Solution:
Given,
y2 = x3
y = x3/2
\(\frac{dy}{dx} = \frac{3}{2} x^{\frac{1}{2}}\)
The arc length formula gives
L = \(\int_1^4 \sqrt{1 + \left ( \frac{dy}{dx} \right )^2 dx}\)
= \(\int_1^4 \sqrt{1 + \frac{9}{4}x dx}\)
If we substitute
u = \(1 + \frac{9}{4}x\), then \(du = \frac{9}{4} dx\)
when x = 4 then u = 10
when x = 1 then u = 13/4
Therefore
L = \(\frac{4}{9} \int_{\frac{13}{4}}^{10} \sqrt{u} \enspace du\)
= \(\left [ \frac{4}{9} . \frac{2}{3} u^{\frac{3}{2}} \right ]_{\frac{13}{3}}^{10}\)
= \(\frac{8}{27} \left [ 10^{\frac{3}{2}} – \left ( \frac{13}{4} \right )^{\frac{3}{2}} \right ]\)
= \(\frac{1}{27} \left ( 80 \sqrt{10} – 13\sqrt{13} \right )\)
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