3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x.
I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you
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?
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
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.
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.
I absolutely understand this now. Thank you so much. I really couldn't have done it without your help Steve.
Glad to help. Gotta watch those typos, especially the copy/paste ones!
To find the area bounded by two curves, you need to find the points of intersection and then integrate the difference between the two curves over that interval.
First, find the points of intersection by setting the two equations equal to each other:
x^3 + x^2 = 2x^2 + 2x
Rearrange the equation to get:
x^3 - x^2 - 2x^2 - 2x = 0
Combine like terms:
x^3 - 3x^2 - 2x = 0
Factor out an x:
x(x^2 - 3x - 2) = 0
Set each factor equal to zero:
x = 0
x^2 - 3x - 2 = 0
Using the quadratic formula, solve for x:
x = (3 ± sqrt(3^2 - 4(-2)))/(2)
x = (3 ± sqrt(17))/2
So the two points of intersection are:
x = 0
x = (3 + sqrt(17))/2
Now, evaluate the integrals to find the area between the curves.
To find the area between the two curves, use the following integral:
A = ∫[a,b] [f(x) - g(x)] dx
where [a,b] is the interval over which the two curves intersect.
In this case, the bounds of integration are 0 and (3 + sqrt(17))/2.
So the integral becomes:
A = ∫[0, (3 + sqrt(17))/2] [(x^3 + x^2) - (2x^2 + 2x)] dx
Simplifying:
A = ∫[0, (3 + sqrt(17))/2] (x^3 - x^2 - 2x^2 - 2x) dx
A = ∫[0, (3 + sqrt(17))/2] (x^3 - 3x^2 - 2x) dx
Integrating term by term:
A = (1/4)x^4 - (x^3) - x^2 evaluated from x = 0 to x = (3 + sqrt(17))/2
A = (1/4)((3 + sqrt(17))/2)^4 - ((3 + sqrt(17))/2)^3 - ((3 + sqrt(17))/2)^2 - (1/4)(0)^4 - (0)^3 - (0)^2
Evaluating this expression gives:
A ≈ 3.1927
So the area bounded by the curves f(x) = x^3 + x^2 and g(x) = 2x^2 + 2x is approximately 3.1927.
Therefore, it seems there was an error in the calculation you provided earlier.