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

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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\i dont really understand these results

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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0something about the constant of integrations?

phi
 4 years ago
Best ResponseYou've already chosen the best response.2Change 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 \[\int\limits_{0}^{t}f(u) du\]

phi
 4 years ago
Best ResponseYou've already chosen the best response.2what 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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0oh so its just forwards and backwards

phi
 4 years ago
Best ResponseYou've already chosen the best response.2yes, 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)

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0makes a lot more sense now , thank you

phi
 4 years ago
Best ResponseYou've already chosen the best response.2part 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
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