I was going over calculus homework with my son the other day, who was frustrated because this problem took him three hours to solve:

Find the arclength between 2 and 5 of the function:

$f(x)$ $=$ $x^5 \over 10$ $+$ $1 \over 6x^3$

...which we solved by taking the first derivative and integrating the "square root of d/dx-squared plus one" over the domain. But, he spun his wheels over the complexity of the indefinite integral and became frustrated.

Assuming he took this to someone near graduation who is effective as a math help TA, How much time would that student mentor need to solve this? What kinds of principles and techniques would he teach my son so he could get there in three minutes instead of 180?

  • $\begingroup$ This is really a question asking "is this indefinite integral too hard to be a reasonable homework question"? You could post your son's labor and ask if there's a better way. Better yet, your son could post. $\endgroup$ – Ethan Bolker May 2 '18 at 23:31
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    $\begingroup$ It’s really just a problem of “failing to notice”: square the derivative and add $1$ and you again get a perfect square — it looks just like the other thing before you added $1$, except for a change from $-1/2$ to $1/2$. Notice that and you can get the thing done in at most $10$ minutes. Failing to notice? Infinite frustration! $\endgroup$ – Lubin May 2 '18 at 23:36
  • $\begingroup$ I was certain it wasn't too hard; Pearson usually stacks the hard ones at the back of the set, not in the middle like this one was. @Lubin, that really suggests that WA's indefinite integral "answer" was not simplified. Does that happen a lot with Wolfram stuff? $\endgroup$ – Rob Perkins May 2 '18 at 23:44
  • $\begingroup$ Three hours is unbelievable! No one noticed the obvious 'square of a binomial' trick!? $\endgroup$ – Christopher Marley May 2 '18 at 23:46
  • $\begingroup$ It makes it clear that these problems have to be carefully arranged to make them doable. $\endgroup$ – Lubin May 3 '18 at 0:22

tick tick tick....

$\int_2^5 \sqrt{1+\frac {dy}{dx}^2} \ dx\\ y = \frac {x^5}{10} + \frac 1{6x^3}\\ \frac {dy}{dx} = \frac 12 x^4 - \frac 12 x^{-4}\\ \sqrt{1+\frac {dy}{dx}^2} = \sqrt {1 + \frac 14x^8 - 2(\frac 12x^4)(\frac 12x^{-4}) + \frac 14 x^{-8}} \\ \sqrt {1 + \frac 12 x^8 - \frac 12 + \frac 14 x^{-8}}\\ \sqrt {\frac 12 x^8 + \frac 12 + \frac 14 x^{-8}}\\ \frac 12 x^4 + \frac 12 x^{-4}$

And now we see that we are going to get a figure that looks very similar to our starting figure.

$\int_2^5 \frac 12 x^4 + \frac 12 x^{-4} \ dx\\ \frac {x^5}{10} - \frac 1{6x^3}|_2^5\\ \frac {5^5}{10} - \frac {2^5}{10} - \frac 1{6(5^3)} + \frac 1{6(2^3)}$

That should take about 10-15 minutes.

If you want to plug it into your calculator, another minute to get. $309.3$

Or if you are anti-calculator

$5^5 - 2^5 = 3125 - 32 = 3093\\ \frac 1{5^3} - \frac 1{2^3} = \frac {2^3 - 5^3}{10^3} = -0.117\\ 309.3 + \frac {0.117}{6} = 309.3195$

It will take a minute or two longer.

