Please help me with this proof.

f(x) is even on [-a, a] and integrable on [0, a]; i.e. there exists int(0,a;f(x)dx).

Prove that int(-a,a;f(x)dx) = 2*int(0,a;f(x)dx).

Thanks for any help,

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- May 15th 2007, 07:18 AMCalculusFanIntegral of even functions
Please help me with this proof.

f(x) is even on [-a, a] and integrable on [0, a]; i.e. there exists int(0,a;f(x)dx).

Prove that int(-a,a;f(x)dx) = 2*int(0,a;f(x)dx).

Thanks for any help, - May 15th 2007, 08:27 AMalinailiescu
Re:

- May 15th 2007, 08:45 AMPlato
In the above, did you assume that Int[-a,0]f(x)dx exsits?

I think that we need to prove that. - May 15th 2007, 08:54 AMalinailiescu
Since I showed that Int[-a,0]f(x)dx=Int[0,a]f(-x)dx, where f is integrable on [0,a], I did prove it, don't I?

- May 15th 2007, 09:00 AMPlato
- May 15th 2007, 10:37 AMCalculusFan
Thanks alinailiescu and Plato for your answers.

But in your development, alinailiescu, when you changed from variable u to x, something hit me:

int(0, a);f(-u)du = int(0, a);f(-x)dx

In my opinion (I'm a newbie) you should substitute -u by x (because you set u = -x) and du by -dx. If not which could be the explanation ?

Thanks a lot for your attention! - May 15th 2007, 10:57 AMThePerfectHacker
You

**cannot**use change of variable. I believe you need the facts that: u=-x is differenciable on (-a,a) and u' is continous on (-a,a) which is true. But you also need to know that f is continous on [-a,a]. The problem does not state that, it just states that f is integrable not necessarily continous.

I will try to do this problem without using the substitution theorem because it is faulty here. - May 15th 2007, 11:32 AMCaptainBlack
Consider any partition P of [-a,0], then there is a corresponding partition

P' = -P of [0,a], and vice versa. Also as f is even the upper and lower

Reimann sums for the partitions P and P' and f are equal, hence if one

limit exists as the length of the longest interval in such partitions goes to

zero so does the other, and hence both integrals exist and are equal.

(You may need to tightent this up a bit)

RonL - May 15th 2007, 11:40 AMThePerfectHacker
- May 15th 2007, 01:10 PMCalculusFan
Thank you all people!

All the reasoning you presented added to my understanding of this defying area of Math. But CapitainBlank's argument put an end to the discussion that I'm having with my mates: we all, finally, agree with the proof.

:) - May 15th 2007, 02:59 PMPlato
Even though it is true that the condition that

*f*be continuous is found in most calculus text and even beginning analysis text, it is sufficient that*f*be integrable with additional conditions on*u*. Jerrold Marsden has a good discussion in his text on analysis.

This should be a warning to all: don’t assume too much.

Clearly RonL’s solution is what was looked for.

I for one mistakenly assumed the question was aimed at a different level. - May 15th 2007, 03:47 PMThePerfectHacker
I am a formalist, that does not apply to me.

---

Anyway, here is what Captain**Blank**was saying.

With my commentaries.

Note, I will use the Darboux Integral.

Which can also be found here.

**Theorem:**Let f be integrable on [0,a] (for a>0) and define g(x) = f(-x) for all x in [0,a]. Then g(x) is integrable on [-a,0] and furthermore its integral is equal to the integral of f on [0,a].

**Proof:**The fact the g is bounded is easily seen. Now follow the attachment. - May 15th 2007, 03:58 PMPlato
I frankly do not understand your point!

The Darboux Integral is equivalent to the Riemann Integral.

Remembering that you are using Ken Ross’s book, look at his chapter of the Integral. Ross says that he has entitled the section ‘Riemann Integral’ ever though he will use the Darboux approach because it is equivalent to that of Riemann. - May 15th 2007, 03:59 PMThePerfectHacker
- May 16th 2007, 07:02 AMcurvatureThis is a well-known result
This well-known result can be found in every college calculus book.