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rocketparrotlet

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I'm falling behind in Calculus, I thought that some of you guys might be able to help me. I am having trouble understanding the fundamental theorem of calculus- where they give you an interval and an equation and you have to find the area. How do you do this?

Any help is appreciated. I need to keep my A in that class.

Thanks,
Mark
 



Hemlock_Mike

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It's been too many years for me but what is interval? Do you mean integral? As I recall -- the first integral gives area.

HMike
 

ZRTMWA

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I'm in calc too, are you in highschool or college calc? We learned the fundamental theorem of calc here in high school calc in the first couple weeks or so but I've forgotten it. Here's the equation for it though:




Check out the "examples" section of the wiki page on the theorem:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
 

Hemlock_Mike

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I was guessing that he was talking about the first integral of an equation between two points to determine area under a portion of a curve.

HMike
 

Byrnz93

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I think I understand your question, but I've been on holidays for some weeks now and I've forgotten nearly everything lol (Actually I'm quite dissappointed in myself, I'm going to revise right now). So I had to look it up to refresh my memory.
If you are talking about the area under a curve, as Hemlock_Mike also thought, then you use antiderivatives and definite intergrals. I can't write down examples to explain it myself, so here's this site Area Under a Curve It will do a better job of explaining.

I seem to recall the equation ZRTMWA posted, gonna read over it now. From what I remember, it does the job as well.
 

Hemlock_Mike

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Byrnz ---

Man -- does that bring back memories!!! That is what I was thinking was needed.

HMike
 

Meatball

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The area under the curve = [F(x+h)-F(x)]/h

F = notation for the anti-derivative of f(x)

x = a point on the x-axis

h= a second point on the x-axis, usually > than x, but not always.


Simple example:

if f(x)=2x

then F(x)=x^2

The area under the curve from 3-5, or [3,5] is equal to

[F(5)-F(3)]/2

= [25-9]/2
= 8 (area under the graph f(x)=2x from 3 to 5)

See? It's not so bad!

Notice that by doing an anit-derivative, you are undoing a derivative. You are now multiplying dx with f(x) instead of taking dy/dx like we do with a derivative. Multiplying these together gives you an answer in squared terms, which is how we get area.

f(x)*dx = F(x) (units squared)

Does that help?

-Tyler
 
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rocketparrotlet

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That helps a lot! Just one question- why is h 2?

They gave us a different equation, one that was more confusing and involved both f(x) and f'(x), but related the two of them instead of the area.

-Mark
 

Byrnz93

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h=2 because it is the difference between the two points. F(x) is where x=3. F(x+h) is where x=5 without +h the second would also equal 3. Therefore to find h, 5-3=2. At least I think that is it (my teacher is gonna be yelling at me so much for forgetting this lol).

Also for the equation they gave you, f(x) is the original equation and f'(x) is the derivative. f(x) could also be thought of as the antiderivative of f'(x), in which case it would be written as F(x).
Hope that helps.
I think I understand the equation you were given, could you explain it?
 

Meatball

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Byrnz93 has got you covered on that one. Just think, if you have two points:

x, and x+h

then the difference between them is h.

x+h-x = h

Are you confused as far as notation? Relating f(x) and f'(x)?

d(F(x)) = f(x)

d(f(x)) = f'(x)

d(f'(x)) = f''(x)

and so on....

I hope we've helped you out at least a little bit!

-Tyler
 

rocketparrotlet

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Got it, thanks! We just got onto the second fundamental theorem of calculus, so I might come on here later with some more questions.

Thank you guys!

-Mark
 
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Mark, I can help you with math and calculus, add me to your MSN if you wish to ask me anything :)
 




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