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anonymous
 one year ago
Im not sure how to integrate this one.. what's the trick?
image and latex coming ...
anonymous
 one year ago
Im not sure how to integrate this one.. what's the trick? image and latex coming ...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[E^{r t} A'[t]  r E^{r t} A[t] = p[t] E^{r t}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it helps I know already that.. A'[t] = r A[t]  p[t] p[t] = r A[t]  A'[t]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits_{0}^{\infty} p(t) E^{r t} dt = ???\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2\[ \begin{align*} &\phantom{=}\int_0^\infty e^{rt}A'(t)re^{rt}A(t)\,dt\\ &=[e^{rt}A(t)]_0^\infty\\ &=A(0)\\ \end{align*} \] \[ \begin{align*} A(0)&=\int_0^\infty p(t)e^{rt}\,dt\\ A(0)&=\int_0^\infty p(t)e^{rt}\,dt \end{align*} \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2If \(\lim_{x\to \infty}A(t)=c\in \mathbb{R}\) then \([e^{rt}A(t)]_0^\infty=A(0)\). If \(\lim_{x\to \infty}A(t)=\pm\infty\) then we have some trouble.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you.. thomas.. I was fishing for ages for a function that would give me the antiderivative to p(t) e^(rt) so I could use the Fundamental Formula but I couldn't see the connection that A(t) e^(rt) was the antiderivative to r A(t) e^(rt)  A'(t) e^(rt) *head slap*! awesome. ...

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2The instructions above just differentiated \(e^{rt}A(t)\). Of course the result of integrating what was differentiated is the thing being differentiated lol!
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