- #36
ObsessiveMathsFreak
- 406
- 8
nrqed said:This is an eye opener for me!
My formation as a physicist has given me the feeling that riemannian sums are the fundamental definition!
I couldn't find my old notes, but I believe this was the way I was introduced to the concept of definite integration. I wish I could better draw pictures, but http://img337.imageshack.us/img337/4536/fundamentalhi1.png .
Let the operation [tex]\int f(x) dx[/tex] be that which finds the family of antiderivatives to f(x), i.e. [tex]\int f(x) dx = F(x) + C \Rightarrow \frac{dF(x)}{dx} = f(x)[/tex], F(x) is the "principal" antiderivative and C is an arbitrary constant.
OK. We want to find the area under the curve of f(x) between the points a and b, with a<b. Denote the function that gives the area between the points a and an arbitrary point x, as [tex]A_a(x)[/tex]. Note x>a. We seek [tex]A_a(b)[/tex] as our final answer. Note that as the area under the function at anyone point is zero, we automatically have [tex]A_a(a) = 0[/tex]
OK, now to examine the function [tex]A_a(x)[/tex] and an arbitrary point. In paticular we want to examine its derivative. Consider the value of the area at [tex]A_a(x)[/tex]. How will the area change as we change x. Let [tex]\Delta x[/tex] be our change in x. Then the area between a and [tex]x + \Delta x[/tex] is given by [tex]A_a(x+\Delta x)[/tex]. The difference between these is the shaded area on the graph [tex]\Delta A[/tex]. Specifically [tex]A_a(x+\Delta x) - A_a(x) = \Delta A[/tex]
Now, look at the area [tex]\Delta A[/tex]. As [tex]\Delta x \rightarrow 0[/tex], we can approximate this area using the area of the trapezium formed by [tex](x,0),(x+\Delta x,0),(x+\Delta x,f(x+\Delta x)),(x,f(x))[/tex]. By the area of a trapezium formula, we obtain [tex]\Delta A \cong \frac{f(x+\Delta x) + f(x)}{2} ((x+\Delta x) - x) \cong \frac{f(x+\Delta x) + f(x)}{2} \Delta x[/tex].
So equating our representations for [tex]\Delta A[/tex], we have;
[tex]A_a(x+\Delta x) - A_a(x) \cong \frac{f(x+\Delta x) + f(x)}{2} \Delta x[/tex]
Dividing by [tex]\Delta x[/tex]
[tex]\frac{A_a(x+\Delta x) - A_a(x)}{\Delta x} \cong \frac{f(x+\Delta x) + f(x)}{2}[/tex]
Now take the limit as [tex]\Delta x \rightarrow 0[/tex] to equate both sides.
[tex]\lim_{\Delta x \rightarrow 0}\frac{A_a(x+\Delta x) - A_a(x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) + f(x)}{2}[/tex]
We can see that the right hand side is the definition of [tex]\frac{d A_a(x)}{dx}[/tex]
It can be seen that the limit of the average becomes;
[tex]\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) + f(x)}{2} = f(x)[/tex]
Therefore we have that:
[tex]\frac{d A_a(x)}{dx} = f(x)[/tex]
And that means, that [tex]A_a(x)[/tex] must be an anti derivative of f(x), [tex]\int f(x) dx[/tex]. i.e.
[tex]A_a(x) = F(x) + C[/tex]
But which C? Well from above, we know that, [tex]A_a(a) = 0[/tex]. So that means;
[tex]A_a(a) = F(a) + C[/tex]
[tex]0 = F(a) + C[/tex]
[tex]\Rightarrow C = - F(a)[/tex]
So we have that the area under the curve f(x) between x and a is given by;
[tex]A_a(x) = F(x) - F(a)[/tex]
Where F(x) is the "principal" antiderivative of f(x). In fact, F(x) can be any antiderivative as the constant differences will cancel. Thus we have that;
[tex]A_a(b) = F(b) - F(a)[/tex]
We traditionally denote [tex]A_a(b)[/tex] as [tex]\int_a^b f(x) dx[/tex] to empahsise that
[tex]F(b) - F(a) = \int f(x) dx \vert_b - \int f(x) dx \vert_a[/tex]. Where [tex]\vert_{d}[/tex] stands for "evaluation at x=d".
Anyway that was how I learned that the area under a curve between a and b is [tex]\int_a^b f(x) dx[/tex]. I only saw the riemannian sum method later, and was initially quite dubious of it. Hopefully this long winded post will be of some use to anyone who gets through it all.
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