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#1 2014-05-21 01:01:20

Shelled
Member
Registered: 2014-04-15
Posts: 29

Properties of Integrals

More integral questions...

So, I think I know what to do overall (apply the properties of integrals to the second expression and then sub in 5 for h(x) ) not too sure how to get there though.
Do I start off with this?

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#2 2014-05-21 01:06:53

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Properties of Integrals

That is not correct. Mostly because h(x-2) is not equal to h(x)-2.

You should take the substitution u=x-2.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#3 2014-05-21 01:23:58

Shelled
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Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

okay. So sub u=x-2, so it's h(u) and take 3 out to the front?
Where would I go from there?

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#4 2014-05-21 01:48:46

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Properties of Integrals

Well, since you changed variables, you also need to change the limits of integration.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#5 2014-05-21 02:21:54

Shelled
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Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

Not really understanding what you mean by that

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#6 2014-05-21 02:26:39

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,436

Re: Properties of Integrals

The old limits are 0 and 4 and 1 and 5. Call them x and solve in u = x - 2.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2014-05-21 02:47:07

Shelled
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Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

As in this?

and then add in the limits?

Last edited by Shelled (2014-05-21 02:47:39)

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#8 2014-05-21 02:51:51

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 86,436

Re: Properties of Integrals

Hi;

Where does it say that h(u) = h(x) = 5?

But first what did you calculate the new limits of integration as?


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#9 2014-05-21 03:18:33

Shelled
Member
Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

Okay, so this then?

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#10 2014-05-21 03:35:45

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,436

Re: Properties of Integrals

That looks right.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#11 2014-05-21 03:59:41

Shelled
Member
Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

What do I do with h?
Do i make it du=hdx

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#12 2014-05-21 04:40:12

Complexity
Member
From: Denmark
Registered: 2013-12-27
Posts: 14

Re: Properties of Integrals

I think it should be du = dx, because u = x-2 => du/dx = 1. However to get any further, I think I personally would need more information about the function h. So I'm just listening from now on smile

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#13 2014-05-21 07:46:41

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,383

Re: Properties of Integrals

hi Shelled,

I don't understand how this question can be done.  Let me explain.  Perhaps someone wiser than me will show why I'm wrong.  Here goes:

You are not told very much about the function h.  All we know is a certain integral.  It's the coloured area on this graph.

HjT6K2W.gif

Now I've made up a shape for h.  We don't know the true shape, but my example will do for explanation purposes.

Then you are asked about the function 3h(x-2).  Now the 3 is no problem.  You get the integral of h(x-2) and multiply it by 3.  So I'll assume we are ok to do that at the end and concentrate on the integral of h(x-2)

So what is this graph like?  h(x-2) when x = 2 will be h(2-2) = h(0).  So when its x coordinate is 2 the y coordinate will be the original graph's y coordinate at x = 0

Similarly at x = 6, h(x-2) = h(6-2) = h(4) so the y coordinate at x=6 will be the same as the y at x=4 on the original graph.  Every point just moves across 2 units to the right.  I've shown that graph too.  Now we can say nothing about that graph outside the interval [2,6]

But you are asked to say what the integral will be from 1 to 5.  How can you ? when you don't know the shape of the graph to the left of x = 2 and you don't know how much area is omitted between 5 and 6.

I can show you either of these:

and

Beyond that I cannot say.  dizzy

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#14 2014-05-21 13:46:43

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,436

Re: Properties of Integrals

Hi;

It is possible to find some answers but I am unable to find a general solution.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#15 2014-05-21 16:13:26

Shelled
Member
Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

Ok, so I asked a tutor about this question today and was told not to bother with it (apparently an older version of the question sheet had been posted up & it was incomplete) neutral

This is the new one:

Is this statement valid?

I've worked a little on it before getting stuck at:

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#16 2014-05-21 18:32:17

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,383

Re: Properties of Integrals

hi Shelled,

That first question probably had a misprint.  The new one can be done.

Make the substitution u = x-2

Differentiate:  du = dx. (Strictly, this is poor terminology but it works smile )

When x = -1, u = -3 and when x = 4, u = 2

So


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#17 2014-05-21 19:46:44

Shelled
Member
Registered: 2014-04-15
Posts: 29

Re: Properties of Integrals

Alright, Thank you smile

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