Integral on domain equals minus integral on the reverse domain?

isalloum4
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Homework Statement



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

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The Attempt at a Solution

 
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Is this a homework question or your own question?
 
I am reading Apostol calculus and I couldn't figure out how the above relation got derived!
 
isalloum4 said:
I am reading Apostol calculus and I couldn't figure out how the above relation got derived!

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?
 
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
 
isalloum4 said:
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

Think about this integral :

##\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?

Now think about this one :

##- \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.
 
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.
 
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.
 
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.
 
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