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#1 Re: Help Me ! » Integration by parts question » 2014-08-25 08:14:40

I see, I now understand why wolfram gave me the answer it did. I had wondered whether it might be because wolfram uses complex values as well, but until you explained I didn't know how to get it to only use real values of x - thanks! big_smile

As for the integration by parts, I suppose I will just have to do a little more practise so that I can spot what to do. Thanks again smile

#2 Help Me ! » Integration by parts question » 2014-08-25 07:05:28

Replies: 3

Having moved on to integration by parts, I've found myself stuck on another question:

I have two problems here. My first is that I get the answer:

Whilst the book gets:

That's no problem, I thought, because:

But when I checked whether this was true in WolframAlpha, it told me that it was false, so I was wondering whether my answer was, in fact, equivalent to the one in the book or not?

My second problem is that I came by this solution by adapting a worked example of

Which I found online.

I did it by setting:

But the problem I have is that I don't really understand why that is the thing to do. In all of the other examples in my book, I've had a product of two functions of x (often f(x) = x). I have then selected one of the functions of x to be u and integrated the other function of x. The result of the integration I then set as v. Here I've got something totally different and the book offers no guidance, unfortunately, so I was wondering if anyone could explain how this works as I may well have similar problems in future.

#3 Re: Help Me ! » Proof Check » 2014-08-23 04:22:59

Oh N.B. for anyone who didn't follow:

I believe. Hence step 3.

#5 Re: Help Me ! » Proof Check » 2014-08-23 04:16:16

Hi bob bundy, suppose we reason thus, based on what we proved before:

#6 Re: Help Me ! » Proof Check » 2014-08-21 10:54:39

Thanks! smile

There's actually a second part of this question and, sadly, I'm totally stuck on that too!!

I reckon this could be done quite easily with integration by parts, but in the textbook this is the last question in an exercise on integration by substitution. Integration by parts is actually the next subsection, so I'm wondering if there's another way?

I don't even know where to start with this. I tried using what I just proved, but it doesn't seem to help - at least not in the very first place. I imagine I'll be able to use it after finding the first integral, but I can't seem to use it to do anything useful to the first integral, since

There's nothing that I can think to substitute.

I did do some algebra to get:

But that does nothing about the x it's multiplied by.

Does anybody have any ideas?

#7 Help Me ! » Proof Check » 2014-08-20 12:35:44

Replies: 6

The last question on integration by substitution in my books asks for a proof. I think I've got it, but obviously the proof's not given in the back, so I could be talking absolute nonsense. I was wondering, therefore, if someone could give it a look over and tell me if I'm right in my approach.

Here's the question:

My approach was as follows:

#9 Re: Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-20 02:05:03

Ah yes, that would have worked very nicely. Much neater!

#11 Re: Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-20 01:55:37

Ta! smile I can show you how I did it, for the sake of interest, but it's nice to know there wasn't a simple method I completely missed:

To be continued...

#12 Re: Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-20 01:41:11

What was your method for solving the four simultaneous equations bobbym? I have now done it, but it took me multiple pages!!

#13 Re: Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-20 01:11:33

bob bundy wrote:

hi Au101,

What's wrong with


I never really knew how to do that, but thanks bob bundy, that way's much easier.

Still, I'm glad I now know how to do the partial fractions - thanks bobbym smile

#14 Help Me ! » Difficult partial fractions and/or difficult integral » 2014-08-19 15:05:52

Replies: 13

Hi again, so the next question that has me completely stumped is:

I have completed the proof, but am stuck on the next half of the question. Because of what it asked me to prove, I attempted to solve the second integral and my method was to try using the substitution:

Here's what I have:

And here is where I get completely stuck. WolframAlpha tells me that this does have a partial fraction expansion, but I can't for the life of me work out how to arrive at it, since neither:


Factorise, meaning that I cannot make either of them zero.

So, either I'm missing a trick and there is something I can do to make these zero, or there's a partial fractions technique I've not learned, or I'm barking up entirely the wrong tree with my substitution.

Any ideas?

#16 Help Me ! » Inverse trig functions » 2014-08-13 05:37:39

Replies: 3

Hello, I'm back again after a while, with a question that has me stumped.

I'm doing an exercise on integration by substitution and was asked to intergrate:

Using the substitution:

So, after doing the integral, I ended up with:

I think, judging by wolframalpha, this is probably not absolutely correct. I allows myself to cheat at one point, for example, by saying:

Which obviously is not true for all values of θ, but this is an A-level textbook I'm using which here is trying to teach integration by substitution, so I assume it expects me to do this.

