Integral on domain equals minus integral on the reverse domain?

[isalloum4]

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

Homework Equations

The Attempt at a Solution

 

[Goa'uld]

Is this a homework question or your own question?

 

[isalloum4]

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

 

[STEMucator]

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?

 

[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

 

[STEMucator]

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.

 

[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.

 

[HallsofIvy]

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.

 

[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.