I hit my goal of getting 25% of the way through Applications of Integrals this week but I’m still not thrilled with the amount of progress I made or my understanding of the material that I got through (i.e. I don’t understand it..). Since I first began KA, I’ve always told myself that I’d do a MINIMUM of an hour of work each day from Tuesday to Saturday. This week I barely passed 5 hours of studying and I feel like I made in a pretty weak effort overall. 😒 I’m struggling to intuitively understand how function’s derivatives and antiderivatives relate to each other and can’t intuitively visualizing what’s going on in my head. I think that I’m getting pretty close to having it figured out but at this point it still seems elusive which is very frustrating. 😠

I began the week working on the first of two exercises in the section Straight Line Motion. Both exercises had to do with using integrals to derive either an objects velocity or position from its acceleration or velocity, respectively. I spent most of Tuesday working on a question that was difficult not because of the calculus (although if I’m being honest I also don’t intuitively understand the calc…), but because the log rules and algebra was tough. Here’s the question:

(Note: I didn’t actually need to find the antiderivative of v(t) to answer this question but, in order to try and get a better feel for what’s going on and also to get some practice, I decided to figure it out just for funsies.)

• v(t) = f(x)
• f(x) = 16 * (1/2)^x – 5
• ∫ f(x) dx = F(x)
  • = ∫ 16 * (1/2)^x – 5 dx
  • = ∫ 16 * (1/2)^x dx – ∫ 5x⁰ dx
  • = 16 * ∫ (1/2)^x dx – ∫ 5x⁰ dx
    • (Note: At this point you need to understand log rules, how to apply, and WHY you need to apply them. You need to turn (1/2)^x into the argument of ln(A) which itself needs to be used as the power on e. By doing this, you can use log rules to move x in front of x * ln(1/2) and, from there, you can use the chain and power rules to find the antiderivative of e^{(x * ln(1/2))}.)
  • = 16 * ∫ e^{(ln((1/2)^x))} dx – ∫ 5x⁰ dx
  • = 16 * ∫ e^{(x * ln(1/2))} dx – ∫ 5x⁰ dx
  • = 16 * ∫ a’(b(x)) dx – ∫ 5x⁰ dx
    • a(x) = e^x
    • a’(x) = e^x
    • b(x) = x * ln(1/2)
    • b’(x) = ln(1/2)
  • = 16 * 1/(ln(1/2)) * ∫ ln(1/2) * a’(b(x)) dx – ∫ 5x⁰ dx
  • = 16 * 1/b’(x) * ∫ b’(x) * a’(b(x)) dx – ∫ 5x⁰ dx
  • = 16 * 1/b’(x) * (a(b(x)) – 5x
  • = 16/ln(1/2) * (e^{x * ln(1/2)}) – 5x
    • (Note: At this step, you use the inverse log rules from before to simplify e^{x * ln(1/2)}.)
  • = 16/ln(1/2) * (e^{ln((1/2)^x)}) – 5x
  • = 16/ln(1/2) * (1/2)^x – 5x
  • = 16/ln(1/2) * 1^x/2^x – 5x
  • = 16/ln(1/2) * 1/2^x – 5x
  • = 16/(ln(2^-1) * 2^x) – 5x
  • = 16/(–ln(2) * 2^x) – 5x
  • = –16/(ln(2) * 2^x) – 5x
  • = –2⁴/(ln(2) * 2^x) – 5x
  • = –2⁴/2^x * 1/(ln(2) – 5x
  • = –2^{4 – x} * 1/(ln(2) – 5x
• F(x) = –2^{4 – x}/(ln(2) – 5x + C

So the antiderivative of 16 * (1/2)^x – 5 is –2^{4 – x}/(ln(2) – 5x + C, but the actual answer to the question was option B) ₀∫³ |v(t)| dt. To get the correct answer on the actual question all you need to know is that distance is the integral of the absolute value of the velocity function. In that sense, this question was easy to answer but, at the same time, I don’t really understand how/why the integral of the absolute value of velocity equals distance.

Here are two more questions from the second exercise where I was asked to find the antiderivative of a(t) and v(t) in order to find the function for a moving particle’s position in the first question and the particles position at t = 1 in the second question:

Question 1

In this question, part of the process is to use to coordinates s(4) = 2 in order to solve for C. Once you solve for C then you know the general equation for s(t), a.k.a. the function for the particle’s position.

Question 2

On Thursday I began the next section, Non-Motion Applications of Integrals. This section helped me connect the dots between how the area underneath a function relates to the antiderivative. Here are two pages from my notes that explain a bit of what I learned:

As you can see from my notes, I believe that if you find the antiderivative of a rate (ex. meters/second) where the rate is measured on the y-axis, you can think of the resulting antiderivative as a function where the y-axis is just the numerator of the rate (i.e. if the rate was meters/second the y-axis of the antiderivative would simply measure meters).  Again, I don’t have an intuitive feel for this why this works (although I know it relates to the FToC) but, at face value, this rule seems simple to understand/memorize. Here’s an example question from this section:

As you can see from my notes, the math itself isn’t too difficult but I have a hard time understanding that the increase in mass is equal to the antiderivative of the function, m'(t) = 2.6/t, which itself is a derivative. This is a good example of what I’m talking about when I say I don’t really understand how/why the FToC works and I can’t visualize what’s going on in my head when trying to work through these kinds of questions. I’m hoping and am pretty confident that with more practice it will eventually start to click for me just like it did in trig when I finally started to intuitively understand the unit circle. 

This coming week I’d like to get through the next two sections of Applications of Integrals (480/1,900 M.P.) which combined have 7 videos and 4 exercises. If I can manage it, I’ll be ~42% of the way through the unit which wouldn’t be too bad after 3 weeks. Lately I’ve been feeling like my progress has been pretty slow and I’m starting to feel a bit antsy to finally get through calculus. It’s been more than 2.5 years of studying math and, as rewarding and gratifying as it’s been, I’m becoming impatient and just want to be able to say I understand calc. I guess on top of learning math, this whole process has been an exercise in developing my patience. 🙄