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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Hi all. I am onto Calculus now. Thing is, I have only done 2 lessons and never studied it before. I'm getting on alright, but I have a really hard question (well, hard to me). It's an extra question, so if I don't do it it isn't a big deal, but I am trying to figure it out.

This (if it works) is a roughly drawn picture. It says:

The figure shows the curve with equation y=5+2x-x^2 and the line with equation y=2. The curve and the line intersect at the points A and B.

a) Find the x-coordinates of A and B.

The shaded region R is bounded by the curve and the line.

b) Find the area of R.

Can someone please take me through step by step how you find the x-Coordinates of A and B? I do have the answers by the way, just don;t know how to get to it. Our teacher gives us the final answers so we can check we have it right, and he marks us on our working.

The thing that's confusing me the most is that the equation isn't in the normal format, and so the x^2 value is negative, which I always thought meant an imaginary number, so I think I've missed something.

As I say, my teacher did say he doesn't mind if I don't do this question, as I have never done calculus before, but I am now intrigued as to how it's done.

Thanks all.

Don't think outside the box. Think there is no box

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,998

Hi amberzak;

For the first problem. Since you know y = 2 then just plug it in to get the points of intersection.

a)

Can you solve that or do you need help?

**In mathematics, you don't understand things. You just get used to them.**

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

I don't want to sound dumb, but could you just put the solving up? I think I have it, but I just need to double check I've done it right. (My biggest problem is confidence)

Don't think outside the box. Think there is no box

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,998

Hi;

Subtract 2 from both sides and turn it around.

Multiply everything by -1

Can you factor that?

**In mathematics, you don't understand things. You just get used to them.**

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Thanks. I didn't do the multiply everything by -1.

So the factoring would be:

(x-3)(x+1) and the points at A and B would be 3 and -1. Is that right?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,998

Correct! Very good!

**In mathematics, you don't understand things. You just get used to them.**

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Okay, so next part of the question. (I can't do the notation on here so I'm just going to write it without the correct notation).

Intergrating 5+2x-x^2.

=5x+2x^2 - x^3

------ ------

2 3

Then supplement the two coordinates in for x

= (5x3 + 2x3^2 - x^3 ) - (5x-1 + 2x-1^2 - -1^3)

------- ------ -------- -----

2 3 2 3

and that's where I get a bit unstuck. The first part is easy enough:

=15 +9 -3) - (this is the bit i am sure I have wrong)

Don't think outside the box. Think there is no box

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

hi amberzak,

The area you have found is for all of the bit below the curve down to the x axis.

The region R is less than that by one rectangle.

Once you subtract that, you should get the required answer.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

I've got

(15 + 18 - 9) - (-5 + 2 + 1/3)

And the rectangle is 4 x 2 = 8.

B

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Could you go through step by step. I'm now completely lost.

The answer on my answer sheet, by the way, is 10 and 2/3

Don't think outside the box. Think there is no box

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

Hi amberzak

Do you know the formula for finding area between graphs of two functions?

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

hi amberzak,

LATER EDIT. THERE IS AN ERROR WITH MY INTEGRATION HERE. SEE POST #20

I'll have a go. I've made a diagram below and shaded some regions.

You question asks for the region I've coloured red.

When you do

you will get the area below the curve down to the x axis. That's what area type integration does.

(Why is a much longer post for another day I think!)

So the integration gives an answer that is too big. Take off 8 for the green rectangle and you should get the right result.

So let's check the integration

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

Hi Bob

That doesn't give a precise answer. And the precise one is even easier than what you did there. You just subtract the two functions (the quadratic and the linear one) and integrate from -1 to 3.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

hi Stefy,

I'm glad you are looking at this too.

I'm getting an answer of 18 and 2/3. Any ideas?

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

Well,you did say to take off 8.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

Sorry for double posting.

You integration is not correct!!!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

Yes, but that was my answer after taking off 8.

ie. 26 and 2/3 take 8

??

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

Look at the post right above your new one.

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Why is the integration not right? Is it because it's missing the denominators for the 2x3^2 and 2x-1^2

(sorry if I'm not making a lot of sense. I've been doing calculus all day)

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,469

You integration is not correct!!!

I see it now. I've spent all day driving and gardening (big tree to cut down) so my brain is not at its best. Here is a correction to post #12.

= 18 and 2/3

less 8 = 10 and 2/3

My apologies for the error before. I need some sleep.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

No,he didn't integrate the term 2x correctly.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
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Ahead of you for the first time. see post 20

Thanks for your help Stefy. I'm off to bed. Bye.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,606

I saw it. I still do not get why the two small parts next to the rectangle don't mess it up.

Here lies the reader who will never open this book. He is forever dead.

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Thanks guys. That's really helpful.

I'll be coming on tomorrow with your questions

Don't think outside the box. Think there is no box

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**amberzak****Member**- Registered: 2012-03-16
- Posts: 80

Anon, I was wondering that as well, actually. I might ask my teacher than tomorrow.

Don't think outside the box. Think there is no box

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