A little confused about integrals

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Discussion Overview

The discussion revolves around the concept of integrals, specifically addressing the confusion regarding the area under a curve, the relationship between integrals and anti-derivatives, and the implications of the Fundamental Theorem of Calculus. Participants explore both theoretical and practical aspects of integration.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that integrals represent the area under the curve of the original function.
  • Others clarify that the area under the curve is closely related to the anti-derivative of the function, referencing the Fundamental Theorem of Calculus.
  • A participant explains that knowing the anti-derivative allows for easier calculation of the area under the curve, as opposed to approximating it through numerous small intervals.
  • Another participant notes that while integration can find the area under a curve, it serves broader purposes beyond this application.
  • There is a discussion about the definition of anti-derivatives, highlighting that an anti-derivative is differentiable while the original function may not be continuous.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the relationship between integrals and anti-derivatives, with some clarifying points while others remain confused. There is no consensus on the necessity of finding anti-derivatives for calculating areas under curves.

Contextual Notes

Some participants express confusion about the definitions and implications of integrals and anti-derivatives, indicating potential gaps in understanding the Fundamental Theorem of Calculus and its applications.

taylor__hasty
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I learned that integrals are finding the area under a curve. But I seem to be a little confused. Area under the curve of the derivative of the function? Or area under the curve of the original function?

If an integral is the area under a curve, why do we even have to find the anti derivative at all?
 
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taylor__hasty said:
I learned that integrals are finding the area under a curve. But I seem to be a little confused. Area under the curve of the derivative of the function? Or area under the curve of the original function?

If an integral is the area under a curve, why do we even have to find the anti derivative at all?

An integral is the area under the curve of the (original) function.

This area is, however, closely related to the anti-derivative of the function. This is the Fundamental Theorem of Calculus.
 
taylor__hasty said:
I learned that integrals are finding the area under a curve. But I seem to be a little confused. Area under the curve of the derivative of the function? Or area under the curve of the original function?

If an integral is the area under a curve, why do we even have to find the anti derivative at all?

The antiderivative (if available) is useful because if ##A(a,z)## is the area under the curve ##y = f(x)## from ##x=a## to ##x=z## (with ##a## some fixed number), and if ##F(x)## is the antiderivative of ##f(x)##, then we have:
\text{area} = A(a,z) = F(z) - F(a)
(Fundamental Theorem of Calculus).

So, if you know the antiderivative function you can use it to calculate the area. That is a lot easier than splitting up the interval ##[a,z]## into 100 billion little intervals and adding up the 100 billion little rectangular areas (which would give a good approximation to the actual area---not necessarily the exact area).
 
taylor__hasty said:
I learned that integrals are finding the area under a curve. But I seem to be a little confused. Area under the curve of the derivative of the function? Or area under the curve of the original function?

If an integral is the area under a curve, why do we even have to find the anti derivative at all?

Integration can be used to find the area under a curve, but that is just one trivial function of the integration tool.

As to anti-derivatives: The definition is that F is the anti-derivative of f if dF/dx = f. This has one important consequence:

F is differentiable. f does not have to be. I fact, f does not even have to be continuous.
 

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