Riemmann/Darboux-Integral Question

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In summary, the first part of the conversation states that if a function F is continuous on [a,b] and its integral with any continuous function G on [a,b] is equal to 0, then F must be equal to 0. The second part asks if this is still true if G(a) = G(b) = 0. The answer is yes, as long as F is only nonzero very near a and b. This can be proven by considering a function G(x) = F(x) * (x-a) * (b-x), which is continuous and only vanishes at the endpoints.
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Homework Statement



(i) Suppose that F is continuous on [a,b] and [tex]\int_a^b FG = 0[/tex] for all continuous functions G on [a,b]. Prove that F = 0.

(ii) Suppose now that G(a) = G(b) = 0. Does it again follow that F must be identically zero?

Homework Equations


Let P be any partition of [a,b]

Upper Darboux
[tex]U(F,P) = \sum_{k=0}^\n sSupF(x_k)(t_k - t_{k-1})[/tex]
Lower Darboux
[tex]L(F,P) = \sum_{k=0}^\n iInfF(x_k)(t_k - t_{k-1})[/tex]

F is integrable iff
inf U(F,P) = U(F) = L(F) = sup L(F,P)

The Attempt at a Solution



For (i) I argued the following.

Let G = F.
Clear H = F*F >= 0 and is continuous.
Let's say H(y) > 0.
So I can find an interval (y - c, y + c) because H is continuous, where H(x) > H(y)/2.
Clearly U(H,P) > H(y)/2*c
Which means U(H) >= H(y)/2*c > 0.

Thus F must be identically zero.

For (ii), I said it still needed to be identically zero, and I'm not certain how the argument needs to be changed. Perhaps if F was only nonzero very near a and b, but because of the continuity of F, wouldn't that still bee a problem?
 
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  • #2
How about G(x)=F(x)*(x-a)*(b-x)? There's nothing special about (x-a)*(b-x), it's just a positive continuous function on [a,b] that vanishes only at the endpoints.
 
  • #3
Thanks Dick, works perfectly. =)
 

FAQ: Riemmann/Darboux-Integral Question

1. What is the Riemann integral?

The Riemann integral, named after mathematician Bernhard Riemann, is a method for computing the area under a curve by dividing the region into smaller rectangles and summing their areas. It is used to calculate the definite integral of a continuous function over a closed interval.

2. How is the Riemann integral different from the Darboux integral?

The Riemann integral and the Darboux integral are two different approaches to calculating the area under a curve. While the Riemann integral uses rectangles with fixed widths, the Darboux integral uses rectangles with varying widths. This allows the Darboux integral to give more accurate results for certain functions.

3. Can any function be integrated using the Riemann or Darboux method?

Both the Riemann and Darboux integrals can be applied to any continuous function over a closed interval. However, for certain functions, the two methods may give different results due to the difference in their approaches.

4. What is the significance of the Riemann and Darboux integrals in calculus?

The Riemann and Darboux integrals are fundamental concepts in calculus and are used to define the definite integral. They allow us to calculate the area under a curve, which has many real-world applications in fields such as physics, engineering, and economics. These integrals also form the basis for other important concepts in calculus, such as the Fundamental Theorem of Calculus.

5. How do the Riemann and Darboux integrals relate to the concept of limits?

The Riemann and Darboux integrals are both based on the concept of limits. The Riemann integral uses the limit of the width of the rectangles approaching zero to calculate the area under a curve, while the Darboux integral uses the limit of the number of rectangles approaching infinity. In both cases, the limit allows for a more precise calculation of the area under the curve.

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