  • $\begingroup$ It's fun watching you edit this on the fly. (numeric solution is 618639/2000, or 309.3195-ish :-) $\endgroup$ – Rob Perkins May 2 '18 at 23:53
  • $\begingroup$ It would be better if I could key it in as well as I see it in my head. Which put my total time closer to 30 minutes $\endgroup$ – Doug M May 2 '18 at 23:58
  • $\begingroup$ Took me just as long as my son to work the problem. :-) $\endgroup$ – Rob Perkins May 3 '18 at 0:04
  • $\begingroup$ I see the difference... the second term in the final integral flips the sign, because you're bringing down a (-4) not a (4) according to the fundamental theory. Fix that and your fractional result looks like mine. $\endgroup$ – Rob Perkins May 3 '18 at 0:27
  • $\begingroup$ Preferred answer, for its completeness and for the way it unfolded in real time, just like how we do math in the real world! $\endgroup$ – Rob Perkins May 6 '18 at 3:09

I just gave it a shot. This is a contrived case which makes it easier to solve than a typical arc length problem. Still, it took me about five minutes to do (writing out each step), so a three minute time is quite unreasonable. Three hours is a lot longer than it should have taken him as well. I would think fifteen minutes might be a reasonable expectation.

The question you should ask him, is after he figured out how to solve it, how long would it take him to do it again from scratch on a fresh sheet of paper.


  • $\begingroup$ I think this time "trick" might be more along the lines of, "When you're stuck, show up for office hours and ask for help." $\endgroup$ – Rob Perkins May 2 '18 at 23:48
  • $\begingroup$ @RobPerkins, I think the key phrase is "three hours to solve it", not "three hours before he gave up". When you figure something out on your own you tend to learn it much better. I hope you gave him an "Attaboy" for sticking to it instead of a rationalization for thinking the homework was too onerous. $\endgroup$ – Cedron Dawg May 3 '18 at 0:12
  • $\begingroup$ I do know he attempted to solve the indefinite integral and match what W-A was showing him. Poor computer got itself lost in trig identities and ended up down a rabbit hole. The proposed solution changes based on how you express the terms under the square root. Maybe we're better off letting computers do the numeric stuff. $\endgroup$ – Rob Perkins May 3 '18 at 0:27
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    $\begingroup$ @RobPerkins, When I was a Calculus TA some 35 years ago, things like that weren't an issue. I think he might have learned a lesson that is even more valuable than the math he was learning. $\endgroup$ – Cedron Dawg May 3 '18 at 0:59
  • $\begingroup$ I have since learned that he gave up after just one hour. No problem, though, because he scored an A on a pretty well-written test that proved to me that he'd learned the material to the college's standards. And then we had a nice talk about how if you cultivate a relationship with a professor instead of treating all of them like forces of nature, you're going to have a better experience in education. Unless the prof is a jerk, in which case, hey, there's ratemyprofessor.com, right? $\endgroup$ – Rob Perkins May 6 '18 at 3:08

$f'(x) = \frac12(x^4-x^{-4})$ -->
$\int \sqrt{1+f'(x)^2} \ dx = \int \sqrt{1+\frac{x^8}4-\frac12+\frac1{4x^8}}$
$= \int \sqrt{\frac{x^8}4+\frac12+\frac1{4x^8}} = \int \sqrt{(\frac{x^4}2-\frac{x^{-4}}2)^2} = \int {x^4\over2} - {1\over(2x^4)} \ dx$

Easy peasy, and if you know this trick, then 3 minutes is a piece of cake!
(A favorite quote from my calculus class is "You should have learned that in pre-calc!" This should've been learned from Algebra 1)

  • $\begingroup$ Ahhh, but you didn't finish it. You still need to plug in the limits and simplify the result. $\endgroup$ – Cedron Dawg May 2 '18 at 23:48
  • $\begingroup$ He got past the roadblock. That idea that "1 plus a square is a complete-able perfect square" is pretty cool and I'm sorry I forgot it. $\endgroup$ – Rob Perkins May 2 '18 at 23:51
  • $\begingroup$ @RobPerkins, Yep, that's what I meant by "contrived" in my answer. Just writing by hand (including a few missing "dx"s) would take three minutes, doing the limit evaluation (without a calculator, LCD style) takes a few more. $\endgroup$ – Cedron Dawg May 2 '18 at 23:55

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