I'm just learning about the convolution integral and am stuck on this example problem:
Given
$x_1(t) = \left \{\begin{array}{lr} 1 : 0 < x < 1 \\ 0: \text{elsewhere} \end{array} \right.,\text{ and}$
$\\ x_2(t) = \left \{\begin{array}{lr} 1 : 0 < x < 1 \\ 2: 1 < x < 2 \\ 0: \text{elsewhere} \end{array} \right.$
Convolve $x_1$ and $x_2$ graphically.
I have approached the problem as follows:
Graph of $x_1$ and $x_2$:
I will do $x_2 * x_1 = \int_0^t\ {x_2(\lambda)x_1(t-\lambda)\ d\lambda}$, and thus:
$\forall t < 0: x_2 * x_1 = 0 \text{, since }x_1(t - \lambda) \text{ and } x_2(\lambda) \text{ do not overlap here. Similarly,} \\ \forall t > 3: x_2 * x_1 = 0. \\ 0 < t < 1: x_2 * x_1 = t \text{, since the product of }x_1(t - \lambda) \text{ and } x_2(\lambda) \text{ is 1 here}$ $\\ \text{and is integrated from 0 to } t. \\ 1 < t < 2: x_2 * x_1 = 2t -2 \text{, since the prodcut of } x_1(t - \lambda) \text{ and } x_2(\lambda) \\ \text{ is 2 here and is integrated from 1 to t. This interval is where I'm having trouble;} \\ \text{ the book tells me I should be getting } t. \\ 2 < t < 3: x_2*x_1 = 6 - 2t \text{, since the product is 2 here and is integrated from } t - 1 \text{ to 2}.$
I haven't illustrated the convlution process graphically here.
I'm grateful for all the help.
When $1<t<2$, the product of $x_1(t-\lambda)$ and $x_2(\lambda)$ is indeed $2$ when $1<\lambda<t$, but the product also equals $1$ when $t-1<\lambda<1$.