What is the arclength of #f(x)=x+xsqrt(x+3)# on #x in [-3,0]#?

1 Answer
Sharkbasket
Mar 25, 2018

#s~~6.54696#

Explanation:

The formula for arc length of a curve defined by #f(x)# is:

#s=int_a^bsqrt(1+f'(x)^2)#

First let's find the derivative #f'(x)#:

#f(x)=x+x(x+3)^(1/2)#

#f'(x)=1+(x+3)^(1/2)+1/2x(x+3)^(-1/2)# (by the product rule)

#f'(x)=1+sqrt(x+3)+x/(2sqrt(x+3))#

Now let's evaluate #f'(x)^2#

#f'(x)^2=[1+sqrt(x+3)+x/(2sqrt(x+3))]^2#

#=1+sqrt(x+3)+x/(2sqrt(x+3))+sqrt(x+3)+(x+3)+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x/(2sqrt(x+3))+(xcancel(sqrt(x+3)))/(2cancel(sqrt(x+3)))+x^2/(4(x+3))#

#=1+2sqrt(x+3)+cancel(2)x/(cancel(2)sqrt(x+3))+(x+3)+cancel(2)(x/cancel(2))+x^2/(4(x+3))#

#=x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+4#

Now you would need to evaluate this (very nasty) integral:

#int_-3^0sqrt(x^2/(4(x+3))+x(2+1/sqrt(x+3))+2sqrt(x+3)+5)#

I punched it into Wolfram|Alpha and got the following result:

enter image source here

So the answer is

#s~~6.54696#

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