What is the arclength of $f (x) = (x - 3) e^{x} - x ln (\frac{x}{2})$ $f(x)=(x-3)e^x-xln(x/2)$ on $x \in [2, 3]$ $x in [2,3]$ ?

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Answer 1 · 2018-07-13T17:53:12

$\approx 6.533320817$ $approx 6.533320817$

We are using the Formula

$int_a^bsqrt(1+f'(x)^2)dx$
$f(x)=(x-3)e^x-xln(x/2)$
we Need the first derivative of $f$

$f'(x)=e^x+e^x(x-3)-ln(x/2)-x*(1/(x/2)*1/2)$

simplifying we get

$f'(x)=e^x(x-2)-ln(x/2)-1$
so we have to calculate

$int_2^3sqrt(1+(e^x(x-2)-ln(x/2)-1)^2)dx$
by a numerical method we get

$approx 6.533320817$

What is the arclength of f(x)=(x-3)e^x-xln(x/2)f(x)=(x−3)ex−xln(x2)f(x)=(x-3)e^x-xln(x/2) on x in [2,3]x∈[2,3]x in [2,3]?