Once again, I didn’t get very far through KA this week. I spent the entire week working through 1 section of the unit Integrals called Integrating With U-Substitution. In my defence, the section was pretty big having 11 videos and 3 articles in it. Even though I didn’t get very far, I did enjoy learning about u-substitution which is what Sal referred to as the ‘reverse chain rule’. I found the algebra pretty satisfying once I started to get the hang of it, but I still feel like I have a ways to go before I completely figure it out. The good thing is that it feels like the type of concept that’s hard to wrap my head around at first but, with a bit more practice, will eventually be pretty easy to understand.

The key to understanding and using u-substitution is to be able to recognize when there’s a composite function, and the derivative of the ‘inner’ function all inside an integral. Confused? This is the way that I think about it:

• ∫ b’(x) * a’(b(x)) * dx

If you can recognize this pattern of functions inside an integral, you can then use u-substitution to find the integral’s antiderivative. (After more than a month of working through this unit, I still don’t know if I’m using the right terminology when talking about integrals and antiderivatives. 😔)

• ∫ b’(x) * a’(b(x)) * dx
  • = a(b(x)) + C

When I first learned about and worked through derivatives, I got in the habit of always labelling composite functions as “a(x), a’(x), b(x)” etc. which is what I also did working through these u-substitution questions. That’s not the way that Sal did it, however. Instead of finding and writing b(x) and b’(x), Sal would use u and du/dx, respectively. Here’s an example of how he does it (and how I’m assuming you’re “supposed” to do it) vs how I do it below:

• ∫(2x + 7)³ dx = ∫ a’(b(x)) dx
  • a(x) = x⁴/4
  • a’(x) = x³
  • b(x) = 2x + 7
  • b’(x) = 2
    • (Note #1: Using my method where I find a(x), b(x) and their derivatives, I start by labelling a’(x), which in this case equals x³, and then using the reverse power rule, which in this case makes a(x) = x³⁺¹/3+1 = x⁴/4.)
    • (Note #2: Even though the integral starts off with only an a’(b(x)) inside it, since b’(x) = 2 you can insert a 2/2 into the integral without changing its value and then factor out a ½ out of the integral leaving you with (½ * ∫2 * (2x + 7)³ dx) which, again, doesn’t change the integral’s value and leaves you with ∫ b’(x) * a’(b(x)) * dx.)
  • ∫ b’(x) * a’(b(x)) * dx = ½ * ∫2 * (2x + 7)³ dx
    • = ½ * a(b(x)) + C
    • = ½ * (2x + 7)⁴/4
    • = (2x + 7)⁴/8
      • (Note: it might be hard to understand here, but the reason why (2x + 7)⁴/4 turns into (2x + 7)⁴/8 is because you’re multiplying the denominator, 4, by 2 because of the ½ in front of a(b(x)).)

Right now, I find it much easier to think through these types of questions in terms of a(x) and b(x) as opposed to u and du/dx. I’m hoping that I’ll feel comfortable using both methods interchangeably with more practice. I’m also a bit worried that my method won’t work if I come up against a more difficult u-sub question but, in my mind anyways, I’m pretty sure my method is the exact same thing as u-substitution so I’m hoping it will continue to work for me. Here’s another example of a question I worked through:

• ∫x * (1/6 (x²) + 1)^1/2 * dx = ∫x * a’(b(x)) dx
  • a’(x) = x^1/2
  • a(x) = x^3/2/(3/2)
    • = 2x^3/2/3
  • b(x) = 1/6 (x²) + 1
  • b’(x) = 2x/6 
    • = x/3
• ∫x * a’(b(x)) dx = 3 * ∫x/3 * a’(b(x)) dx
  • = 3 * ∫b’(x) * a’(b(x)) dx
    • = 3 * a(b(x)) + C
    • = 3 * 2(1/6 (x²) + 1)^3/2/3 + C
    • = 2(1/6 (x²) + 1)^3/2 + C

After working through a bunch of u-sub questions using indefinite integrals, I then was shown how to use u-sub with definite integrals. This helped me clarify that a definite integral has bounds on it whereas an indefinite has no bounds on it. (Something that, in retrospect, seems pretty obvious.) I don’t have the slightest intuition as to how/why the formula for u-sub on a definite integral works, but this is it:

In the above photo, in the first step where you see ∫ and x₁ and x₂, the x’s stand for the left and right bound of the definite integral. I don’t understand why, but in order to solve these questions you run both x-values through b(x) and then use their resulting values as the bounds. (As I go through my notes right now to explain this, I’m realizing I REALLY don’t understand the notation behind this or how it works, in general.) Here’s a question I worked through:

• ₁∫² 15x²(x³ – 7)⁴ * dx
  • a’(x) = x⁴
  • a(x) = x⁵/5
  • b(x) = x³ – 7
  • b’(x) = 3x²
• ₁∫² 15x²(x³ – 7)⁴ * dx
  • = 5 * ₁∫² 3x²(x³ – 7)⁴ * dx
  • = 5 * ₁∫² b’(x) * a’(b(x)) * dx
  • = 5 * _b(1)∫^b(2) a’(x) * d(x)
  • = 5 * _{((1)^3 – 7)} ∫^{((2)^3 – 7)} a’(x) * dx
  • = 5 * _-6∫¹ a’(x) * dx
  • = 5 * A(x) _-6|¹
  • = (5 * A(x₂)) – (5 * A(x₁))
  • = (5 * (1)⁵/5) – (5 * (-6)⁵/5)
  • = (5/5) – (5 * -7,776/5)
  • = 1 – (-7,776)
  • = 1 + 7,776
  • = 7,777

I find this entire thing very confusing and don’t understand how/why it works at all. As always, I’m hoping if I keep working through these types of questions and get better at going through the process and using the formula, eventually the intuition for what’s going on will come to me. 

I only have 5 videos and 6 exercises remaining in the unit Integrals (2,080/3,200 M.P), so I’m hoping I can get through it all by the end of this week so I can start the unit test the following week. If that happens and I’m able to pass the unit test in Week 120, I’ll have made it through Integrals by Dec. 20^th which will I’ll be happy about. Looking further ahead, it’d be nice to get through this subject, Calculus 1, by February or March which would then give me 5/6 months to get through Calculus 2 before reaching the 3-year mark of me working on KA. When I first started using KA, I naively thought I’d be able to learn calculus and everything leading up to it in a year. Nonetheless, considering all the work I’ve had to do, going from arithmetic to calculus in 3 years would still be pretty good!