Question about fundamental thm calc part 1

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In summary, the conversation discusses the differentiability of the function g defined by g(x) = ∫f(t)dt on the interval [a,b] and the possible discrepancy with the statement that g' = f. It is explained that differentiability is often only defined on open intervals and if one wishes for g to be differentiable on [a,b], there needs to be a definition for differentiability on closed intervals.
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Miike012
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If f is continuous on [a,b] then the function g defined by
g(x) = ∫f(t)dt a <= x <= b is continuous on [a,b] and differentiable on (a,b) and g' = f

Question...

If f is continuous on [a,b], then why is g only differentiable on (a,b)?
This does not make sence... if g' = f g should be diff on [a,b] after making the first statement "f is continuous on [a,b]"
 
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  • #2
The thing is that differentiability is often only defined on open intervals. This is likely what your text has done.

If you want g to be differentiable on [a,b], then you need to have a definition for differentiability on closed intervals.

Mod note: as this is not a homework problem, I mover it to the math forums.
 

FAQ: Question about fundamental thm calc part 1

What is the fundamental theorem of calculus part 1?

The fundamental theorem of calculus part 1 states that if a function is continuous on a closed interval and has an antiderivative, then the definite integral of that function over that interval can be calculated by evaluating the antiderivative at the endpoints of the interval.

How is the fundamental theorem of calculus part 1 used?

The theorem is used to simplify the process of calculating definite integrals by providing a direct relationship between integrals and antiderivatives. It allows for the evaluation of definite integrals without having to use limit notation or Riemann sums.

What is an antiderivative?

An antiderivative, also known as an indefinite integral, is a function whose derivative is equal to the original function. In other words, it is the reverse process of differentiation. For example, the antiderivative of 2x is x^2 + C, where C is a constant.

How does the fundamental theorem of calculus part 1 relate to the area under a curve?

The fundamental theorem of calculus part 1 states that the definite integral of a function over a closed interval is equal to the area under the curve of the function on that interval. This allows for the calculation of areas under curves without having to use geometric methods.

Are there any restrictions on the use of the fundamental theorem of calculus part 1?

Yes, there are certain conditions that must be met in order for the theorem to be applicable. The function must be continuous on the interval and have an antiderivative. Additionally, the interval must be closed and finite. If any of these conditions are not met, the theorem cannot be used.

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