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if∫0,3f(x)dx=12, ∫0,6f(x)dx=42,find ∫3,6(2f(x)−3)dx=12.

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- Thread starter lab-rat
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- #1

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if∫0,3f(x)dx=12, ∫0,6f(x)dx=42,find ∫3,6(2f(x)−3)dx=12.

- #2

Mark44

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This is a little difficult to read. I think this is what you're being asked:

if∫0,3f(x)dx=12, ∫0,6f(x)dx=42,find ∫3,6(2f(x)−3)dx=12.

[tex]\int_0^3 f(x)dx = 12[/tex]

[tex]\int_0^6 f(x)dx = 42[/tex]

[tex]\text{Find }\int_3^6 (2f(x) - 3)dx = 12[/tex]

I used LaTeX to format what you wrote, but I don't think the last equation is correct. I believe it should be

[tex]\text{Find }\int_3^6 (2f(x) - 3)dx[/tex]

Your book should have some theorems about definite integrals. That's where you need to be looking.

- #3

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The thing is, it that last equation is correct. It is =12. Which is why it doesn't make sense to me.

- #4

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Have you calculated the value of the integral?

- #5

Mark44

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1) Given the first two equations, the value of the integral in the third equation couldn't be 12. I get a significantly larger number.

2) The wording of the problem is weird, especially the word "find". It would make sense to find the value of the integral, or to show that the integral's value was some particular number. As used in this problem, it is like saying find x = 2. What is the exact wording of this problem? Are you translating from some other language?

- #6

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- #7

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- #8

Mark44

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Your book should have some theorems about definite integrals. That's where you need to be looking.

- #9

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We don't have a text book for this class.

- #10

gb7nash

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if∫0,3f(x)dx=12, ∫0,6f(x)dx=42,find ∫3,6(2f(x)−3)dx=12.

If this is your problem word for word, then you need to get confirmation on what the problem is. Like Mark said, the conclusion to this "problem" is false. I'm guessing the problem is to find [itex]\int_{3}^6 (2f(x) - 3)dx[/itex], but nobody here will know unless you find out what the true problem statement is.

- #11

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I have looked through all of my notes and can't find anything to get me started...

- #12

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The first step is to recognise what the 2 integrals that are given, are given for. You are told that

[tex]\int^{6}_{0}f(x)dx=42[/tex]

And that

[tex]\int^{3}_{0}f(x)dx=12[/tex]

Now these are obviously going to be needed to solve the problem, try to find a way to use them.

- #13

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Is this correct? If so I don't know how 2f(x)-3 will affect the integral without knowing what the actual function is... I must be missing something

- #14

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Is this correct? If so I don't know how 2f(x)-3 will affect the integral without knowing what the actual function is... I must be missing something

Yes that is correct.

So now that you know

[tex]\int^6_3f(x)dx = 30 [/tex]

You have everything you need to integrate it.

- #15

Mark44

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There is no integration that takes place, since the function is unknown and unknowable. This is a matter of substitution only.Yes that is correct.

So now that you know

[tex]\int^6_3f(x)dx = 30 [/tex]

You have everything you need to integrate it.

The piece that the OP is missing is this:

[tex]\int_a^c f(x)~dx = \int_a^b f(x)~dx + \int_b^c f(x)~dx [/tex]

This is the theorem I was referring to back in post #2, but since the class is being presented without a textbook, that makes it harder to look up a theorem in the book.

- #16

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There is no integration that takes place, since the function is unknown and unknowable. This is a matter of substitution only.

The piece that the OP is missing is this:

[tex]\int_a^c f(x)~dx = \int_a^b f(x)~dx + \int_b^c f(x)~dx [/tex]

This is the theorem I was referring to back in post #2, but since the class is being presented without a textbook, that makes it harder to look up a theorem in the book.

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]

- #17

hunt_mat

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- #18

Mark44

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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?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]

- #19

gb7nash

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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.

- #20

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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??

- #21

gb7nash

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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]?

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...

- #22

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- #23

Curious3141

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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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