The Fundamental Theorem of Calculus

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Homework Statement


I'm having trouble wrapping my head around this concept. I understand integration and differentiation individually. I even understand the algebraic manipulations that reveals their close relationship. However, the typical geometric interpretation of a 1-D curve being the derivative of the area function below it seems odd to me. I'm trying to get an intuitive understanding of how a 1-D curve, say A'(x) = f(x) , is the derivative of an area function, A(x). I think I might know where my source of confusion lays: When dealing with derivatives, I'm used to visualizing in terms of tangent lines drawn to a 1-D curve. It seems weird to apply it to an area function, which is more irregular polygon than it is 1-D curve. I hope my concerns make sense.

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The Attempt at a Solution

 
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Googling doesn't help ?
 
try googling

mit Session 47: Introduction of the Fundamental Theorem of Calculus

and watch that
 
+ S[x+dx] = S[x] + f(x)dx. Obvious from geometry. => f(x)=S'(x)
 
William White said:
try googling

mit Session 47: Introduction of the Fundamental Theorem of Calculus

and watch that
Great video, but it didn't quite answer my question. I might not have been able to articulate my question that well over the internet. Oh, well. Thanks, anyhow.
 
I think the missing link between integration and differentiation is primitiv(-isation,-ization, -ation, ..., I don't know). It is the 'reciprocal' operation to differentiation: ##\phi## is a primitive of a piecewise continuous fonction ##f## if and only if ##\phi## is differentiable on the domain of ##f## and ##\phi' = f##.
The set of all primitives of ##f## vary by an additive constant. So all primitives of ##f## have the form ##\phi = \phi_0 + C##, where ##\phi_0## is a particular primitive of ##f##.
It happens that the function ##\phi_0(x) :=\int_{x_0}^x f(t) dt ## is a primitive of ##f## wherever ##f## is continuous (just show that if ## f## is continuous in ##a##, ## \frac{\phi_0(x) - \phi_0(a)}{x-a} \rightarrow f(a) ## as ## x\rightarrow a## ). I think it explains a little bit the link between integration and differentiation.The link between the area under the curve of ##f##, defined on ##[a,b]##, and its integral, comes from the construction of the integral of piecewise continuous functions from the integral of step-functions on ##[a,b]##, which are homogenous to an area
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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