What is the arc length of $f (x) = (\frac{4 x^{5}}{5}) + (\frac{1}{48 x^{3}}) - 1$ $f(x)=((4x^5)/5) + (1/(48x^3)) - 1$ on $x \in [1, 2]$ $x in [1,2]$ ?

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What is the arc length of $f (x) = (\frac{4 x^{5}}{5}) + (\frac{1}{48 x^{3}}) - 1$ $f(x)=((4x^5)/5) + (1/(48x^3)) - 1$ on $x \in [1, 2]$ $x in [1,2]$ ?

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Answer 1 · 2016-04-04T19:23:25

$\frac{47651}{1920} ≅ 24.818$ $47651/1920~=24.818$

Explanation:

$f (x) = \frac{4 x^{5}}{5} + \frac{1}{48 x^{3}} - 1$ $f(x)=(4x^5)/5+1/(48x^3)-1$
$f' (x) = 4 x^{4} - \frac{1}{16 x^{4}}$ $f"'"(x)=4x^4-1/(16x^4)$

$L = \int_{a}^{b} \sqrt{1 + {[f' (x)]}^{2}} \cdot d x$ $L=int_a^b sqrt(1+[f"'"(x)]^2)*dx$

$L = \int_{1}^{2} \sqrt{1 + {(4 x^{4} - \frac{1}{16 x^{4}})}^{2}} \cdot d x$ $L=int_1^2 sqrt(1+(4x^4-1/(16x^4))^2)*dx$
$L = \int_{1}^{2} \sqrt{1 + 16 x^{8} - \frac{1}{2} + \frac{1}{256 x^{8}}} \cdot d x$ $L=int_1^2 sqrt(1+16x^8-1/2+1/(256x^8))*dx$
$L = \int_{1}^{2} \sqrt{16 x^{8} + \frac{1}{2} + \frac{1}{256 x^{8}}} \cdot d x$ $L=int_1^2 sqrt(16x^8+1/2+1/(256x^8))*dx$
$L = \int_{1}^{2} \sqrt{{(4 x^{4} + \frac{1}{16 x^{4}})}^{2}} \cdot d x$ $L=int_1^2 sqrt((4x^4+1/(16x^4))^2)*dx$
$L = \int_{1}^{2} (4 x^{4} + \frac{1}{16 x^{4}}) \cdot d x$ $L=int_1^2 (4x^4+1/(16x^4))*dx$
$L = (\frac{4}{5} x^{5} - \frac{1}{48 x^{3}}) {∣ ∣ ∣}_{1}^{2}$ $L=(4/5x^5-1/(48x^3))|_1^2$
$L = \frac{4 \cdot 32}{5} - \frac{1}{48 \cdot 8} - (\frac{4}{5} - \frac{1}{48})$ $L=(4*32)/5-1/(48*8)-(4/5-1/48)$
$L = \frac{128 - 4}{5} - \frac{1 - 8}{384} = \frac{124}{5} + \frac{7}{384} = \frac{47616 + 35}{1920}$ $L=(128-4)/5-(1-8)/384=124/5+7/384=(47616+35)/1920$
$L = \frac{47651}{1920} ≅ 24.818$ $L=47651/1920~=24.818$

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What is the arc length of $f (x) = (\frac{4 x^{5}}{5}) + (\frac{1}{48 x^{3}}) - 1$ $f(x)=((4x^5)/5) + (1/(48x^3)) - 1$ on $x \in [1, 2]$ $x in [1,2]$ ?

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What is the arc length of $f (x) = (\frac{4 x^{5}}{5}) + (\frac{1}{48 x^{3}}) - 1$ $f(x)=((4x^5)/5) + (1/(48x^3)) - 1$ on $x \in [1, 2]$ $x in [1,2]$ ?