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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

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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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

You should take the substitution u=x-2.

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

The knowledge of some things as a function of age is a delta function.

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

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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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

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

The knowledge of some things as a function of age is a delta function.

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Not really understanding what you mean by that

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

As in this?

and then add in the limits?

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

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Okay, so this then?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

That looks right.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

What do I do with h?

Do i make it du=hdx

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

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,979

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.

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.

Bob

Children are not defined by school ...........The Fonz

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

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

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)

This is the new one:

Is this statement valid?I've worked a little on it before getting stuck at:

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,979

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 )

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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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Alright, Thank you

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