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- Thread starter Gear300
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- #2

rock.freak667

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[tex]\int x^N dx =\frac{x^{N+1}}{N+1} + C[/tex]

so you can just use that definition.

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Dick

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[tex]\int x^N dx =\frac{x^{N+1}}{N+1} + C[/tex]

so you can just use that definition.

That's not a definition. It's a theorem. The proof that integration corresponds to antidifferentiation is called the "Fundamental Theorem of Calculus".

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rock.freak667

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ah well I learned something today

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differentiate it!

- #6

HallsofIvy

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Differentiation is a "direct" problem because we are given a formula for it. Integration, or "anti-differentiation" is defined only as the inverse of differentiation. A result of that fact is that the anti-derivatives of most integrable functions cannot be given in terms of elementary functions.

Of course, we also know that there are an infinite number of other functions having that same derivative. fortunately, we can prove, using the mean value theorem, that if two functions have the same derivative, they must differ only by a constant- we know that any anti-derivative of [itex]x^n[/itex] must be of the form [itex](1/(n+1))x^{n+1}+ C[/itex] where C is a constant.

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I see. A lot of information given, thanks.

- #8

Gib Z

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Is there any mathematical proof for the antiderivative or did they simply say that it reverses differentiation and came up with it just by looking at the method for derivation?

When you say "anti derivative" it implies, the process of differentiation reversed. From the rest of your sentence I think you may have instead wanted to use the word "integral". The Only reason we can use those words interchangeably is because of our knowledge of the Fundamental theorem of Calculus, but remember they were originally two different things.

Now if that quote was asking if there is another way to prove [tex]\int^b_a x^n = \frac{b^{n+1} - a^{n+1}}{n+1}[/tex], then yes. Riemann sums.

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