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

1 Answer
Rui D.
Apr 4, 2016

#47651/1920~=24.818#

Explanation:

#f(x)=(4x^5)/5+1/(48x^3)-1#
#f"'"(x)=4x^4-1/(16x^4)#

#L=int_a^b sqrt(1+[f"'"(x)]^2)*dx#

#L=int_1^2 sqrt(1+(4x^4-1/(16x^4))^2)*dx#
#L=int_1^2 sqrt(1+16x^8-1/2+1/(256x^8))*dx#
#L=int_1^2 sqrt(16x^8+1/2+1/(256x^8))*dx#
#L=int_1^2 sqrt((4x^4+1/(16x^4))^2)*dx#
#L=int_1^2 (4x^4+1/(16x^4))*dx#
#L=(4/5x^5-1/(48x^3))|_1^2#
#L=(4*32)/5-1/(48*8)-(4/5-1/48)#
#L=(128-4)/5-(1-8)/384=124/5+7/384=(47616+35)/1920#
#L=47651/1920~=24.818#

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