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# Silly quick integration question watch

1. For the curve , the graphing software gave me a negative area for x = - 6 to -1. I'm guessing you must take the absolute value?

My second question is..

the software also said the area is 0 for x = -3 to 3, and im guessing it added the positive with the negative. You need to integrate from -3 to 0, then 0 to 3, take absolute values and then add them, right?
2. 1) Yes
2) Yes, or just integrate between limits 0 and 3, then double your answer.
3. 1) Yes.

2) That or integrate between 0 and 3 and just double it.

*Edit*

The guy above copied me, not the other way round .
4. Use the trapezium rule
5. (Original post by samiz20891)
Use the trapezium rule
Why would you do that? You can integrate the function in your head.
6. I used simpson's rule. anyways, thanks but i got one more question

For , (or -2, -3),etc there is no area for say, -2 to 0, or 0 to 2, but there is an area for -3 to -2, or 2 or 3?

What about for -2 to 3, does that exist? I'm guessing no?
7. You can't integrate over asymptotes, ie. where the function is discontinuous over the specified domain.
8. Economist1 is right. But I don't know what you mean by "there is no area from 0 to 2" - there clearly is and it's in fact infinitely large.
9. (Original post by Swayum)
Economist1 is right. But I don't know what you mean by "there is no area from 0 to 2" - there clearly is and it's in fact infinitely large.

I understand not integrating through the asymptote. What i meant by 0 to 2 or -2 to 0 was that because there is an asymptote, the graph would never touch the y axis , therefore you could say there is no area or there is an infinite area. I assume the former is more preffered
10. (Original post by G O D I V A)
I understand not integrating through the asymptote. What i meant by 0 to 2 or -2 to 0 was that because there is an asymptote, the graph would never touch the y axis , therefore you could say there is no area or there is an infinite area. I assume the former is more preffered
The area is infinite.

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