What is the arc length of #f(x)=-xln(1/x)-xlnx# on #x in [3,5]#?

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Answer 1 · 2017-03-13T03:45:08

#2#

Use the rule #log(a^b)=blog(a)# to simplify the function:

#f(x)=-xln(x^-1)-xln(x)#

Bringing the #-1# out:

#f(x)=xln(x)-xln(x)#

#f(x)=0#

This is the straight line #y=0#. Thus its arc length on #x in [3,5]# is just the line segment with length #2#.

Using #f(x)=0# we can apply the arc length formula for #f# on #x in [a,b]# for the same result:

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

#L=int_2^4sqrt(1+0)dx#

#L=int_2^4dx#

#L=x]_2^4#

#L=2#