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Questions about definite integrals |
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| Sep14-12, 03:34 PM | #1 |
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Questions about definite integrals
While reading my calc book, I have developed a few questions about the situations in which definite integrals can exist. I've thought about these questions, and I feel that if I am able to answer some of them, I can make some other problems much easier, such as testing for convergence of a definite integral.
1) If a definite integral diverges, can we conclude that it does not exist? 2) If a definite integral does not exist, can we conclude that it diverges? 3) If [itex] f(a) [/itex] is not defined, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist? 3) If [itex] f(a) [/itex] is not defined, but but [itex]\lim_{x→a+}f(x) [/itex] is, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist? 5) If [itex] f(a) [/itex] is defined, but [itex]\lim_{x→a+}f(x) [/itex] does not, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist? I assume all these questions can be answered using the definition of the definite integral, along with the definition of continuity, but how exactly? Thanks! Note: Assume that b>a for the above integrals BiP |
| Sep14-12, 08:55 PM | #2 |
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An integral is just a fancy way of doing a sum.
So. 1. yes and no - depends what you mean by "exist". 2. no. Can you think of where diverging integrals may come from - say - in Nature? 3. no. f(x) only needs only to exist in (a,b) not [a,b] to be integrable. 4. Also answered by 3. 5. Can you come up with an example to illustrate this? I think you are describing a situation where f is not a function or is not peicewise continuous within (a,b). Generally, in order to be integrable, f has to be a piecewise-continuous function in the range of the integration. Look these up. |
| Sep14-12, 09:43 PM | #3 |
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| Sep14-12, 09:46 PM | #4 |
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Questions about definite integralsAlso, I have heard of something called Lipschitz continuity, does that have any bearing on this problem? Also, for a Riemann integral, does the function have to be continuous in (a,b) or in [a,b] ? wikipedia proves the FTC assuming continuity in [a,b] but I assume that is superfluous condition and that continuity in (a,b) suffices to determine the existence of a riemann integral? Thanks! BiP |
| Sep14-12, 11:23 PM | #5 |
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The simplest way to create an undefined value is to divide by zero.
If I define a function u(x): u=0 for x≤0 and u=1 for x>0 then f(x)=1/u(x-a) would be undefined for x≤a ... but I can still do the integral from a to b if b>a. (It's b-a). I'm pretty sure you can do the rest. |
| Sep14-12, 11:50 PM | #6 |
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$$\int_a^bf(x)dx = \lim_{N\rightarrow \infty} \frac{b-a}{N} \sum_{n=0}^N f(\frac{b-a}{N}n + a)$$... but if I just bring the sum in each end by one unit ... in the limit N → ∞ it's the same. $$\int_a^bf(x)dx = \lim_{N\rightarrow \infty} \frac{b-a}{N} \sum_{n=1}^{N-1} f(\frac{b-a}{N}n + a)$$ ... though there is danger in that kind of argument of shaving a bit of each end until there's nothing left. I can also take the sum to within a small distance ##\epsilon << (b-a)/N## of the limits ... so I do the sum in range ##[a+\epsilon, b-\epsilon]##, instead of ##[a,b]##, and see what happens in the limit ##\epsilon \rightarrow 0##. $$\int_a^bf(x)dx = \lim_{N\rightarrow \infty}_{\epsilon \rightarrow 0}\frac{b-a-2\epsilon}{N} \sum_{n=0}^N f(\frac{b-a-2\epsilon}{N}n + a + \epsilon)$$ I'm sure AlephZero can refine this argument. Few beginning courses are even this explicit anyway. |
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