Does the Fundamental Theorem of Calculus Yield a Constant for F(x) = Constant?

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The discussion centers on the application of the Fundamental Theorem of Calculus regarding the function F(x) when F(x) is a constant. It is established that the derivative of a constant function is zero, leading to the conclusion that the definite integral of DF(x)/dx over any interval yields zero. However, the anti-derivative of DF(x)/dx results in F(x) + C, where C is an arbitrary constant. This clarifies that while the definite integral evaluates to zero, the indefinite integral retains the constant term.

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I am embarrassed to ask this question. I have seen PhD's not know the basic answer to questions. I have been humbled. Could someone help please. I am writing a paper and need the answer to a question.

For the Fundamental Theorem for integrals of derivatives

[itex]\int[/itex](DF(x)/dx)dx = [F(x)][itex]^{a}_{b}[/itex]

I did not put in the limits on the first integral, too hard to use this system.

If F(x) = Constant, does the Fundamental Theorem give a constant.

I know the derivative is 0, but the equation I am using is at a min. for this constant, so it should give 0. The answer for x^2 is just x^2 but it is continuous and has a derivative. Questions always come up no matter what level you are at. Another PhD could not answer this. Call me a dummy, I do not care.

Would the answer be a that Constant. I could not find the answer anywhere.
 
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No, the answer for the definite integral is 0, for F(x)=constant.

The ANTI-derivative of DF/Dx, for F(x)=constant is, however, F(x)+Constant.
 

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