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Definite integral

 
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Jan18-12, 11:27 AM   #18
 
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Definite integral


Quote by Bread18 View Post
You still have to integrate it, perhaps evaluate would have been a better word, but either way, he now was to integrate
[tex]\int^6_32f(x)-3dx[/tex]
Evaluate is a better word, and that will involve substitution (i.e., replacing specific integrals by their known values). Integration is not a part of this problem. How could it be, since f(x) is not known?
Jan18-12, 04:02 PM   #19
 
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Quote by Mark44 View Post
Integration is not a part of this problem. How could it be, since f(x) is not known?
Well, technically it still is, just not on f(x). You still have to integrate 3.
Jan19-12, 06:35 PM   #20
 
All right, well this is probably wrong but here is what I got!

I took out the 2, which I didn't I could do which is probably what confused me the most..
I'm having trouble with the symbols right now but I essentially put the 2 in front of the integral of the function and substracted the integral of 3. Which gave me 60-3x... Is that right??
Jan19-12, 07:05 PM   #21
 
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Quote by lab-rat View Post
but I essentially put the 2 in front of the integral of the function and substracted the integral of 3.
That sounds right. So you rewrote your problem as [itex]2 \int_{3}^6 f(x)dx - \int_{3}^6 3 dx[/itex]?

Quote by lab-rat View Post
Which gave me 60-3x... Is that right??
The first part is fine, but the bolded part doesn't make any sense. If you're working with a definite integral...
Jan19-12, 07:41 PM   #22
 
Yes that is how I wrote it. How should I solve the second part? I've been looking everywhere... The only thing I could find was that it's supposed to be the anti derivative.. Well 3x is the antiderivative of 3 isn't it?
Jan19-12, 08:05 PM   #23
 
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Quote by lab-rat View Post
Yes that is how I wrote it. How should I solve the second part? I've been looking everywhere... The only thing I could find was that it's supposed to be the anti derivative.. Well 3x is the antiderivative of 3 isn't it?
Yes, so substitute the upper bound into that expression to get a value. Then substitute the lower bound into that expression to get a value. Then take the first value minus second value, and that's the definite integral.

Remember [itex]\int_a^b f(x)dx = F(b) - F(a)[/itex], where F(x) is the antiderivative of f(x).
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