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marytheshade
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let r be the region enclosed by the graph of f(x)=1/x^2 g(x)=e^x and the line x=1 and x=k where k>1 find the limit as k approaches infinity of A(k).
A(k) we derived before and I believe that it is the integral from 1 to k ((1/k)+(E^k))
 one year ago
 one year ago
marytheshade Group Title
let r be the region enclosed by the graph of f(x)=1/x^2 g(x)=e^x and the line x=1 and x=k where k>1 find the limit as k approaches infinity of A(k). A(k) we derived before and I believe that it is the integral from 1 to k ((1/k)+(E^k))
 one year ago
 one year ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.1
your formula for the area is not quite right
 one year ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
ok i'm not sure what I did wrong
 one year ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
I mean initially it was integral from 1 to k (1/2x^2e^x)dx but when you differentiate wouldn't you get what I got?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you can't take the derivative, the area is a function of the definite integral
 one year ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
woops i meant anti derivative
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what do you get after integrating?
 one year ago

marytheshade Group TitleBest ResponseYou've already chosen the best response.0
Well integrating the equation I got what I said before, ((1/k)+(E^k))...i didn't try to solve it any further
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you did not evaluate correctly\[A(x)=\int_1^k\frac1{x^2}e^{x}dx=\left.\frac1x+e^{x}\right_1^k=\left(\frac1k+e^{k}\right)\left(1+\frac1e\right)\]
 one year ago
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