So obviously I then need to give the answer in terms of x, which is where things go a little wrong. This is what I've done:

Which obviously gives me a final answer for the integral of:

But the book gives:

This seems so close to what I have that I think I must be missing something, but I can't work out what. Wolframalpha says that the two aren't equivalent, so I don't quite know why the two answers are different.

#17 Re: Help Me ! » Logic Question » 2013-11-09 15:20:04

It's been a while and you should wait for someone to confirm, but as far as I know yes, everything you've written is fine smile

#18 Re: Help Me ! » counting problem to spell MATH » 2013-09-29 08:39:44

I got twelve by following it round, almost as if it were a clock, but then I noticed that there are other ways (e.g. left, up, left). I'm sure there's a way to be sure (this isn't really my area of expertise) but just a cursory attempt at a brute force solution seems to suggest to me that 16's a pretty good answer.

#19 Re: Help Me ! » quick logic question » 2013-08-11 02:06:20

I'm not sure that's necessarily true. NN could be a woman and, therefore, the mother of "the son of NN". The father of "the son of NN" would then be her husband/spouse/whatever. Then the son of the father of NN would be NN's brother and, therefore, the uncle of "the son of NN" and the brother-in-law of "the father of the son of NN".

Presumably the use of the word "the" excludes the other possibility that "the father of NN" had many sons, and - therefore - one of them was NN's brothers.

#20 Re: Help Me ! » Change your subject. » 2013-08-05 08:24:13

Hmmmm....on second thought, maybe? I'm not sure. My original thinking was that when the first ± is +, so is the second one and when the first ± is -, so is the second one. So, for example:

When the first ± is +, so is the second one and when the first ± is -, so is the second one. So we have:

Hence the need for a ∓ sign. In this case we would have +(+5) or -(-5) and only these options. But it seems reasonable to be able to say:

So, yes, I think you're right. Ignore my second post. The first one still stands, though, I think smile

But yes, anonimnystefy is right. Sorry for confusing the matter further, the important thing to note is √x is always the positive, principal root, hence the need for the ± sign before the radical sign.

#21 Re: Help Me ! » Change your subject. » 2013-08-05 06:40:07

Another way to think about it is if √25 = ± 5, then the ±'s would cancel each other out. You would have:

#22 Re: Help Me ! » Change your subject. » 2013-08-05 06:37:24

I would imagine it dates back to early geometry. If, say, you're trying to calculate the length of a hypotenuse, you're not interested in the negative values. I imagine this general precedence of the principal square root was incorporated into the notation when it was defined. Because, of course, we define our notation to be useful to us and easy to work with. But, without realising it, you've always been using the convention whenever you've gone:

If it weren't for the fact that the √ sign only referred to the principal value, you wouldn't need the ±, that would be implied by the √. Then you could just write:

It's just the way we learn to think about it conceptually smile

#23 Re: Help Me ! » Change your subject. » 2013-08-05 06:19:39

I agree, it was sloppy phrasing on my part as well. Although, bobbym, I think your post #35 was correct. 9 does have two square roots (the principal root being 3, the other being -3) the problem is that the notation √9 gives us the principal square root. This is why we have to write the ± sign before the radical.

Thus ± √9 = ±3 and √9 = 3.

But "the square root of nine is plus, or minus, three" is correct (or, perhaps I should say "the square roots of 9 are..."). It is a problem of formal notation that we have, I believe - if I have understood everything correctly.

#24 Re: Help Me ! » Change your subject. » 2013-08-05 05:53:19

anonimnystefy wrote:


only, but both 3 and -3 satisfy the equation

Excellent technical observation, we should, really, say that:

However, it is true to say that the square root of 9 is plus or minus 3. The problem we have is that the sign √ refers to the principal square root only.

This does make the thing a little harder to understand, though tongue

Suffice it to say that when we square root both sides of an equation, we must include the ± sign, as an equation of the form:

Has two solutions.

#25 Re: Help Me ! » Change your subject. » 2013-08-05 05:03:36

Yep, because a negative number times a negative number is a positive number, so:

Try it:

Therefore, when we 'undo' the squaring, by square rooting, the answer could be positive, or negative. The square root of 4 is either 2, or -2. We can't tell.


Edit: this is why bobbym's original solution had two answers


Which is the same as:

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