Find the area of the region between y=X^2-2x and the x axis for the interval from 0 to 2.

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Find the area of the region between y=X^2-2x and the x axis for the interval from 0 to 2.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Integrate [f(x) - g(x)] on the given interval (in this case g(x) = 0 because it's the x axis): \[\int\limits_{0}^{2} (x ^{2}-2x) dx\] \[[(x ^{3}/3)-(2x ^{2}/2)]\] (from 0 to 2, I don't know how to make the editor do stacked #'s) \[[((2)^3/3))-(2)^2] - [0]\]
thank you!
ok same question but the second part is to find the area for the interval from 0 to 3

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Same method, just plug "3" in for x instead of 2 in the last step: \[[((3)3/3))−(3)2]−[0]\]
Oops, that came out all wrong! Here it is again: \[[((3)^3/3))−(3)^2]−[0]\]
yea for some reason i'm getting the wrong answer :S this method should give me 0 right?
No, it should equal some number greater than or equal to 0, since it's an area (you can get negative if the area is below the x-axis). So for the first one: \[[((2)^3/3))−(2)^2]−[0] = [(8/3) - 4] = -4/3\] It's probably should have made it more clear that f(x) = (x^2-2x) and g(x) = 0. Sorry if that was confusing!

Not the answer you are looking for?

Search for more explanations.

Ask your own question