I didn’t pass the Applications of Integrals unit test this week. I ended up getting 3 of 19 questions wrong on my first attempt and I’m now sitting at 2 wrong of 9 attempted on my second attempt. I said last week that I’d be glad to do the test again in order to practice and get a stronger grasp on integrals which is true but the frustrating part is that, of the 5 questions I’ve got wrong, I didn’t get a single question wrong because I didn’t understand the calculus, just because I simply didn’t understand what the question was asking. It’s pretty annoying that I’m going to have to redo the test at least one more time considering that, so far, I’ve understood how the calculus works on every single question. Nonetheless, it’s still good practice for me and will help give me an even stronger grasp on integrals.

Below are 6 questions that came up in my ~1.5 attempts at working through the test this week. Some of them I got correct and some of them I got wrong. 

Question 1

This was the first question I did on Tuesday morning (it was question 6/19) and I got this one correct. After having worked through the ‘washer method’ types of questions for so long, I had to stop to think through how the area of R being rotated around the y-axis would be calculated with a single πr² as opposed to π((r₂)² – (r₁)²). For the most part, I’m now able to visualize how each step in the process of these types of questions works which makes it all seem much more straightforward in my head than it used to.

Question 2

This was the 7^th question of the unit test from my first attempt which I ended up getting wrong because I didn’t understand what I was being asked. Because the question at the end uses the word “displacement”, I thought the question was asking me how far the particle was from its starting point which is why I chose the second answer. The question was actually just asking how far the particle travelled between seconds 6 to 9 which was the first answer. I understand what ₆∫⁹ v(t) dt means and what the 5 + ₆∫⁹ v(t) dt means and the difference between the two, I just didn’t understand what the question was asking…

Question 3

This was the 11^th question from my first attempt at the test which I ended up getting correct. I was happy with my work on this question for a few reasons, one being that I found this question difficult to understand what was being asked of me but still managed to figure it out. I don’t remember going through a question like this in any of the exercises I’ve done throughout Applications of Integrals so I was happy that I was able to make sense of what was being asked of me find the solution. I was also happy that 1) I was able to figure out the Log rules necessary to come up with the correct solution and 2) that I was able to do all the arithmetic without using a calculator. Now that I think about it, these types of questions have me using arithmetic, algebra, and calculus all within one question. It’s pretty satisfying that I’m able to fluently work through all the different forms of math to solve these questions.

Question 4

I actually lied when I said I knew how to work through the calculus on all of the questions that came up on the test. This was question 19 of my first attempt at the unit test which I got wrong because I couldn’t remember what the process was in order to solve it. From the photo of my notes, the 5 steps needed to solve this question are:

• Step 1 – Find the antiderivative of a(t) which equals v(t) + C:
  • a(t) = –cos(t)
    • v(t) = ∫ a(t) dt
      • = ∫ –cos(t) dt
      • = –sin(t) + C
• Step 2 – Using the given that v(π) = 2, solve for C in a(t) = v(t) + C = –sin(t) + C:
  • v(t) + C = –sin(t) + C
    • v(π) = 2 = –sin(π) + C
      • 2 = (0) + C
      • 2 = C
• Step 3 – Find the antiderivative of v(t) which equals s(t) + C:
  • v(t) = –sin(t) + 2
    • s(t) = ∫ (–sin(t) + 2t⁰) dt
      • = cos(t) + 2t + C₂
• Step 4 – Using the given that s(π/2) = 3π, solve for C₂ in v(t) = s(t) + C₂ = cos(t) + 2t + C₂:
  • s(t) + C₂ = cos(t) + 2t + C₂
    • s(π/2) = 3π = cos(π/2) + 2(π/2) + C₂
      • 3π = (0)+ 2(π/2) + C₂
      • 3π = π + C₂
      • 2π = C₂
• Step 5 – Knowing that s(t) = = cos(t) + 2t + 2π, solve for s(0):
  • s(0) = cos(0) + 2(0) + 2π
    • = 1 + 2(0) + 2π
    • = 2π + 1

Question 5

This was the first question I got wrong on my second attempt at the unit test. I accidently forgot to screen shot the actual question, but it was something along the lines of, “The rate of change in Jim’s bank account over 3 months was xyz. What does ₀∫¹ r(t) dt mean?” (This obviously wasn’t the exact question but, whatever the question was, it was phrased in a confusing way.) As you can see, I chose the second answer which was incorrect. I understand that ₀∫¹ r(t) dt = –$100 means that from the time interval 0 to 1 month the total amount of money in the account decreased by $100, I just didn’t fully comprehend the question and what it was asking me.

Question 6

Again, this was a question where I fully understood the calculus/algebra/arithmetic but I didn’t read the question properly and therefore came up with the wrong answer. I plugged the functions into Desmos and saw that they intersected at x = 1 and x = 6 and I assumed I was trying to find the area between the functions within that interval. I didn’t notice that the question asked me to find the area between x = 1 and x = 4 until after I submitted my answer. As you can see from my notes, I used x = 6 as the upper bound and actually did all the math correct but obviously reached the wrong answer given I used the wrong upper bound.

I’m REALLY hoping I won’t have to do the unit test more than two more times this coming week. As I’ve said multiple times throughout my last few posts, I’m definitely ready to get through Applications of Integrals (1,840/1,900 M.P.) and start on the Calc 1 Course Challenge so that I can get through it and finally (FIN 👏🏼 A 👏🏼 LLY 👏🏼) start on Calc 2. PLEASE GOD, GIVE ME THE STRENGTH TO GET THROUGH THIS!!!! 🙏🏼🙏🏼🙏🏼