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anonymous
 3 years ago
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))
anonymous
 3 years ago
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))

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1your formula for the area is not quite right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i'm not sure what I did wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1you can't take the derivative, the area is a function of the definite integral

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0woops i meant anti derivative

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1what do you get after integrating?

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

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