Calculus - Area Between Curves

Fairly certain I am making a stupid mistake somewhere along the way, but no matter how many variations I attempt I can not get the right answer.

y=x+1
y=9-x^2

x=-1, x=2

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

Alright, so I know it is A = Integral (From A to B) upper function - lower function

So A = Integral (From -1 to 2) (9-x^2) - (x-1)
A = Integral (From -1 to 2) (-x^2-x+8)

I then integrate;

(-x^3/3 - x^2/2 + 8x) | -1 to 2

Then I do F(-1) - F(2).

Question

Calculus - Area Between Curves

Fairly certain I am making a stupid mistake somewhere along the way, but no matter how many variations I attempt I can not get the right answer.

y=x+1
y=9-x^2

x=-1, x=2

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

Alright, so I know it is A = Integral (From A to B) upper function - lower function

So A = Integral (From -1 to 2) (9-x^2) - (x-1)
A = Integral (From -1 to 2) (-x^2-x+8)

I then integrate;

(-x^3/3 - x^2/2 + 8x) | -1 to 2

Then I do F(-1) - F(2).

anonymous · Answer

But, after doing F(-1) - F(2) I can never seem to get the right answer. Must be doing something wrong and I assume it is a basic, silly mistake. Thanks for any help!

anonymous · Answer

Calculus 2 I presume?

anonymous · Answer

Yes, Cal 2

anonymous · Answer

I am in calculus three. I hated calculus two. haha

anonymous · Answer

Do you have the right answer but you aren't sure how to work it?

anonymous · Answer

I will try to solve it if you know the answer, otherwise I may give you an incorrect answer and I don't wanna do that.

anonymous · Answer

Yes. The answer is in the back of the book and I am fairly sure I knew how to work it (I put almost all of the steps for working it in the question), but when I work it out I can NEVER get the answer in the back of the book, which is 32/3

anonymous · Answer

Ok Turing, now I did that - and got my intersections of 2.37 and -3.37 ,but isnt the question asking you to analyze from -1 to 2?

anonymous · Answer

Archtype I will try to work it for you, but let me suggest this website to you. It is the best math site out there. http://tutorial.math.lamar.edu/ go to cal 2 notes and it shows you how to work it.

anonymous · Answer

Haha, that is how I found this website Jmtrapasso. Thanks a lot for the link. I actually worked most of the problem out using that site (thats where I got my formulas), but even using that can not get the correct answer. Ill post the algebra I did -

anonymous · Answer

Do you go to Lamar? :O

anonymous · Answer

$$\int\limits_{-1}^{2}(-x^2-x+8)$$
=-(1/3)x^3 - (1/2)x^2 + 8x | -1, 2

= [-(1/3)(-1)^3 - (1/2)(-1)^2 + 8(-1)] - [-(1/3)(2)^3 - (1/2)(2)^2 + 8(2)] 
= [1/3 - 1/2 - 8] + [8/3 + 2 - 16]

= Negative number, not right!

anonymous · Answer

No, I don't go to Lamar, just a great set of notes.

turingtest · Answer

the way it's worded doesn't really make it clear,but it does seem you're right, lets see what I get:$$\int\limits\limits_{-1}^{2}(9-x^2)-(x+1)dx=\int\limits\limits_{-1}^{2}-x^2-x+8dx$$$$=-x^3/3-x^2/2+8$$
evaluated from -1 to 2
(-8/3-2+8)-(-1/3-1+8)=7/3-1=4/3
is 4/3 not correct?

anonymous · Answer

Ok, I am a BIT of an idiot, as I was looking at the answer to the wrong problem.

The answer is supposed to be 19.5, which might be what I got. Need to figured it out again.

anonymous · Answer

Turning you left off an x on the 8, i think?

anonymous · Answer

Well, I got 19.49 if you completely ignore the negative signs.... which isn't really possible. I got -8.16 + 11.3333

anonymous · Answer

I am not getting 19.5 :(

turingtest · Answer

me neither...
and you're right about the x, but still can't get it

anonymous · Answer

Alright. Thanks a lot for your help guys. Ill probably just move on and ask for help in class. I really appreciate the effort and I will probably be back again soon ;) Hopefully I can give you both a medal.

anonymous · Answer

My guess is the book is wrong. That happens more than often. Wish I could be of more help. Goodluck!

anonymous · Answer

I got the right answer if you are still here.

anonymous · Answer

I am!

anonymous · Answer

We were doing bottom function - top function. Flip them around and you get 8x - 1/3(x^3) - 1/2 (x^2) after you integrate. Then plug in the (2) - (1) and you get 19.5

anonymous · Answer

(2) - (-1) *

anonymous · Answer

That is really weird. Not sure how the heck x+1 can be considered the top function.

anonymous · Answer

$$\int\limits_{-1}^{2?}  9 - x^2 - x -1$$

anonymous · Answer

But I just started working another problem and again I am getting the wrong numerical answer. I am going to try flipping them and see if that gives the right answer.

anonymous · Answer

lol, yeah, that was the problem on this new one as well. I am clueless how those are the top equations. I mean... I am looking at graphs of both problems and it is fairly obvious that it was being done right.

anonymous · Answer

Dawkins gives an example of were sqr root of x is the top and x^2 is the bottom. It can be really confusing at times! Like in calculus III we are doing double integrals in 3D. It makes me cry at night :(

anonymous · Answer

I should create a new thread to give you another medal :) I'm really happy that one problem is fixed. But that is ridiculouse (RE: Dawkins)

anonymous · Answer

And kill me now (double integrals... 3d....)

anonymous · Answer

Well I am going to log off for now. Very glad I could help you! Was nice to meet you, and if you have any other questions please feel free to message me ;)

anonymous · Answer

You too! Thanks a lot, and good luck with your Cal III adventures!

anonymous · Answer

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