# Subdividing an integral into a sum of integrals over a given interval

1. Jun 30, 2014

### "Don't panic!"

How does one prove the following:
$$\int^{c}_{a} f\left(x\right)dx = \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx$$
where $f\left(x\right)$ is continuous in the interval $x\in \left[a, b\right]$, and differentiable on $x\in \left(a, b\right)$.

My approach was the following:

Given that $\int^{b}_{a} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)$, where $x^{\ast}_{i} \in\left[x_{i},x_{i+1}\right]$, we have that
$$\int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) + \lim_{n\rightarrow \infty} \frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)$$
$$\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \left[\frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)+\frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)\right] = \lim_{n\rightarrow \infty} \frac{1}{n}\left[\left(b-a\right) +\left(c-b\right) \right]\sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)$$
$$\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) = \int^{c}_{a} f\left(x\right)dx$$

However, I have a feeling that this isn't quite correct?!

2. Jun 30, 2014

### micromass

You have made a notational error:

You have said

$$\int_a^b f(x)dx = \lim_{n\rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^n f(x^*_i)$$

which is fine. But then you also do

$$\int_b^c f(x)dx = \lim_{n\rightarrow \infty} \frac{c-b}{n} \sum_{i=1}^n f(x^*_i)$$

However, this is wrong. You cannot use the same notation $x^*_i$ in both equations. Indeed, the x_i^* of the integral $\int_a^bf(x)dx$ are not the same $x_i^*$ of the other integral. So you need to denote it as something else such as

$$\int_b^c f(x)dx = \lim_{n\rightarrow \infty} \frac{c-b}{n} \sum_{i=1}^n f(y^*_i)$$

Also, your points $x_i^*$ should be depending on $n$, since if $n$ changes, then so do these points. So you should probably call them $x_{i,n}^*$.

3. Jun 30, 2014

### "Don't panic!"

Ah ok, thanks for pointing that out. How should I proceed with the proof from that point then?

4. Jul 1, 2014

### HallsofIvy

What you can do is choose a refinement of every partition of [a, c] by adding the point "b". Then you can separate the sum from a to b and the sum from b to c.

5. Jul 1, 2014

### "Don't panic!"

Ah ok, so I should start with $\int_{a}^{c} f\left(x\right) dx = \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n}\sum_{i=1}^{n} f\left(x_{i,n}^{\ast}\right)$, and then redefine the point $x_{i,n}^{\ast}$ by adding the point "b", such that I can split the sum up on the RHS? (i.e. choose a set of points, $x_{i,n}^{\ast} \in\left[ x_{i},x_{i+1} \right]$ in the interval $x\in \left[a,b\right]$, and then another one, $y_{j,n}^{\ast} \in\left[ y_{j},y_{j+1} \right]$ in the interval $x\in \left[b,c\right]$)

Last edited: Jul 1, 2014