Question

\[\begin{align*}
\\\int_0^tf(t-u)\text du=\int_0^tf(u)\text du\\
\end{align*}\]?

Answer 1

\[\text{(i)}\]
\[\begin{align*}
\\&\int_0^tf(t-u)\text du
\\\text{let } v=t-u\\
\text dv=-\text du\\
\\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\
&=\int_t^0f(v)\cdot -\text dv\\
&=\int_0^tf(v)\text dv\\
\end{align*}\]
\[\text{(ii)}\]
\[\begin{align*}
\\&\int_0^tf(t-u)g(u)\text du\\
\\\text{let } v=t-u\\
\text dv=-\text du\\
\\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\
&=\int_t^0f(v)g(t-v)\cdot -\text dv\\
&=\int_0^tf(v)g(t-v)\cdot \text dv\\
\end{align*}\]

Answer 2

\i dont really understand these results

Answer 3

which one?

Answer 4

either ?

Answer 5

\[\begin{align*}
\\\int_0^tf(t-u)\text du=\int_0^tf(u)\text du\\
\end{align*}\]?

Answer 6

Do you follow the math?

Answer 7

yes

Answer 8

maybe

Answer 9

something about the constant of integrations?

Answer 10

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
\[\int\limits_{0}^{t}f(u) du\]

Answer 11

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

Answer 12

oh so its just forwards and backwards

Answer 13

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)

Answer 14

makes a lot more sense now ,
thank you

Answer 15

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

Answer 16

yes