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A Few Quick Questions: Integral Notation and Integrals |
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| Dec19-06, 04:36 PM | #1 |
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A Few Quick Questions: Integral Notation and Integrals
1. The problem statement, all variables and given/known data
The integral (2x) dx between x= -2 and 1 and indefinate integral (dx)/((x+3)^2) 2. Relevant equations None really. 3. The attempt at a solution I am working through practice problems for my final later this week. I have come across a few integral problems where the answer comes out as a negative number. I have no problems doing these questions I am just having an issue understanding how an area can be negative. And for the second one does this simply mean that it is integral (1/((x+3)^2) dx?? Thanks |
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| Dec19-06, 04:51 PM | #2 |
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[tex] \int_\pi^{2\pi}sin(x)dx=[-cos(x)]_\pi^{2\pi}=-2 [/tex] Looking at the graph of the sine function, we see that the given integral is in the region where the function is negative, and so this gives rise to the "negative" area under the curve. |
| Dec19-06, 05:22 PM | #3 |
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Thanks,
So when you are asked to find the area under a curve between two points and the function happens to be both above and below the x-axis on that interval. You are really being asked to find the difference between the area above and below the graph to the x axis? I guess I shouldn't be thinking in terms of area then. |
| Dec19-06, 05:47 PM | #4 |
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A Few Quick Questions: Integral Notation and Integrals[tex] \int_0^{2\pi}sin(x)dx[/tex] Because the area above the graph is the same as the area below the graph, you immediately know the integral evaluates to 0, without doing any integration. |
| Dec19-06, 05:48 PM | #5 |
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