Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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.

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

[tex]

\int_{a}^{b} [f(x) - g(x)] dx

[/tex]

3. 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Calculating integrals of area between two functions, involving absolute values

**Physics Forums | Science Articles, Homework Help, Discussion**