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marytheshade

  • 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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  1. TuringTest
    • 3 years ago
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    your formula for the area is not quite right

  2. marytheshade
    • 3 years ago
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    ok i'm not sure what I did wrong

  3. marytheshade
    • 3 years ago
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    I mean initially it was integral from 1 to k (1/2x^2-e^-x)dx but when you differentiate wouldn't you get what I got?

  4. TuringTest
    • 3 years ago
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    you can't take the derivative, the area is a function of the definite integral

  5. marytheshade
    • 3 years ago
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    woops i meant anti derivative

  6. TuringTest
    • 3 years ago
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    what do you get after integrating?

  7. marytheshade
    • 3 years ago
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    Well integrating the equation I got what I said before, ((1/k)+(E^-k))...i didn't try to solve it any further

  8. TuringTest
    • 3 years ago
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    you did not evaluate correctly\[A(x)=\int_1^k\frac1{x^2}-e^{-x}dx=\left.-\frac1x+e^{-x}\right|_1^k=\left(-\frac1k+e^{-k}\right)-\left(-1+\frac1e\right)\]

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