What is the arclength of $f(x)=xcos(x-2)$ on $x in [1,2]$ ?

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Answer 1 · 2018-04-22T01:14:11

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the solution is

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

$f'(x)=cos(x-2)-xsin(x-2)$

$L=int_1^2sqrt(1+[cos(x-2)-xsin(x-2)]^2)*dx$

$L=int_1^2sqrt(1+cos^2(x-2)-2xcos(x-2)*sin(x-2)+x^2sin^2(x-2))*dx$

then complete the steps manually

What is the arclength of f(x)=xcos(x-2)f(x)=xcos(x-2) on x in [1,2]x in [1,2]?