- #1
Ulagatin
- 70
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Hi All,
I'm in Year 10 (doing pure mathematics one step below pre-tertiary level) at the moment, and I thought I would work myself through a mathematics text to get myself to a good solid position to study higher mathematics (I want to be a physicist). I was inspired by one of Ian Stewart's books (Taming the Infinite: The story of mathematics), and so based on this inspiration, I decided to self-study mathematics. I happened upon a copy of Apostol's "One Variable Calculus".
I am roughly 90 pages through, did well with sets, did reasonably with sequences too I think (and wrote my first proof but not sure if it was right in the end - I was happy though), but once I hit the example problems for solving integrals, I was stuck.
So, I have a few example problems that I have to share, which I will take a stab at, and so any of you willing to help out can show me the correct method in solving integrals (I'll only post the first at the moment).
The first problem:
[tex]
\int_{-1}^{3} [x] dx
[/tex]
Theorem 1.6: [tex]
\int_{a}^{c} s(x) dx + \int_{c}^{b} s(x) dx + \int_{b}^{a} s(x) dx = 0
[/tex]
Theorem 1.2: [tex]
\int_{a}^{b} [s(x) + t(x) dx] = \int_{a}^{b} s(x) dx + \int_{a}^{b} t(x) dx = 0
[/tex]
Theorem 1.5: [tex]
\int_{a}^{b} s(x) dx < \int_{a}^{b} t(x) dx
[/tex]
First problem:
Let s(x) = 0 and t(x) = [x] satisfying the comparison theorem for [x] > 0.
Let -1 be represented by 'a', 1 by 'b' and 3 by 'c'.
So, we then have the following:
[tex]
\int_{a}^{c} t(x) dx + \int_{b}^{c} t(x) dx + \int_{a}^{b} t(x) dx = 0
[/tex]
Now, there is a range of 4 in the interval [a, c], or numerically [-1, 3] for the first integral. A range of two in the second integral exists, as the interval is [b, c] represented numerically by [1, 3]. And finally the third integral covers the interval [a, b], numerically [-1, 1].
My idea was that, if [x] = 1 (whatever the square brackets may mean), that as the integral in this problem is from -1 to 3, the solution would be 4, given that it is a step function. The answers in the book tell me that the solution to the integral is actually 2. Would this mean that [x] is equal to 1/2? If I understand correctly, as the integral is telling me the area of the function, that [x] is actually the y-coordinate of the step function?
If this is the case, then to get 2, it must be (if my understanding of the method is correct) that [x] = 1/2, and so over an interval of 4, the function is intersecting the y-axis at [x] = 0.5. This would obviously mean 0.5*4 = 2.
Is this understanding correct in any shape or form? I am sure it is not! I have heard of the "integral of 1/cabin" joke, and so do natural logs come into play here?
Any help whatsoever is greatly appreciated, and so, thanks in advance!
I know some of you might think I'm crazy to try and learn from Apostol at this stage, but I am sure an understanding of a few key points should allow me to be able to (hopefully) get through this text.
Cheers,
Ulagatin
I'm in Year 10 (doing pure mathematics one step below pre-tertiary level) at the moment, and I thought I would work myself through a mathematics text to get myself to a good solid position to study higher mathematics (I want to be a physicist). I was inspired by one of Ian Stewart's books (Taming the Infinite: The story of mathematics), and so based on this inspiration, I decided to self-study mathematics. I happened upon a copy of Apostol's "One Variable Calculus".
I am roughly 90 pages through, did well with sets, did reasonably with sequences too I think (and wrote my first proof but not sure if it was right in the end - I was happy though), but once I hit the example problems for solving integrals, I was stuck.
So, I have a few example problems that I have to share, which I will take a stab at, and so any of you willing to help out can show me the correct method in solving integrals (I'll only post the first at the moment).
Homework Statement
The first problem:
[tex]
\int_{-1}^{3} [x] dx
[/tex]
Homework Equations
Theorem 1.6: [tex]
\int_{a}^{c} s(x) dx + \int_{c}^{b} s(x) dx + \int_{b}^{a} s(x) dx = 0
[/tex]
Theorem 1.2: [tex]
\int_{a}^{b} [s(x) + t(x) dx] = \int_{a}^{b} s(x) dx + \int_{a}^{b} t(x) dx = 0
[/tex]
Theorem 1.5: [tex]
\int_{a}^{b} s(x) dx < \int_{a}^{b} t(x) dx
[/tex]
The Attempt at a Solution
First problem:
Let s(x) = 0 and t(x) = [x] satisfying the comparison theorem for [x] > 0.
Let -1 be represented by 'a', 1 by 'b' and 3 by 'c'.
So, we then have the following:
[tex]
\int_{a}^{c} t(x) dx + \int_{b}^{c} t(x) dx + \int_{a}^{b} t(x) dx = 0
[/tex]
Now, there is a range of 4 in the interval [a, c], or numerically [-1, 3] for the first integral. A range of two in the second integral exists, as the interval is [b, c] represented numerically by [1, 3]. And finally the third integral covers the interval [a, b], numerically [-1, 1].
My idea was that, if [x] = 1 (whatever the square brackets may mean), that as the integral in this problem is from -1 to 3, the solution would be 4, given that it is a step function. The answers in the book tell me that the solution to the integral is actually 2. Would this mean that [x] is equal to 1/2? If I understand correctly, as the integral is telling me the area of the function, that [x] is actually the y-coordinate of the step function?
If this is the case, then to get 2, it must be (if my understanding of the method is correct) that [x] = 1/2, and so over an interval of 4, the function is intersecting the y-axis at [x] = 0.5. This would obviously mean 0.5*4 = 2.
Is this understanding correct in any shape or form? I am sure it is not! I have heard of the "integral of 1/cabin" joke, and so do natural logs come into play here?
Any help whatsoever is greatly appreciated, and so, thanks in advance!
I know some of you might think I'm crazy to try and learn from Apostol at this stage, but I am sure an understanding of a few key points should allow me to be able to (hopefully) get through this text.
Cheers,
Ulagatin