Math Is Fun Forum

yage

Member

Registered: 2012-01-14

Posts: 2

Struggling with integration

Hi all
I'm a 24 year old college student and am learning calculus over my holiday break. I studied a subject last semester which taught calculus up to partial derivatives, but it was VERY poorly taught (to the point where the university is evaluating the results of all students of the subject).  I absolutely love math and I would like to study calculus and its applications in my spare time (particularly astronomy applications)

I am struggling with a few things, in particular integration by substitution. The professor showed an example of a very basic function but did not explain when this method is used, or why. From what I can gather, it is used when the function also contains its own derivative.
I had enlisted the help of a maths tutor because I hadn't studied math in about 7 years (and never any calculus).
Anyway here is my question.

I need to integrate the function f(x) = 2x/(5+x^2). Now, if we allow u(x) = 5+x^2, then du/dx = 2x. We move 2 out of the integral, but what happens to the x as the numerator? I am to understand it equals 1, as the derivative of u is removed from the integral. The integral is now 1/u which is ln (5+x^2) + C.

Now, I don't understand why exactly the derivative is gone! My maths tutor had shown me a method, but I don't understand exactly how it works. The method is as follows:
f(x)= 2x/(5+x^2) dx
u(x)= 5+x^2, du/dx= x  --> du = x^2/2
=2 ∫ 1/(5+x^2) dx^2/2
=2/2 ∫ 1/(5+x^2) dx^2
= ∫ 1/(5+x^2) dx^2+5
= ∫ 1/u du
= ln|u|
=ln(5+x^2) + C

His reasoning was that we had to match the dx on the RHS with the function, and then balance it out with the coefficient outside the integral sign. I've never heard of this, and my professor or tutors at university never mentioned any such method. Actually, they haven't even explained the meaning of 'dx' on the RHS so I am really lost.
I know this is a long post but I appreciate your time and would really appreciate any help.

jamestisaac

Member

Registered: 2012-01-14

Posts: 1

Re: Struggling with integration

Please help to fiind the next number in the series below

2, 7, 19, 45, ---

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Struggling with integration

Hi jamestisaac;

One possibility for the next number is 99

Please start a new thread when you want to ask a question, when you tack it on to another post as you have done you bury that post.

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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.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,546

Re: Struggling with integration

hi yage

Welcome to the forum.

Integration is a summation process.  The integral sign comes from an elongated 's' for sum.

If you can split a problem into the sum of an infinite number of small steps, each of which has an infinitesimally small 'bit' then essentially you have an integration.

For example, a typical use is to find the area of a shape.  The small bits are rectangles with height 'y' and width a tiny bit of 'x', usually written 'dx'.  It is important to realise that 'dx' is a single algebraic symbol, meaning a little bit of x, not two symbols, 'd' and 'x'.

So

In words you could describe this as 'add up lots of infinitesimally thin rectangles each with height y and width dx'.

Some people only ever use integration to find areas so they start to think this is all it is for.  But you can get volumes, works out centres of gravity ........ lots more too.

It is possible to show that integration is the reverse of differentiation, so if you can do that, there's a chance you can integrate a function.

Now to your problem.

Yes, you're right.  We have the derivative of x squared here, so substitution is a possible way to do this one.  It's not the only way, neither is this the only time when substitution may work, but it certainly is ok here.

What you do is choose a new function, let's say 'u', and replace every x bit with an appropriate 'u' bit.  (If the integral has x limits, you have to convert them to u limits.)

So, if u = 5 + x^2, we've got to convert the 'dx' into a 'du'

You do that by using this:

But we have u in terms of x, so we can only differentiate the other way round

You can invert this to get:

In substitutions where you have the derivative of the function 'u' that's helpful because it allows us to ignore changing the 2x component at all.  Here's why:

So the 2x top and bottom, conveniently cancel out to leave:

and you've already shown you know what to do with that.

Hope that helps. 

Bob

ps.  I've been studying the way your tutor has shown you.  The key step is

This is a sound mathematical step, now I've sorted out the 2 and the brackets but I've never seen substitutions done this way, and I'm not surprised it has left you confused.  It left me confused for a while.  In fact, I was going to write that I thought your tutor had made a mistake, but then I worked out what had been done.  It uses calculus at a more advanced level than you want at the moment and I wouldn't use it myself. I've been teaching substitution my way for 44 years now (gosh, is it really that long!) and it has worked for countless students so I recommend it.  Just make sure you change everything from x into u, and if the integral becomes easier, you're well on the way!

Last edited by Bob (2012-01-14 21:04:05)

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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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

yage

Member

Registered: 2012-01-14

Posts: 2

Re: Struggling with integration

Hi all thanks for your responses.

I have gone back and rehashed over the differentiation module of calculus to get a better handle on integration, and I totally get it now. My professor told us to memorise the rules for integration by substitution/parts etc without explaining WHY or HOW the laws work, which really ticked me off because I want to understand and not memorise.

It's almost like a light bulb switched on in my head and I feel so silly for not understanding such an trivial concept of calculus!
Thank you all again

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Struggling with integration

Hi;

It is nice to get such a good answer to your question!

My professor told us to memorise the rules for integration by substitution/parts etc without explaining WHY or HOW the laws work, which really ticked me off because I want to understand and not memorise.

On the surface your professor sounds lazy but remember he has a limited time to cover a lot of material. Of course understanding is much better than memorization but in a practical sense if you do not have the time then memorizing some rules will get you through. Mathematics demands that you do a lot of research on your own like you have done by bringing the question here. Also, I do not recommend getting ticked off at your professors, it is an extremely unproductive mental state.

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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.