3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x.

Question

I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you

Answer 1

The curves intersect st (-1,0) and (0,0) so the area is

∫[-1,0] (x^3-x^2) - (2x^2+2x) dx
= 1/4 x^4 - x^3 - x^2 [-1,0]
= -1/4

Was there some rotation involved?

Answer 2

No there wasn't. This is exactly the problem our substitute gave us, and I was having a hard time figuring out why that was so.

Do they not intersect at (2, 0) as well?

I was told this answer is incorrect as well (-1/4). I'm perplexed. :S

Answer 3

Don't think so.
f(2) = 4
g(2) = 12

If the answer involves pi, there has to be some rotating involved. That answer must be for some other problem.

Check with your instructor. Something's amiss here.

Answer 4

I took a look at Reiny's solutions. He does assume you are revolving the region, so that's bogus.

I see I made a typo. The curves do also intersect at (2,12)

If you want algebraic (signed) area,
∫[-1,2] (x^3+x^2) - (2x^2+2x) dx
= -9/4

If you want total unsigned area, then you need

∫[-1,0] (x^3-x^2) - (2x^2+2x) dx
+ ∫[0,2] (2x^2+2x)-(x^3+x^2) dx
= 5/12 + 8/3
= 37/12

So, if you got 5/12, it looks like you only figured the left-hand side of the area.

Answer 5

I absolutely understand this now. Thank you so much. I really couldn't have done it without your help Steve.

Answer 6

Glad to help. Gotta watch those typos, especially the copy/paste ones!