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 k1 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))

Question

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))

turingtest · Answer

your formula for the area is not quite right

anonymous · Answer

ok i'm not sure what I did wrong

anonymous · Answer

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?

turingtest · Answer

you can't take the derivative, the area is a function of the definite integral

anonymous · Answer

woops i meant anti derivative

turingtest · Answer

what do you get after integrating?

anonymous · Answer

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

turingtest · Answer

you did not evaluate correctly$$A(x)=\int_1^k\frac1{x^2}-e^{-x}dx=\left.-\frac1x+e^{-x}\right|_1^k=\left(-\frac1k+e^{-k}\right)-\left(-1+\frac1e\right)$$