Proof Involving Absolute Value of Integral

[turophile]

Here's the problem:

Show that \(\displaystyle \left|\int_{a}^{b}f\right|\leq\int_a^b|f|\). (In case that notation doesn't make sense, here is it in words: Show that the absolute value of the integral of f from a to b is less than or equal to the integral of the absolute value of f from a to b.)

My question:

Any hints on getting started with this one?

 

[mmm4444bot]

Understanding why the inequality holds is a good start, before writing a proof. Have you gotten that far? (If not, it has to do with signed areas between the curve of f and the x-axis, and whether we change the curve to |f(x)| before integrating versus integrating first and taking the absolute value of the result.) Once you know what's going on, perhaps, you could take cases.

 

[Wally]

Proofs are normally in the context of the course being given since an outsider can't know what theorems or other tools are available for use in proving something. However, this proof probably goes back to the definition of the definite integral as a Riemann sum. It is reasonable to say that the summation of SIGMA a[sub:3513ceb7]i[/sub:3513ceb7] is less than or equal to a summation of SIGMA | a[sub:3513ceb7]i[/sub:3513ceb7] |.

 

[BigGlenntheHeavy]

\(\displaystyle This \ should \ help.\)

\(\displaystyle Evaluate \ \int_{0}^{2}|2x-1|dx\)

\(\displaystyle First \ note \ |2x-1| \ = \ 2x-1. \ x \ \ge \ 1/2 \ and \ |2x-1| \ = \ -(2x-1) \ if \ x \ < \ 1/2.\)

\(\displaystyle Hence, \ \int_{0}^{2}|2x-1|dx \ = \ \int_{0}^{1/2}-(2x-1)dx \ + \ \int_{1/2}^{2}(2x-1)dx \ = \ 5/2.\)

\(\displaystyle Now, \ what \ is \ the \ value \ of \ \int_{0}^{2}(2x-1)dx?\)

\(\displaystyle Does \ \int_{1/2}^{2}|2x-1|dx \ = \ \int_{1/2}^{2}(2x-1)dx, \ If \ so. \ why?\)

\(\displaystyle See \ graph.\)

[attachment=0:16jog3ie]aaa.jpg[/attachment:16jog3ie]

 

[turophile]

Thanks for all of these ideas. To give some context, this problem comes from a section in the chapter on the integral titled "Properties of the integral; antiderivatives". The Riemann integral is introduced a few sections after this one. The properties introduced in this section are (assuming f and g continuous on [a, b]):

(a) \(\displaystyle \int_a^b(f+g)=\int_a^bf+\int_a^bg\)

(b) \(\displaystyle \int_a^b(f-g)=\int_a^bf-\int_a^bg\)

(c) \(\displaystyle \int_a^bcf=c\int_a^bf\)

(d) If \(\displaystyle f\geq0\) on [a, b] then \(\displaystyle \int_a^bf\geq0\).

(e) If \(\displaystyle f\geq{g}\) on [a, b] then \(\displaystyle \int_a^bf\geq\int_a^bg\).

(f) Also, a definition is given: assuming a < b, \(\displaystyle \int_b^af=-\int_a^bf\)

So I think that the proof probably needs to use at least one of these. I'm thinking (d) will probably be needed for this proof, though I'm not quite seeing how to use it.

 

[turophile]

BGTH's example is very helpful. Working it out on paper, and using 0 to 1 as the limits of integration, I can see that \(\displaystyle \left|\int_0^1f\right|=0\) but \(\displaystyle \int_0^1\left|f\right|=1/2\). So this demonstrates the principle in that case. But how do I go about applying one or more of (a) through (f) in my previous post to write a general proof for this problem?

 

[BigGlenntheHeavy]

\(\displaystyle Good \ question \ turophile, \ where \ are \ our \ purists \ when \ we \ need \ them?\)

 

[Deleted member 4993]

Thanks for all of these ideas. To give some context, this problem comes from a section in the chapter on the integral titled "Properties of the integral; antiderivatives". The Riemann integral is introduced a few sections after this one. The properties introduced in this section are (assuming f and g continuous on [a, b]):

(a) integral{a to b} (f + g) = integral{a to b} f + integral{a to b} g

(b) integral{a to b} (f - g) = integral{a to b} f - integral{a to b} g

(c) integral{a to b} cf = c integral{a to b} f

(d) If f >= 0 on [a, b] then integral{a to b} f >= 0.

(e) If f >= g on [a, b] then integral{a to b} f >= integral{a to b} g.

(f) Also, a definition is given: assuming a < b, integral{b to a} f = - integral{a to b} f

So I think that the proof probably needs to use at least one of these. I'm thinking (d) will probably be needed for this proof, though I'm not quite seeing how to use it.

You are using the theorem above where f(x) become negative

 

[turophile]

Using mmm4444bot's suggestion, I'll proceed with the proof by breaking things up into cases.

1. First, consider f ? 0 on [a, b]. Then by (b), above, the integral of f from a to b ? 0. Also, since f ? 0 on [a, b], then f = |f| on [a, b], so the integral of f from a to b equals the integral of |f| from a to b, and the absolute value of the integral of f from a to b equals the integral of |f| from a to b. So, for f ? 0 on [a, b], the absolute value of the integral of f from a to b is less than or equal to the integral of the absolute value of f from a to b.

2. Now consider f ? 0 on [a, b]. By (f) and (b), above, we can reverse the limits of integration so that f ? 0 on [b, a], and repeating the discussion in (1), we see that that the absolute value of the integral of f from a to b is less than or equal to the integral of the absolute value of f from a to b.

3. So now I need to deal with the situation where f is neither (A) greater than or equal to 0 on [a, b] nor (B) less than or equal 0 on [a, b] (as in BGTH's example). And in that example, the absolute value of the integral of f from a to b is less than the integral of the absolute value of f from a to b. But I'm not sure how to show this generally. Should I make a general example where f is less than zero an arbitrary number of times on [a, b] and, consider the integral of f to be the sum of the integrals of partitions of [a, b], where the partitions are made to separate out the intervals where f is less than zero from those where it is greater than or equal to zero, and then show that the absolue value of the sum of the integrals of the partitions of f is less than or equal to the sum of the integrals of the partitions of the absolute value of f?

Or is there a better way to go about this proof? Any additional guidance would be much appreciated.