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UnkleRhaukus
\[\begin{align*} \\\int_0^tf(t-u)\text du=\int_0^tf(u)\text du\\ \end{align*}\]?
\[\text{(i)}\] \[\begin{align*} \\&\int_0^tf(t-u)\text du \\\text{let } v=t-u\\ \text dv=-\text du\\ \\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\ &=\int_t^0f(v)\cdot -\text dv\\ &=\int_0^tf(v)\text dv\\ \end{align*}\] \[\text{(ii)}\] \[\begin{align*} \\&\int_0^tf(t-u)g(u)\text du\\ \\\text{let } v=t-u\\ \text dv=-\text du\\ \\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\ &=\int_t^0f(v)g(t-v)\cdot -\text dv\\ &=\int_0^tf(v)g(t-v)\cdot \text dv\\ \end{align*}\]
\i dont really understand these results
\[\begin{align*} \\\int_0^tf(t-u)\text du=\int_0^tf(u)\text du\\ \end{align*}\]?
something about the constant of integrations?
Change of variables is worth knowing how to do. \[\int\limits_{0}^{t}f(t-u) du\] we can let v= t-u this means u= t-v (solve for u) assuming t is constant, take the derivative of both sides du = -dv next, the limits are in terms of u, u=0 to u=t using v= t-u, we find the limits in terms of v are v= t-0= t, and v= t-t=0 replace t-u with v, replace du with -dv \[\int\limits_{t}^{0}f(v) (-dv)\] Finally, we can reverse the order of integration (going from t to 0 with -dv is the same as going 0 to t with +dv \[\int\limits_{0}^{t}f(v) dv\] and as v is just a dummy variable of integration, re-write it (if you like) as \[\int\limits_{0}^{t}f(u) du\]
what is all means is f(t-u) integrated from 0 to t is the same as integrating f(u) from 0 to t You can think of the f(t-u) as starting at the right side (at t) and (because of -u) integrating back to 0, and f(u) is done "normally" from 0 to t
oh so its just forwards and backwards
yes, just the where you start and what direction... but it is the same end-points and the same function. We have to be careful of the sign (going right to left gives the negative of going left to right)
makes a lot more sense now , thank you
part ii is showing that convolution can be calculated "either way" flip f, and slide f past g or flip g and slide g past f