This week I got through quite a bit of content and made some solid progress towards grasping integrals and with it calculus, in general. I only finished 3 of the 4 exercises in Integrals but, nonetheless, I’m still really happy with what I was able to accomplish this week. My big news is that I FINALLY started to wrap my head around the Fundamental Theorem of Calculus after weeks of working on it, however my grasp on it is still pretty tenuous. Nonetheless, feeling like I’ve gotten over the hump of understanding the FToC was a huge relief and a big confidence booster. I struggled a bit with the 3 exercises I worked through this week which was tempering, but overall I’m very happy with what I learned this week and feel like, after weeks of feeling stagnant, I’m finally making some headway towards understanding calculus. WOO!!! 🥳 😭

The turning point for me this week in understanding the FToC came from watching this video:

I’d actually watched this video before, but the concept of what’s being taught in this video finally clicked for me this week. Here are the notes I took from this video that explain why the area underneath a function is related to the slope of that function’s antiderivative:

I think I could have done a better job of explaining this proof of the FToC in my notes. To paraphrase Step 4, I said, “the average height of f’(x) is 2/π because I proved that to be the average slope on f(x)” which is true but I think it could be explained a bit more clearly. However, I’m still having a bit of a hard time understanding this concept so it’s hard for me to put it into words. I do think that my notes above give a relatively decent explanation of why the FToC is true. After going through this proof, one thing I’m now once again finding confusing is derivatives and how the sum, power, and chain rules seem to magically create a function that indicates the slope of its antiderivative at every point.

On Wednesday I got back to working through the remaining exercises in Integrals starting with one called Finding Derivatives with Fundamental Theorem of Calculus. The process to solve the questions in this exercise was really easy but I spent a bit of time trying to think through exactly what was happening. Here’s an example question:

When working through these questions, I would sometimes close my eyes and think about the FToC proof that I went through above and think about a function in an integrand turning into its antiderivative and then flipping back to the antiderivative’s derivative, i.e. the function that was in the integrand in the first place. I still find it confusing, but thinking through these questions this way seems to help.

The following exercise I worked through, Finding Derivatives with Fundamental Theorem of Calculus: Chain Rule, was very similar except that the upper bound of the integral would be a function itself. I learned that when this is the case, you need to think of the integral as a composite function where the upper bound is the inner function and the rest of the integral is the outer function. Here’s a page from my notes that shows the steps to solve these types of questions:

• Step 1 – Write out the integral
  • F(x) = ₁∫^sin(x) (2t – 1) dt
• Step 2 & 3 – Rewrite each part of the integral as its own function using Lagrange notation:
  • g(x) = ₁∫^x (2t – 1) dt
  • g’(x) = 2x – 1
  • h(x) = sin(x)
  • h’(x) = cos(x)
    • (Note: In the photo of my notes, I have g(x) and h(x) switched to represent the opposite functions. When dealing with composite functions, I always like to nest the functions in alphabetical order, ex. a(b(x)) or in this case g(h(x)).)
• Step 4 – Note that F(x) equals g(h(x)):
  • F(x) = ₁∫^sin(x) (2t – 1) dt
  • g(h((x)) = ₁∫^sin(x) (2t – 1) dt
  • F(x) = g(h(x))
• Step 5 – Using the chain rule, evaluate F’(x), i.e. g’(h(x)) * h’(x):
  • F’(x) = g’(h(x)) * h’(x)
    • = (2(sin(x)) – 1) * cos(x)

Once I got the hang of how to label the upper bound as the inner function of the integral and find their respective derivatives separately, these questions became very straightforward to solve. Here are two examples from my notes:

I got through that exercise on Thursday and then made an attempt at the following exercise, U-Substitution Definite Integrals, but got stumped on the first question. The question asked me to evaluate the integral ₀∫^1/4 8x/√(1 – 4x²). I thought I was supposed to turn √(1 – 4x²) into u, including the square root, but I was actually supposed to only turn (1 – 4x²) into u leaving it as √u. Here’s the question and my notes on how I eventually solved it after getting it wrong:

I ended up getting through that exercise on Friday and felt pretty good about my grasp on u-sub by the end of the day. I still need a LOT more practice with these types of questions before I feel completely confident with them, but I definitely have a much stronger understanding of them now than I did a few weeks ago. Here’s another question I worked through in this exercise:

I finished the week attempting the final exercise I left that I need to redo in Integrals which is called Integration Using Completing the Square. This exercise kicked my ass and I wasn’t able to get through it before the end of the week. Here’s the first question I was given that I failed miserably at:

As you can see from the screen shots of the question, not only did I need to know how to use the completing-the-square method to solve this question, but I also needed to know that the antiderivative of 1/√(k² – x²) is arcsin(x/k). I learned how that derivative works at some point in the past but this week I couldn’t remember it when I tried to work through this question. After I got the question wrong, I Googled arcsin(x)’s derivation and was able to figure it out. Here’s it’s derivation and also the derivation of arcsin(x/k):

At the end of the week, I discovered something that seemed like a light coming from the end of the integral tunnel. After feeling a bit depressed that I had no clue that the antiderivative of arcsin(x/k) is 1/√(k² – x²), I Googled, “derivatives to memorize for integration” and found the following picture:

Looking at the list of derivatives and antiderivatives, I realized I’m pretty close to having these key functions figured out and memorized. Part of what’s been so difficult about learning math is that I never know where the end of a subject is which leaves me feeling like it could go on forever. Seeing this list of derivatives and antiderivatives gives me hope that, once I wrap my head around the bottom trig functions, I’m the verge of fully understanding derivatives and integrals. SO CLOSE!! 😄

This coming week I’m hoping it won’t take me too long to get through the exercise Integration Using Completing the Square. I want to make sure I take enough time to (hopefully) get an intuitive feel for how to solve the questions in that exe raise, however. Once I get through it, I’ll be able to restart the Integrals (3,070/3,200 M.P.) unit test which I REALLY hope I will pass on the next attempt. Regardless of whether I pass the unit test this week or not, when I eventually get through this unit it will be a huge accomplishment for me. I’m certain this is the longest I’ve ever worked on a single unit before now and it has been pretty demoralizing at times. The end of this unit is in sight though and once I finally get there I’m sure I’ll be MUCH closer to understanding calculus. 🤓