# Integral on domain equals minus integral on the reverse domain?

1. Aug 17, 2013

### isalloum4

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

Why integral from a to b of f(x) equals "minus (-)" integral from b to a of f(x)??? when a<b or a>b

2. Relevant equations

3. The attempt at a solution

2. Aug 17, 2013

### Goa'uld

Is this a homework question or your own question?

3. Aug 17, 2013

### isalloum4

I am reading Apostol calculus and I couldn't figure out how the above relation got derived!

4. Aug 17, 2013

### Zondrina

You want this I believe :

$\int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx$

Is it that you're not understanding how to prove it or is it conceptually confusing?

5. Aug 17, 2013

### isalloum4

Exactly that I meant! It is conceptually confusing ( even though I thought I am good at that) and I don't know where it came from!
Many thanks for help

6. Aug 17, 2013

### Zondrina

$\int_{a}^{b} f(x) dx$ where you assume $a≤b$.

In words, it represents the area under the curve when we integrate f(x) from a to b. Right?

$- \int_{b}^{a} f(x) dx$ now you assume $a≥b$.

This one will represent the area under the curve when we integrate f(x) from b to a. The negative sign 'flips' the answer around so you will get the same answer as if you integrated from a to b.

I hope that clears that up a bit.

Do you know about Riemann sums at all? The mesh of a partition? Any definitions you have would be great.

7. Aug 17, 2013

### isalloum4

Thanks for your time! But I still don't understand the geometric representation for this. And, both integrals are related to each other only when a<b for both the positive and the negative.

8. Aug 17, 2013

### HallsofIvy

Staff Emeritus
If I remember correctly, Apostol defines the integral $\int_a^b f(x)dx$ through "Riemann sums" only for a< b, the defines $\int_b^a f(x)dx= -\int_a^b f(x)dx$.

9. Aug 17, 2013

### Goa'uld

Here is a simple way to look at this. When $\frac{d}{dx}F(x)=f(x)$, $\int_{a}^{b}f(x)dx=F(b)-F(a)$. Also, $\int_{b}^{a}f(x)dx=F(a)-F(b)=-(F(b)-F(a))$. Therefore, $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$.