anonymous
  • anonymous
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))
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
your formula for the area is not quite right
anonymous
  • anonymous
ok i'm not sure what I did wrong
anonymous
  • anonymous
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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
you can't take the derivative, the area is a function of the definite integral
anonymous
  • anonymous
woops i meant anti derivative
TuringTest
  • TuringTest
what do you get after integrating?
anonymous
  • anonymous
Well integrating the equation I got what I said before, ((1/k)+(E^-k))...i didn't try to solve it any further
TuringTest
  • TuringTest
you did not evaluate correctly\[A(x)=\int_1^k\frac1{x^2}-e^{-x}dx=\left.-\frac1x+e^{-x}\right|_1^k=\left(-\frac1k+e^{-k}\right)-\left(-1+\frac1e\right)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.