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

Question

1242 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-12T20:23:36

#approx 17.060431259#

Given #f(x)=(x^2+24x+1)/x^2#
we differentiate this function with respect to #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#