What is the arclength of $f (x) = \frac{x^{2} + 24 x + 1}{x^{2}}$ $f(x)=(x^2+24x+1)/x^2$ in the interval $[1, 3]$ $[1,3]$ ?

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Answer 1 · 2018-07-12T20:23:36

$\approx 17.060431259$ $approx 17.060431259$

Given $f (x) = \frac{x^{2} + 24 x + 1}{x^{2}}$ $f(x)=(x^2+24x+1)/x^2$
we differentiate this function with respect to $x$ $x$ using the Quotient rule

$(u/v)'=(u'v-uv')/v^2$

so we get

$f'(x)=((2x+24)x^2-(x^2+24x+1)2x)/x^4$
cancelling $xne 0$

$f'(x)=(2x^2+24x-2x^2-48x-2)/x^3$

$f'(x)=(-2(12x+1))/x^3$

so we get

$int_1^3sqrt(1+(-2(12x+1)/x^3)^2)dx$
by a numerical method we get $17.06031259$

What is the arclength of f(x)=(x^2+24x+1)/x^2 f(x)=x2+24x+1x2f(x)=(x^2+24x+1)/x^2 in the interval [1,3][1,3][1,3]?