How do you find the lengths of the curve #y=intsqrt(t^-4+t^-2)dt# from [1,2x] for the interval #1<=x<=3#?

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Answer 1 · 2018-04-11T09:22:37

#= 7/3#

Arc length is:

# s= int _{a}^{b} sqrt {1+(y')^{2}} dx qquad triangle#

For function:

#y=int_(1)^(2x) sqrt(t^-4+t^-2)dt#

...use Liebnitz's Rule, specifically tailored here:

#(d)/(dx) int _{u(x)}^{v(x)} \ f(t) \ dt = f(v)\ v'(x) - f(u) \ u'(x)#

#= sqrt((2x)^-4+(2x)^-2)* 2 = 2 sqrt(1/(16x^4)+1/(4x^2)) #

So #triangle# becomes:

# s= int _{1}^{3} sqrt {1+1/(x^2)+1/(4x^4)} dx #

# s= int _{1}^{3} sqrt { (1+1/(2x^2))^2} dx #

# s= [ x - 1/(2x) ]_{1}^{3} = 7/3#