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Find the arc length of the graph of the function f(x)=5 \sqrt{x^{3}} from x= 5 to x=6
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so i know the arclength formula is \[\int\limits_{5}^{6}\sqrt{1+f'(x)^2}\]
and f(x)=5x^(3/2) f'(x)=\[15/2\sqrt{x}\]
i get to \[\int\limits_{5}^{6}\sqrt{1+225/4(x)}dx\]
do a substitution on the inside of the sqrt u=1+225x/4
du=225/4dx?
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yep 4du/225=dx
Don't forget to change the limits too.
then you evaluate your a and b with u
\[\frac{4}{225} \int\limits_{1+\frac{225(5)}{4}}^{1+\frac{225(6)}{4}}\sqrt{u} du\]
does it then go to \[4/225\int\limits_{282.25}^{338.5}\sqrt{u}du\]
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to\[4/225(2u ^{3/2}/3)\left| \right|\]
from a->b
last part is to plug in and subtract
thanks
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