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\[\begin{align*}
\\\int_0^tf(tu)\text du=\int_0^tf(u)\text du\\
\end{align*}\]?
 one year ago
 one year ago
\[\begin{align*} \\\int_0^tf(tu)\text du=\int_0^tf(u)\text du\\ \end{align*}\]?
 one year ago
 one year ago

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\text{(i)}\] \[\begin{align*} \\&\int_0^tf(tu)\text du \\\text{let } v=tu\\ \text dv=\text du\\ \\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\ &=\int_t^0f(v)\cdot \text dv\\ &=\int_0^tf(v)\text dv\\ \end{align*}\] \[\text{(ii)}\] \[\begin{align*} \\&\int_0^tf(tu)g(u)\text du\\ \\\text{let } v=tu\\ \text dv=\text du\\ \\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\ &=\int_t^0f(v)g(tv)\cdot \text dv\\ &=\int_0^tf(v)g(tv)\cdot \text dv\\ \end{align*}\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\i dont really understand these results
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\begin{align*} \\\int_0^tf(tu)\text du=\int_0^tf(u)\text du\\ \end{align*}\]?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
something about the constant of integrations?
 one year ago

phiBest ResponseYou've already chosen the best response.2
Change of variables is worth knowing how to do. \[\int\limits_{0}^{t}f(tu) du\] we can let v= tu this means u= tv (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= tu, we find the limits in terms of v are v= t0= t, and v= tt=0 replace tu 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, rewrite it (if you like) as \[\int\limits_{0}^{t}f(u) du\]
 one year ago

phiBest ResponseYou've already chosen the best response.2
what is all means is f(tu) integrated from 0 to t is the same as integrating f(u) from 0 to t You can think of the f(tu) 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
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
oh so its just forwards and backwards
 one year ago

phiBest ResponseYou've already chosen the best response.2
yes, just the where you start and what direction... but it is the same endpoints and the same function. We have to be careful of the sign (going right to left gives the negative of going left to right)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
makes a lot more sense now , thank you
 one year ago

phiBest ResponseYou've already chosen the best response.2
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
 one year ago
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