I did a good job making up for my lack of progress last week by getting through 15 videos, 2 articles, and 5 exercises this week! In total I made it through ~3.5 sections which I’m really happy about. I learned a lot of things this week, most of which was pretty easy to understand but, nonetheless, will still take a bit of time before I fully understand the concepts. There were a few things I learned that I didn’t understand at all, however. That said, I’m confident that I’ll be able to figure it all out with a bit more practice. 💪🏼

As a brief side note, one of the first things I learned early in the week was the way to denote a Riemann Sum with n number of subdivisions which is “R(n)“. For example, if you’re taking a Riemann Sum using 10 subdivisions you’d denote it with R(10). 

In the bottom half of the photo, you can see the full Riemann Sum notation which states:

• R(n) = _i=1Σⁿ Δx * f(x_i)
  •  Δx = (b – a)/n
    • “(b – a)” refers to the interval being looked at.
    • “n” refers to the number of subdivisions.
  • f(x_i) = a + (Δx * i)
    • In this equation, “a” refers to the space between the y-axis and the start of the interval.
    • “(Δx * i)” refers to each x-value being used in the index to calculate that specific bar’s height for the Riemann Sum. 

Early in the week I learned that a definite integral IS a Riemann sum where n approaches infinity. (I spent a lot of time covering how the notation of a R.S. relates to the notation of a definite integral, but I haven’t been shown how/why you’re able to integrate n -> ∞ into the notation which, at this point, I find really confusing. In any case, here’s a screen shot from KA and a picture from my notes that shows how the two relate to each other:

The section where this was covered, Defining Integrals with Riemann Sums, had one exercise in it which had questions that gave me a definite integral or a Riemann Sum and asked me to rewrite whatever was given to me in the alternate form. Once I got the hang of it, I found this exercise pretty straightforward. Here’s an example of a question I worked through:

I find it useful to remember that when I’m looking at a Riemann Sum, I’m looking at an equation that states area = width * height (as long it’s a rectangular R.S. and not a trapezoidal one). Remembering this makes it easier to determine that 5/n would be the width of each rectangle and ln(2 + 5i/n) would be the height of each rectangle. That’s a bit besides the point, however, since what the question is really asking is what the interval of the integral is. Since the height of each rectangle is ln(2 + 5i/n), this means that the interval starts at 2 since ln(2 + 5i/n) = ln(a + xi/n). Knowing that a = 2 lets you then figure out what b equals:

• Width = Δx = (b – a)/n
  • (b – a)/n = 5/n
    • b – 2/n = 5/n
    • b – 2 = 5
    • b = 7

After working through that section, I then learned about what Sal referred to as the Fundamental Theorem of Calculus, which I was hyped about because it sounds really important. This, however, was the one thing I learned this week that I haven’t wrapped my head around yet. The general idea of this theorem is that if you have an integral, for example F(x) = _a∫^x f(t)dx, the derivative of the integral at x IS f(x):

As I said, right now I don’t really know why this theorem works. I spent a good amount of time this week trying to think through it and I’m pretty sure I figured out how/why it works but I’m not certain. Here are two pages from my notes that describe why I think the derivative of F(x) (a.k.a. F'(x)) is f(x). (FYI, when I wrote out these two pages of notes I wasn’t planning on adding them to this post which is why the writing isn’t very legible. Also, I should mention that it’s conventional for integrals to be denoted with a capital letter, i.e. F(x).)

Because integrals describe the area between a function and the x-axis, if the function is positive and above the x-axis, as you move along the function in the positive x-direction more positive area will be ‘added’ to the sum of the total area. Conversely, if the function is below the x-axis, as you move along the function in the positive x-direction more negative area will be ‘added’ to the sum of the total area. I’m almost certain that this is correct and the reason why F'(x) = f(x) but, as you can tell from my terrible explanation of it, I still don’t have it completely figured out. Even though I don’t intuitively understand how this theorem works, I was able to get through the questions I was given pretty easily once I figured out the process to solve them:

• Q. What’s the derivative of F(x) = ₂∫^2x √(15 – t) * dt?
  • (Note: The function being looked at is f(t) = √(15 – t), but because you’re taking the derivative at point b of the interval, a.k.a. 2x, you need to consider this as a function inside a function where 2x replaces t and, therefore, you must use the chain rule to solve the derivative.)
  • Step 1 – Identify each section of the integral and label each section accordingly:
    • F(x) = ₂∫^2x √(15 – t) * dt
      • F’(x) = a(x) = √(15 – x)
        • (Note: This is the Fundamental Theorem of Calculus at play here. To find the derivative of F(x), you simply replace the variable t with x.)
      • b(x) = 2x
      • b’(x) = 2
  • Step 2 – Use chain rule to solve for the derivative:
    • F(x) = _a∫^b(x) f(t) * dt
    • F’(x) = a(b(x)) * b’(x)
      • = √(15 – 2x) * 2
      • = 2√(15 – 2x)

I found working through these questions simple once I got the hang of the process but, overall, I still find these questions hard to wrap my head around. In the last article I read about the Fundamental Theorem of Calculus I was given the following chart which sums up the relationship between the properties of a function in an interval and how they relate to the integrals derivative:

(Note: The screen shot uses the term ‘antiderivative’ and I’m not sure if that’s just the term used for the derivative of an integral or if there’s a difference between a derivative and an antiderivative.)

The final thing I went through this week was very straightforward and was simply that you can scale an integral by multiplying it by any given value. Here’s a page from my notes that explains this concept and highlights the notation used when scaling an integral:

I’m now 28% of the way through Integrals (880/3,200 M.P.) so I’m still hoping/planning to get through the unit before the New Year. I just realized that I’m 69% of the way through Calculus 1 (11,720/17,100 M.P. – also, nice 😏) and when I’m finished with Integrals I’ll be ~82% of the way through Calc 1 with only 2 more units remaining, Differential Equations (which sounds pretty hard) and Applications of Integrals. I started Calculus 1 around March 20^th of this year so I’m hoping I can get through the entire subject before the 1-year mark. Given how long it’s taking me to get through this unit, I think getting through the final 2 units and finishing off the course challenge in that time frame will be a tall order, but I’m going to try!