Let $a,b\in\mathbb C$ and $c\in[a,b]$. Let $f$ be continuous on $[a,b]$. Use the definition to show that \begin{equation} \int_{[a,b]}f=\int_{[a,c]}f+\int_{[c,b]}f \end{equation}
Note: You should stick to the definition, which gives, e.g., $\int_{[a,b]}f=\int_0^1 f(a+t(b-a))\cdot (b-a)dt$.
Ok. My first step is noting that $[a,b]$ is actually the line segment connecting the complex numbers $a$ and $b$, which is the image of the curve \begin{equation} \gamma(t)=a+(b-a)t, \quad 0 \le t \le 1 \end{equation} which explains where the definition of $\int_{[a,b]}$ comes from, using \begin{equation} \int_\gamma f = \int_a^b f(\gamma(t))\gamma'(t)dt \end{equation}
Now here is the start of my proof: \begin{align*} \int_{[a,c]}f+\int_{[c,b]}f&=\int_0^1 f(a+t(c-a))\cdot (c-a)dt+\int_0^1 f(c+t(b-c))\cdot (b-c)dt \\ &=(c-a)\int_0^1 f(a+t(c-a))dt+(b-c)\int_0^1 f(c+t(b-c))dt \end{align*}
I'm not great with manipulating integrals, so I appreciate any help you can offer. Thanks!
What the theorem is saying, in essence, is that you can subdivide the segment and add the integrals over the two pieces, and the result is the same as the integral over the whole segment.
The definition of $\int_{[a,b]}f$ allows you to write
$$\begin{eqnarray} \int_{[a,b]}f &= & \int_0^1 f(a+t(b-a)) (b-a)\; dt \\ & =& \int_0^h f(a+t(b-a)) (b-a)\; dt + \int_h^1 f(a+t(b-a)) (b-a)\; dt \end{eqnarray}$$
where $h$ is real and $0 < h < 1.$ Now see if you can find a suitable value of $h$ for given values of $a$, $b$, and $c$ that makes the two intgrals in the sum above equal to the two integrals in the sum you have already written.