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:

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

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

$L=int_2^4dx$

$L=x]_2^4$

$L=2$

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