I didn’t get too far this week on KA. I got through 4 exercises and 2 videos but unfortunately that was it. I started the week working on the first of three exercises in the section Integrating with U-Substitution and easily finished the first 2 exercises in about 15 minutes. The third exercise then took me about 3 days to get through. 😑 By the end of it, I definitely felt like I had a better grasp on u-substitution and on definite and indefinite integrals, in general. So, although I didn’t get as far as I’d hoped I would, I’m pretty happy with the progress I was able to make this week.

Last week I mentioned that my method of working through these u-sub questions is by denoting the composite functions inside the integrals as a(x), b(x), c(x), etc. Then, instead of using the notation u and du/dx, I use b(x) and b’(x), respectively. Answering the questions this way has worked for me so far and I’m pretty sure my method is essentially the exact same thing as using u and du/dx but I could be wrong. In any case, my plan is to continue doing it this way until I reach a point where it doesn’t work, if that ever happens.

Here are 2 example questions I worked through in the third exercise of Integrating with U-Substitution:

            Question 1:

• ∫3x²(x³ + 1)⁶ dx = ∫ b’(x) * a’(b(x)) * dx
  • A(x) = x⁷/7
  • a'(x) = x⁶
  • b(x) = x³ + 1
  • b’(x) = 3x²
• ∫ b’(x) * a’(b(x)) * dx = A(b(x)) + C
  • = (b(x))⁷/7 + C
  • = (x³ + 1)⁷ + C

Question 2: 

• ₀∫^π/3 sin(x)cos³(x) * dx = ₀∫^π/3 sin(x) * a’(b(x)) * dx
  • A(x) = x⁴/4
  • a'(x) = x³
  • b(x) = cos(x)
  • b’(x) = -sin(x)
• ₀∫^π/3 (-1/1) b’(x) * a’(b(x)) * dx = (-1) ₀∫^π/3 b’(x) * a’(b(x)) * dx
  • = (-1) (A(b(x)) ₀|^π/3
  • = (-1) (((cos(π/3))⁴/4) – cos(0))⁴/4) 
  • = (-1) ((1/2)⁴/4 – (1)⁴/4)
  • = (-1) ((1/16)/4 – (1/4)
  • = (-1) ((1/16 * 1/4) – (1/4)
  • = (-1) (1/64 – 16/64)
  • = (-1) (-15/64)
  • = 15/64

As far as I can tell, the only difference between evaluating an indefinite integral vs evaluating a definite integral is that you apply the fundamental theorem of calculus (ie. A(b(x₂) – (A(b(x₁)) when solving a definite integral.

After finally getting through the third exercise in Integrating with U-Substitution, I began the following section, Integrating with Long Division and Competing the Square, on Saturday. I made it through the first exercise pretty easily which had questions where I needed to use polynomial long division to simplify the integrand. As a side note, I just learned that an integrand is the function that’s within the integral sign that is being evaluated:

Below are 2 example questions I worked through from this exercise. For both questions, I’ve sectioned off the polynomial long division I worked through in order to simplify the function inside the integrand. I don’t know how I would be able to write out polynomial long division here on Microsoft Word so I’m not going to bother doing the math below these questions.

            Question 1

            Question 2

As a final side note, I was given a question in the third exercise of Integrating with U-Substitution where a’(x) = sec(x)tan(x). This question really threw me off because I hadn’t seen an integral with a’(x) equalling two trig functions. Below there are two photos where the first photo explains that a’(x) = sec(x)tan(x) and that it took me an hour to figure that out and the second going through the math behind finding the derivative of sec(x) which equals sec(x)tan(x):

This coming week I’ll be starting on the second exercise in the section Integrating with Long Division and Competing the Square which deals with integration using the ‘completing the square‘ method. There was one video in this section that showed how to do this which completely went over my head so I have a feeling this exercise is going to take me at least a few days to get through. If that’s the case, I think it might be tough for me to finish off the unit Integrals (2,400/3,200 M.P.) before the start of the New Year. I still think it’s possible, but it’s probably going to be a photo finish! 💨 📸 🏁