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Definite integral

by lab-rat
Tags: definite, integral
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gb7nash
#19
Jan18-12, 04:02 PM
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Quote 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.
lab-rat
#20
Jan19-12, 06:35 PM
P: 44
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??
gb7nash
#21
Jan19-12, 07:05 PM
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Quote 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 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...
lab-rat
#22
Jan19-12, 07:41 PM
P: 44
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?
Curious3141
#23
Jan19-12, 08:05 PM
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Quote 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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