How can I integrate the function (http://www.wolframalpha.com/input/?i=rect(t)-rect(t-1) ):
$$rect(t)-rect(t-1)$$
in an intuitive way, or without using heaviside function? Until -0.5 the integral is 0, after I have a straight line $t$ until 0.5 and after?
Thank you very much.
According to Wolfram Mathworld, the "rect" function $\Pi$ is defined by $$ \Pi(t) := \begin{cases} 1 & \text{if $|t|<\frac{1}{2}$}, \\ \frac{1}{2} & \text{if $|t| = \frac{1}{2}$, and} \\ 0 & \text{if $|t| > \frac{1}{2}$.} \end{cases} $$ This actually makes it very easy to integrate, if you keep in mind that the integral represents the (signed) area bounded between the curve, the limits of integration, and the horizontal axis.
With $f(t) := \Pi(t) - \Pi(t-1)$, the integral of $f$ is always going to represent the area of a rectangle, or the sum of the areas of two rectangles. For example, $$ \int_{0}^{2} f(t)\,\mathrm{d}t $$ is the area under the first bump, bounded on the left by the line $x=0$, and above the second bump. The area under the first bump is a rectangle with width $\frac{1}{2}$ and height 1, hence it has area $\frac{1}{2}$. The area above the second graph is a rectangle with width 1 and height $-1$, hence it has area $-1$. Thus $$ \int_{0}^{2} f(t)\,\mathrm{d}t = \frac{1}{2} - 1 = -\frac{1}{2}.$$ Visually, this looks like
The Desmos worksheet from which that image was created can be found this link.