } $$
$$ = \displaystyle{ (8/27) (1 + (9/4)(4))^{3/2} - (8/27) (1 + (9/4)(0))^{3/2} } $$
$$ = \displaystyle{ (8/27) (10)^{3/2} - (8/27) (1)^{3/2} } $$
$$ = \displaystyle{ (8/27) 10^{3/2} - 8/27 } $$
$$ = \displaystyle{ (8/27) ( 10^{3/2} - 1 ) } $$
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