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:

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

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

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

How do you find the lengths of the curve y=intsqrt(t^-4+t^-2)dty=intsqrt(t^-4+t^-2)dt from [1,2x] for the interval 1<=x<=31<=x<=3?