I fell short of reaching my goal for this week which was to get through two sections, but I’m still very happy with the effort I put in and what I was able to learn/get through. Even though it took me a long time to get through the material, I felt like I had a firm grasp on what was covered in the videos and exercises. The exercises had me use a lot of algebra which, as I’ve mentioned, is the type of math I enjoy the most since it’s sort of like solving little puzzles. The calculus I used is starting to feel similar in that it too is starting to feel like solving little puzzles, especially now that I have a better grasp on the fundamental rules, formulas, etc. Calculus feels much less intimidating/daunting to me now than it used to which feels like a huge accomplishment. 💪🏼 😤
I started the week working on a new section titled Area: Vertical Area Between Curves. The gist of what I learned is that you can find the area between two functions and within a certain interval by subtracting the value of the lesser definite integral from the value of the greater definite integral within that interval. Here’s a screen shot from the third video of this section and a picture of my notes that explains this concept:
It’s probably difficult to interpret but my notes show three separate but related concepts. The first shows that the area between two functions along a specific interval which are both positive can be calculated by finding the integral of each function and subtracting the smaller value from the greater value. The second and third concept are the exact same in that, to find the area between the two functions along a specific interval even when one function is positive and the other is negative AND when both functions are negative, you simply subtract the lesser function’s integral from the greater function’s integral and, boom, you found the area.
The majority of the rest of my week spent working through the three exercises from this section. Here are 4 example questions I worked on:
Question 1
- h(x) = x2 + 2
- H(x)-4|2 = -4∫2 h(x) dx
- = -4∫2 x2 + 2x0 dx
- = x3/3 + 2x -4|2
- = ((2)3/3 + 2(2)) – ((–4)3/3 + 2(–4))
- = (8/3 + 4) – (–64/3 – 8)
- = (8/3 + 12/3) – (–64/3 – 24/3)
- = (20/3) – (–88/3)
- = 20/3 + 88/3
- = 108/3
- = 36
Question 2
- f(x) = (x – 1)1/3
- X-intercept = lower bound
- f(x) = 0
- 0 = (x – 1)1/3
- 03 = ((x – 1)1/3)3
- 0 = x – 1
- 1 = x
- F(x) 1|k = 12
- F(x) 1|k = 1∫k f(x) dx
- = 1∫k (x – 1)1/3 dx
- a(x) = 3x4/3/4
- a’(x) = x1/3
- b(x) = x – 1
- b’(x) = 1
- = 1∫k (1) * (x – 1)1/3 dx
- = 1∫k b’(x) * a’(b(x)) dx
- = a(b(x)) 1|k
- = 3(x – 1)4/3/4 1|k
- 12 = (3(k – 1)4/3/4) – (3(1 – 1)4/3/4)
- = (3(k – 1)4/3/4) – (0)
- 12 * 4/3 = (3(k – 1)4/3/4) * 4/3
- 16 = (k – 1)4/3
- 163/4 = ((k – 1)4/3)3/4
- 8 = k – 1
- 9 = k
- = 1∫k (x – 1)1/3 dx
- F(x) 1|k = 1∫k f(x) dx
Question 3
- f(x) = x + 1
- g(x) = 25/x2 + x
- G(X) – F(x) 1|5 = ?
- (Note: You have to graph g(x) and f(x) in order to see which function is greater. That being said, this whole time I’ve been wondering why you can’t simply say the area would be equal to the absolute value of the difference between the two functions, i.e. |F(x) – G(x)|. It seems likely to me, however, that if this did work that Sal would have mentioned it. 🤔)
- G(X) 1|5 = 1∫5 g(x) dx
- = 1∫5 (25/x2 + x) dx
- = 1∫5 (25x–2 + x) dx
- = –25x–1 + x2/2 1|5
- = –25/x + x2/2 1|5
- = (–25/5 + 52/2) – (–25/1 + 12/2)
- = (–5 + 25/2) – (–50/2 + 1/2)
- = (–10/2 + 25/2) – (–49/2)
- = (15/2) – (–49/2)
- = 15/2 + 49/2
- = 64/2
- = 32
- F(X) 1|5 = 1∫5 f(x) dx
- = 1∫5 x + 1x0 dx
- = x2/2 + x 1|5
- = (52/2 + 5) – (12/2 + 1)
- = (25/2 + 10/2) – (1/2 + 2/2)
- = 35/2 – (3/2)
- = 32/2
- = 16
- G(X) – F(x) 1|5 = 32 – 16
- = 16
Question 4
(Note: In these types of questions, the key word is “bound”. What the question is asking is what is the area between the enclosed space between the y-axis and the two functions. The first step is to find the x-coordinate where the functions cross over each other.)
- f(x) = x2 + 3
- g(x) = 2x + 6
- G(X) – F(X) 0|a = ?
- x2 = a = g(x) = f(x)
- g(x) – f(x) = 0
- 2x + 6 – (x2 + 3) = 0
- 2x + 6 – x2 – 3 = 0
- –x2 + 2x + 3 = 0
- –x2 – x + 3x + 3 = 0
- –x(x + 1) + 3(x + 1) = 0
- (x + 1)( –x + 3) = 0
- 0’s @ [x = –1] and [x = 3]
- x2 = a = 3
- 0’s @ [x = –1] and [x = 3]
- G(X) – F(X) 0|3 = ?
- G(x) 0|3 = 0∫3 g(x) dx
- = 0∫3 2x + 6x0 dx
- = x2 + 6x 0|3
- = (32 + 6(3)) – (2(0)3 + 6(0))
- = (9 + 18) – (0)
- = 27
- F(x) 0|3 = 0∫3 f(x) dx
- = 0∫3 x2 + 3x0 dx
- = x3/3 + 3x 0|3
- = (33/3 + 3(3)) – (03/3 + 3(0))
- = (27/3 + 9) – (0)
- = 9 + 9
- = 18
- G(x) 0|3 = 0∫3 g(x) dx
- G(X) – F(X) 0|3 = 27 – 18
- = 9
I made it to the next section, Area: Horizontal Area Between Curves, on Saturday. I made it through the only 2 videos of the section but haven’t started the exercise. So far what I’ve learned in this section is a bit tricky for me to understand but it also seems like I’ll be able to figure it out with a bit of practice. I’m going to wait until next week’s post to explain what I learned in this section since I’m sure I’ll have a better understanding of it after I work through the exercise.
I’m now 38% of the way through Applications of Integrals (720/1,900 M.P.). In total there are 14 videos and 10 exercises left for me to get through in this unit. I’m hoping I can finish this unit in the next 3 weeks which would give me a good shot at getting through the Calculus 1 Course Challenge by the end of May. If that happens, I’ll end up being a month behind my original goal of completing it by the start of May but, nonetheless, I’ll still be happy. Calculus is taking me way longer to get through than I anticipated which is disappointing but, at the same time, I’m reminding myself that calculus is a much tougher/denser subject than algebra, geometry, etc. and also that it typically takes MUCH longer than 3 years to learn calculus.