Proof that sum of integrals of same function on adjacent intervals equals the integral over both intervals I am seeking a proof that
$$\int_a^c f(x) dx + \int_c^b f(x) dx  = \int_a^b f(x) dx$$
Please use Stewart's (Calculus: Early Transcendentals, 2016, 8e, p. 378) definition of the definite integral:

This is what I have:
Suppose 
$$u=\int_a^c f(x) dx = \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{c-a}{n})$$ 
and 
$$v=\int_c^b f(x) dx = \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n})$$
Then 
$$u+v = \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n}) + \lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{c-a}{n}) $$
$$=\lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n} + \frac{c-a}{n})$$
$$=\lim \limits_{n \to \infty} \sum_{i=0}^n f(x_i^*) (\frac{b-a}{n})$$
$$=\int_a^b f(x) dx$$
Edit: As @Matthew Conroy pointed out, the $f(x_i^*)$ cannot be factored out in the sum because the values of $x_i^*$ are different for each definite integral. So, this proof is invalid.
 A: Each integral is the difference of two primitives:
$$\int^c_a f(x) dx = F(c) - F(a)$$.
Therefore, the LHS becomes $F(c) - F(a) + F(b) - F(c) = F(b) - F(a)$, which is the RHS.
A: Given $\epsilon >0$, let $N$ be sufficiently large that all three of
$|\int_a^b f(x) dx - \sum_{i=0}^n f(x_i^*) (\frac{b-a}{n})|,$
$|\int_a^c f(x) dx - \sum_{i=0}^n f(x_i^*) (\frac{c-a}{n})|,$
$|\int_b^c f(x) dx - \sum_{i=0}^n f(x_i^*) (\frac{b-c}{n})|$
are less than $\epsilon$  for all $n>N$ and all choices of the $x_i^*$.
Now consider any $n$ which is greater than both $\frac{b-a}{c-a}N$ and $\frac{b-a}{b-c}N$. Let $k$ be the integer part of $\frac{c-a}{b-a}n$, then $k\ge N$. 
Consider a division of $[a,b]$ into $n$ subintervals and a division of $[a,c]$ into $k$ subintervals. For $i\le k$, the two $i$th subintervals of each are
$[a+(i-1)\frac{b-a}{n},a+i\frac{b-a}{n}]$ and $[a+(i-1)\frac{c-a}{k},a+i\frac{c-a}{k}]$.
By definition, $k+1>\frac{(c-a)n}{b-a}\ge k$ and so $1\ge \frac{(b-a)k}{(c-a)n}>\frac{k}{k+1}>\frac{k-1}{k}\ge \frac{i-1}{i}.$
Therefore $a+i\frac{b-a}{n}>a+(i-1)\frac{c-a}{k}\ge a+(i-1)\frac{b-a}{n}$ and so the two $i$th subintervals have a non-zero intersection. Choose $x_i^*$ to be in this intersection. 
Define the remaining $n-k$ $x_i^*$s in an analogous manner on $[c,b]$, with $y_i^*=x_{i+k+1}^*.$
Then $\sum_{i=0}^n f(x_i^*) (\frac{b-a}{n})=\sum_{i=0}^k f(x_i^*) (\frac{b-a}{n})+\sum_{i=k+1}^n f(x_i^*) (\frac{b-a}{n})$
$$=\frac{k(b-a)}{n(c-a)}\sum_{i=0}^k f(x_i^*) (\frac{c-a}{k})+\frac{(n-k)(b-a)}{n(b-c)}\sum_{i=0}^{n-k-1} f(y_i^*) (\frac{b-c}{n-k}).$$
Note that $\frac{k(b-a)}{n(c-a)}$ lies between $1$ and $\frac {k}{k+1}$ and therefore tends to $1$ as $n$ tends to infinity. Similarly for $\frac{(n-k)(b-a)}{n(b-c)}$.
Thus we can prove that $|\int_a^c f(x) dx + \int_c^b f(x) dx  - \int_a^b f(x) dx|$ is smaller than any given positive number and is therefore zero.
A: 
Lemma
We use the following important equivalent form of integral:$$\int_a^bf(x)dx=\lim_{n\to \infty} \sum_{i=s_n}^{t_n}f(x_i)\Delta x$$where $\Delta x={k\over n}$ for some arbitrary $k>0$, $x_i=i\Delta x$ and $${s_n:\Bbb N\to \Bbb N\\t_n:\Bbb N\to \Bbb N\\\lim_{n\to \infty}{t_n\over n}={b\over k}\\\lim_{n\to \infty}{s_n\over n}={a\over k}}$$a proof can be found at the end of the answer.

Using the lemma, let $\Delta x={c-a\over n}$ and $x_i=i\Delta x$. Therefore by defining $N_a=\lfloor{a\over c-a}n\rfloor$ and $N_b=\lfloor{b\over c-a}n\rfloor$ we obtain$$\int_a^b f(x)dx=\lim_{n\to \infty}\sum_{i=N_a}^{N_b}f(x_i)\Delta x\\\int_b^c f(x)dx=\lim_{n\to \infty}\sum_{i=N_b+1}^{n}f(x_i)\Delta x$$which are true since $$\lim_{n\to\infty}{N_a\over n}={a\over c-a}\\\lim_{n\to\infty}{N_b\over n}={b\over c-a}$$Finally
$$
\int_a^b f(x)dx+\int_b^c f(x)dx{=\lim_{n\to \infty}\sum_{i=N_a}^{N_b}f(x_i)\Delta x+\sum_{i=N_b+1}^{n}f(x_i)\Delta x\\=\lim_{n\to \infty}\sum_{i=N_a}^{n}f(x_i)\Delta x\\=\int_a^cf(x)dx}$$Hence the proof is complete $\blacksquare$
Proof of the lemma
Since it is clear that $$\int_a^bf(x)dx=\int_0^{b-a}f(x+a)dx$$(because of the invariance of area under shifting), we may without loss of generality assume that $a=0$. To complete the proof of the lemma we must show that $$\lim_{n\to \infty}\sum_{i=\lfloor{an\over k}\rfloor}^{s_n}f(x_i)\Delta x=0\\\lim_{n\to \infty}\sum_{i=t_n}^{\lfloor{bn\over k}\rfloor}f(x_i)\Delta x=0$$since the function $f(x)$ is bounded, we may write $$m<f(x)<M$$ for some $m,M\in\Bbb R$ therefore
$$
-{k|m|\over n}\left|s_n-\lfloor{an\over k}\rfloor\right|<
\sum_{i=\lfloor{an\over k}\rfloor}^{s_n}f(x_i)\Delta x< {k|M|\over n}\left|s_n-\lfloor{an\over k}\rfloor\right|
$$
since both the limits tend to zero by our assumption, then $$\lim_{n\to \infty}\sum_{i=\lfloor{an\over k}\rfloor}^{s_n}f(x_i)\Delta x=0$$and similarly$$\lim_{n\to \infty}\sum_{i=t_n}^{\lfloor{bn\over k}\rfloor}f(x_i)\Delta x=0$$This completes the proof of the lemma $\blacksquare$
