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

1 Answer
Steve M
Nov 15, 2016

# f(x) = (x-2)/x^2 #

# :. f'(x) = ( (x^2)(1) - (x-2)(2x) ) / (x^2)^2 #
# :. f'(x) = ( x^2 - 2x^2 +4x ) / x^4 #
# :. f'(x) = ( 4x-x^2 ) / x^4 #
# :. f'(x) = ( 4-x ) / x^3 #

Arc Length is given by:
# L = int_(-2)^(-1) sqrt (1+f'(x)^2) dx #
# :. L = int_(-2)^(-1) sqrt (1+(( 4-x ) / x^3)^2) dx #
# :. L = int_(-2)^(-1) sqrt (1+( 4-x )^2 / x^6) dx #
# :. L = int_(-2)^(-1) sqrt ((x^6+( 4-x )^2) / x^6) dx #
# :. L = int_(-2)^(-1) sqrt (x^6+( 4-x )^2) / x^3 dx #

This definite integral does not have an elementary intrinsic solution and would need to be solved numerically, using either a computer or estimated using the Trapezium Rule or Simpson's Rule

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