#### rocketparrotlet

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

Thanks,

Mark

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- Thread starter rocketparrotlet
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Any help is appreciated. I need to keep my A in that class.

Thanks,

Mark

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HMike

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Check out the "

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

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HMike

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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.

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Byrnz ---

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

HMike

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

HMike

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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

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

Last edited:

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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

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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

Hope that helps.

I think I understand the equation you were given, could you explain it?

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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

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Thank you guys!

-Mark