- #1
Ulagatin
- 70
- 0
Hi all,
I'm currently preparing for pre-tertiary mathematics, studying from Apostol's "One-Variable Calculus". I have just begun to work on the theory of integration of trigonometric functions, but I found that with the last set of exercises (on finding area between two functions, over some interval) I wasn't able to complete them (the issue was with absolute value functions).
Any assistance would be greatly appreciated.
Take one, for example, with these functions (over the interval [-1, 1]):
[tex] f(x) = |x| [/tex]
[tex] g(x) = x^2-1 [/tex]
I recognise that over this interval, [tex] f(x) > g(x) for all x [/tex], so the answer should be equal to the integral of the difference f(x) - g(x) with respect to x over the previously said interval.
How do I tackle this from here? My graphics calculator software tells me the area is equal to 2.33..., but I'm not sure how to arrive at this value.
[tex]
\int_{a}^{b} [f(x) - g(x)] dx
[/tex]
To show that I can work these integrals, I first show another example which may assist to solve the original example.
[tex] f(x) = x(x^2 - 1) [/tex]
[tex] g(x) = x [/tex]
The interval we must evaluate this integral on is as follows: [tex] [-1, 2^{1/2}] [/tex].
Over the interval [0, 1], f(x) > g(x). Over the interval [tex] [0, 2^{1/2}] [/tex], g(x) > f(x).
So to formalise this now, we have the following:
[tex]
\int_{-1}^{0} [x(x^2 - 1) - x] dx + \int_{0}^{2^{1/2}} [(x - x^3 - x)] dx [/tex]
[tex] = -\int_{0}^{-1} [x^3 - x - x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = -\int_{0}^{-1} [x^3 - 2x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = -\left[ \frac{x^4}{4} - {x^2} \right]_{0}^{-1} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = [-\frac{-1^4}{4} + -1^2] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = [-\frac{1}{4} + 1] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = \frac{3}{4} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = \frac{3}{4} + \left[ \frac{-x^4}{4} + {x^2} \right]_{0}^{2^{1/2}} [/tex]
[tex] = \frac{3}{4} + [-\frac {(2^{1/2})^4}{4} + (2^{1/2})^2] = \frac{3}{4} +[-\frac {4}{4} + 2] [/tex]
[tex] = \frac{3}{4} - 1 + 2 = \frac{3}{4} + 1 [/tex]
[tex] = \frac{3+4}{4} = \frac{7}{4} = 1.75. [/tex]
Now, I believe that, perhaps with the absolute value example, you can substitute values in, similarly to this example I've worked. First, I note that both these functions are monotonically decreasing over the interval [-1, 0] and increasing over [0, 1].
So, I'll try:
[tex] \int_{-1}^{0} [f(x) - g(x)] dx + \int_{0}^{1} [f(x) - g(x)] dx [/tex]
[tex] = -\int_{0}^{-1} [|x| - (x^2 - 1)] dx + \int_{0}^{1} [|x| - (x^2 - 1)] dx [/tex]
[tex] = \left[ \frac{|x^2|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{-1} + \left[ \frac{|x^2|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{1} [/tex]
[tex] = 0 - [\frac{1}{2} - (\frac{-1^3}{3} +1)] + [\frac{1}{2} - (\frac{1}{3} - 1)] [/tex]
[tex] = [-\frac{1}{2} + (-\frac{1^3}{3} + 1)] + [\frac{1}{2} - \frac{1}{3} + 1] [/tex]
[tex] = - \frac{1}{2} - \frac{1}{3} + 1 + \frac{1}{2} - \frac{1}{3} + 1 = \frac{4}{3} [/tex]
Clearly, I'm out by 1. Where have I gone wrong? Any assistance would be simply grand. I tried with the first integral being positive, but I ended with the same result mysteriously enough. I just don't understand!
Cheers,
Davin
I'm currently preparing for pre-tertiary mathematics, studying from Apostol's "One-Variable Calculus". I have just begun to work on the theory of integration of trigonometric functions, but I found that with the last set of exercises (on finding area between two functions, over some interval) I wasn't able to complete them (the issue was with absolute value functions).
Any assistance would be greatly appreciated.
Homework Statement
Take one, for example, with these functions (over the interval [-1, 1]):
[tex] f(x) = |x| [/tex]
[tex] g(x) = x^2-1 [/tex]
I recognise that over this interval, [tex] f(x) > g(x) for all x [/tex], so the answer should be equal to the integral of the difference f(x) - g(x) with respect to x over the previously said interval.
How do I tackle this from here? My graphics calculator software tells me the area is equal to 2.33..., but I'm not sure how to arrive at this value.
Homework Equations
[tex]
\int_{a}^{b} [f(x) - g(x)] dx
[/tex]
The Attempt at a Solution
To show that I can work these integrals, I first show another example which may assist to solve the original example.
[tex] f(x) = x(x^2 - 1) [/tex]
[tex] g(x) = x [/tex]
The interval we must evaluate this integral on is as follows: [tex] [-1, 2^{1/2}] [/tex].
Over the interval [0, 1], f(x) > g(x). Over the interval [tex] [0, 2^{1/2}] [/tex], g(x) > f(x).
So to formalise this now, we have the following:
[tex]
\int_{-1}^{0} [x(x^2 - 1) - x] dx + \int_{0}^{2^{1/2}} [(x - x^3 - x)] dx [/tex]
[tex] = -\int_{0}^{-1} [x^3 - x - x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = -\int_{0}^{-1} [x^3 - 2x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = -\left[ \frac{x^4}{4} - {x^2} \right]_{0}^{-1} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = [-\frac{-1^4}{4} + -1^2] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = [-\frac{1}{4} + 1] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = \frac{3}{4} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx [/tex]
[tex] = \frac{3}{4} + \left[ \frac{-x^4}{4} + {x^2} \right]_{0}^{2^{1/2}} [/tex]
[tex] = \frac{3}{4} + [-\frac {(2^{1/2})^4}{4} + (2^{1/2})^2] = \frac{3}{4} +[-\frac {4}{4} + 2] [/tex]
[tex] = \frac{3}{4} - 1 + 2 = \frac{3}{4} + 1 [/tex]
[tex] = \frac{3+4}{4} = \frac{7}{4} = 1.75. [/tex]
Now, I believe that, perhaps with the absolute value example, you can substitute values in, similarly to this example I've worked. First, I note that both these functions are monotonically decreasing over the interval [-1, 0] and increasing over [0, 1].
So, I'll try:
[tex] \int_{-1}^{0} [f(x) - g(x)] dx + \int_{0}^{1} [f(x) - g(x)] dx [/tex]
[tex] = -\int_{0}^{-1} [|x| - (x^2 - 1)] dx + \int_{0}^{1} [|x| - (x^2 - 1)] dx [/tex]
[tex] = \left[ \frac{|x^2|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{-1} + \left[ \frac{|x^2|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{1} [/tex]
[tex] = 0 - [\frac{1}{2} - (\frac{-1^3}{3} +1)] + [\frac{1}{2} - (\frac{1}{3} - 1)] [/tex]
[tex] = [-\frac{1}{2} + (-\frac{1^3}{3} + 1)] + [\frac{1}{2} - \frac{1}{3} + 1] [/tex]
[tex] = - \frac{1}{2} - \frac{1}{3} + 1 + \frac{1}{2} - \frac{1}{3} + 1 = \frac{4}{3} [/tex]
Clearly, I'm out by 1. Where have I gone wrong? Any assistance would be simply grand. I tried with the first integral being positive, but I ended with the same result mysteriously enough. I just don't understand!
Cheers,
Davin