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Old September 18th, 2007, 04:00 PM
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Default Intergrals

I am a bit stuck on this:
(note: | = intergration symbol)

2|6-3|6

Now i think im right at that point. Not sure what to do from there?
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Old September 18th, 2007, 04:03 PM
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I don't get very well what you want.

Use some of LaTeX.

See the LaTeX Help forum.
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Old September 18th, 2007, 04:05 PM
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\left[ 2\int_3^5 6\right] - \left[ 3\int_3^5 6\right]

does that make more sense?
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Old September 18th, 2007, 04:10 PM
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Yeah a lot.

It's very easy, since

\int dx=x+k, then \int_a^b dx=b-a

Does that make sense?
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Old September 18th, 2007, 04:13 PM
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wait so i say 5-3? what about the 6 and 3 and 2?
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Old September 18th, 2007, 04:22 PM
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Quote:
Originally Posted by taurus View Post
wait so i say 5-3? what about the 6 and 3 and 2?
2\int_3^5 6 - 3\int_3^5 6

You're integrating with respect to what?

We'll assume dx.

Thus, 2\cdot [6(5) - 6(3)] - 3\cdot [6(5)-6(3)] = (2\cdot 12) - (3\cdot 12) = 24 - 36 = -12
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Old September 18th, 2007, 04:51 PM
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wait i think i got the question wrong, suppose we have:
\int_3^5 f(x) dx = 6, \int_3^6 f(x) dx = 5, \int_3^5 g(x) dx = 6

now it asks to find the value of the following integrals:
\int_3^5 2f(x) -3 g(x) dx =
??
?? any ideas

Last edited by taurus; September 20th, 2007 at 03:47 AM.
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