Splitting up an interval of integration

["Don't panic!"]

How does one prove the following relation?

[tex]\int_{a}^{b}f(x)dx= \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx [/tex]

Initially, I attempted to do this by writing the definite integral as the limit of a Riemann sum, i.e.

[tex] \int_{a}^{b}f(x)dx= \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k})[/tex]

Where [itex] x^{\ast}_{k}\in\left[x_{k}, x_{k+1} \right] [/itex].

Then,

[tex] \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx= \\ = \lim_{n\rightarrow\infty}\frac{(c-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) +\lim_{n\rightarrow\infty}\frac{(b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(c-a)+ (b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k} = \int_{a}^{b}f(x)dx[/tex]

But I have a feeling that this isn't correct?!

 

[ShayanJ]

Your feeling is correct!
The two Riemann sums are not the same because they're taken in different intervals and so, in general, the terms aren't equal and so you can't factor them!
Consider two functions h(x) and k(x) defined as:
[itex]
h(x)=\left\{ \begin{array}{cc} f(x) \ \ \ \ x \in (a,c) \\ 0 \ \ \ \ otherwise \end{array} \right.
[/itex]
and
[itex]
k(x)=\left\{ \begin{array}{cc} f(x) \ \ \ \ x \in [c,b) \\ 0 \ \ \ \ otherwise \end{array} \right.
[/itex].
Its obvious that [itex] f(x)=h(x)+k(x) [/itex] when [itex] x \in (a,b) [/itex]. Use this for the proof!

 

["Don't panic!"]

Ah, ok. thanks for your help.

Would this be correct then?

[tex] \int_{a}^{b}f(x)dx = \int_{a}^{b}\left(h(x)+k(x)\right)dx = \\ = \int_{a}^{b}h(x)dx +\int_{a}^{b}k(x)dx = \int_{a}^{c}h(x)dx +\int_{c}^{b}k(x)dx = \int_{a}^{c}f(x)dx +\int_{c}^{b}f(x)dx [/tex]

As [itex] h(x)[/itex] is zero outside the interval [itex](a,c)[/itex] and so will provide no further contributions after this point in the integral, and similarly for [itex]k(x)[/itex].

 

[ShayanJ]

Yes, that's it!

 

["Don't panic!"]

Ok, cool. Thanks very much for your help with this, much appreciated!