How Do I Integrate the Last Part of This Arc Length Equation?

[chaotixmonjuish]

I'm working on this problem

x^5/6+1/(10x^3) [1,2]

and I got the equation:

sqrt(1+(5x^4/6-3/10^4)^2) or
sqrt(1+25x^8/36+9/100x^8-1/2)

I'm not sure how to integrate the last part, is there some sort of obvious substitution I'm missing?

 

[CompuChip]

Is the equation you get
$$\sqrt{1+ \frac{25 x^8}{36} + \frac{9}{100}x^8-\frac{1}{2}}$$
If so, it's easy to integrate (just collect polynomials and integrate term by term.
If not, please be a little more clear (use TeX if possible, or else at least put brackets when writing things like a/bc, which can be read as (a/b)*c or a/(b*c)).

 

[HallsofIvy]

CompuChip, that's clearly not what is intended because he did use parentheses to indicat that the x³ in the original function was in the denominator.

Let's check: y= x⁵/6+ 1/(10x³)= x⁵/6+ (1/10)x^-3 has derivative y= (5/6)x⁴- (3/10)x^-4. Squaring that, (y')²= (25/36)x⁸- 2(5/6)(3/10)+ (9/100)x^-8= (25/36)x⁸- (1/2)= (9/100)x^-8. Adding 1 to that just changes the "-1/2" to "+1/2" and since the first was a square, this is also a square with the sign changed: (a- b)²= a²- 2ab+ b² while (a+b)= a²+ 2ab+ b²: 1+ (y')²= (25/36)x⁸+ (1/2)+ (9/100)x^-8= [(5/6)x⁴+ (3/10)x^-4]² and so $\sqrt{1+ (y')^2}= \sqrt{[(5/6)x^4+ (3/10)x^{-4}]^2}= (5/6)x^4+ (3/10)x^{-4}$.
That should be easy to integrate.


(Tip: in general length integrals are very difficult. That's why calculus professors (who are not actually monsters) like tricks like the above that ensure the integral will be easy. It's a good idea to always check to see if you have a "perfect square"!)

 

[dexteronline]

By putting your function in my Arc Length online applet, i get arc length of 5.36
Applet Link
http://www.thinkanddone.com/ge/ArcLength.html"

http://www.thinkanddone.com/ge/ArcLength.aspx"

 

[HallsofIvy]

But that doesn't answer his question, does it?

 

[dexteronline]

Yes it doesn't answer for integral to be found. Yet I thought he may want to know the length of the curve