# Definite integral

by lab-rat
Tags: definite, integral
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P: 805
 Quote by Mark44 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.
 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??
HW Helper
P: 805
 Quote by lab-rat 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 $2 \int_{3}^6 f(x)dx - \int_{3}^6 3 dx$?

 Quote by lab-rat 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...
 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?
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P: 2,640
 Quote by lab-rat 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 $\int_a^b f(x)dx = F(b) - F(a)$, where F(x) is the antiderivative of f(x).

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