Riemann Integrability of the indicator function-- small clarification 
Suppose $a\le s<t\le b$. Define $f:[a, b]\mapsto \mathbb{R}$ by
$$f(x) = \begin{cases}
1 \; \; s<x<t \\
0 \; \; \text{otherwise} 
\end{cases}$$
Prove that $f$ is Riemann integrable on $[a, b]$ and that $\int_a^b f = t-s$.

I am trying to construct a proof of this statement, but I don't see how the assertion $\int_a^b f = t-s$ is true. If we think of $\int_a^b f$ in terms of area under a curve, I would say that $\int_a^b f = t-s$ would hold iff $f(x) = 1$ for when $s\le x\le t$ (replacing strict inequalities with non-strict ones) and $0$ otherwise. Can someone please explain how $\int_a^b f = t-s$ holds given how $f$ is defined originally?
Note: Please don't give a proof of the statement since that is my HW assignment. I'd appreciate just some intuitive clarifications/ explanations.

Notice that $$\int_a^bfdx=\int_{a}^sfdx+\int_s^tfdx+\int_t^bfdx=\int_s^t1dx$$ When the function it is the value $1$, what the integral really means, it's the lenght of the interval. Or in terms of area, it means the area of a rectangle of length $1$ and width $t-s$.

