What is the arc length of $f (x) = 6 x^{\frac{3}{2}} + 1$ $f(x)=6x^(3/2)+1$ on $x \in [5, 7]$ $x in [5,7]$ ?

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Answer 1 · 2018-03-19T17:41:15

$L = \frac{4}{243} (568 \sqrt{142} - 203 \sqrt{406})$ $L=4/243(568sqrt142-203sqrt406)$ units.

$f (x) = 6 x^{\frac{3}{2}} + 1$ $f(x)=6x^(3/2)+1$

$f'(x)=9sqrtx$

Arc length is given by:

$L=int_5^7sqrt(1+81x)dx$

Integrate directly:

$L=2/243[(1+81x)^(3/2)]_5^7$

Hence

$L=4/243(568sqrt142-203sqrt406)$

What is the arc length of f(x)=6x^(3/2)+1 f(x)=6x32+1f(x)=6x^(3/2)+1 on x in [5,7]x∈[5,7]x in [5,7]?