anonymous
  • anonymous
Hey, could someone please help me with the following question: Find the area of the bounded region with the given functions y=1/x, y=x, y=1/4x. My book gives the answer ln(2), but I don't know how. When I attempted to solve the problem I made the integral from 0 to 2, based off my graph and integrated the (x-1/x-0.25x)dx? How should I combine the given equations to find the area?
MIT 18.01 Single Variable Calculus (OCW)
  • 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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1328490010040:dw| |dw:1328490384526:dw|
anonymous
  • anonymous
This is provided the area is that found in the first quadrant.
anonymous
  • anonymous
Thank you a ton- your response was super helpful.

Looking for something else?

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

More answers

anonymous
  • anonymous
Quick question- why does the integral have yo be broken up- the area is all above the x-axis?
anonymous
  • anonymous
Never mind I see why- the function that is on top changes from [0,2] , so the integral end points has to change to reflect that switch.
anonymous
  • anonymous
yep!

Looking for something else?

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