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Calculus! For beginners, with a ‘no theorems’ approach!

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blogfast25 - 26-3-2016 at 13:42

Quote: Originally posted by aga  

The "lim delta t->0" part is a mystery.

Basically i do not know how it is applied to the delta x/delta v part, so cannot compute the whole thing.

Sorry if my Density appears to be increasing.


You aren't supposed to be able to compute:

$$\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}$$

... because I haven't taught the limit theorem. But I have taught you the rules of derivation, so that you can compute:

$$v(t)=\frac{dx(t)}{dt}$$

... simply by applying the rules to the function x(t), which in the exercise was specified:

$$x(t)=3t^2+2t+4$$

[Edited on 26-3-2016 by blogfast25]

aga - 27-3-2016 at 01:00

Quote: Originally posted by blogfast25  
You aren't supposed to be able to compute:
$$\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}$$

Oh. That'd explain that then.

$$v(t)=\frac{dx(t)}{dt}$$
So does that mean, in words, that v (as a function of time) is equal to the difference in x (as a function of time) divided by the difference in time ?

$$x(t)=3t^2+2t+4$$
then you're saying that the first derivative of x (as a function of time) =v(t) ?
$$v(t) = 6t+2$$

If that's right, still struggling to see how that relates to this part
$$v(t)=\frac{dx(t)}{dt}$$
I was off in the realms of
$$v(t) = \frac{x_2(t_2)-x_1(t_1)}{t_2-t_1}$$
$$v(t) = \frac{(3t_2 ^2+2t_2+4)-(3t_1 ^2+2t_1+4)}{t_2-t_1}$$
$$v(t) = \frac{3t_2 ^2+2t_2-3t_1 ^2-2t_1}{t_2-t_1}$$
$$v(t) = \frac{3(t_2^2-t_1^2) +2(t_2-t_1)}{t_2-t_1}$$
$$v(t) = \frac{3(t_2^2-t_1^2)}{t_2-t_1} + \frac{2(t_2-t_1)}{t_2-t_1}$$
$$v(t) = \frac{3(t_2^2-t_1^2)}{t_2-t_1} + 2$$
and getting very lost indeed.

Hang on, so if
$$v(t) = 6t+2$$
$$a(t) = 6$$

[Edited on 27-3-2016 by aga]

blogfast25 - 27-3-2016 at 06:10

$$v(t)=6t+2$$

and:

$$a(t)=6$$

... are both correct. From a kinetic PoV, speed is the first derivative of position (in time) and acceleration is the first derivative of speed (in time).

Regarding:

$$v = \frac{x(t_2)-x(t_1)}{t_2-t_1}$$

That would be the average value of v over the interval t<sub>1</sub> to t<sub>2</sub>.

To find the instantaneous speed v(t) we need to take the limit:

$$v(t)= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$$
$$v(t)= \lim_{\Delta t \to 0} \frac{x(t+\Delta t)-x(t)}{\Delta t}$$

Let's work out the numerator first:

$$x(t+\Delta t)-x(t)=3(t+\Delta t)^2+2(t+\Delta t)+4 - (3t^2+2t+4)$$


$$= 3(t^2+2t\Delta t+(\Delta t)^2)+2t+2\Delta t+4-3t^2-2t-4$$
$$=3t^2+6t\Delta t+3(\Delta t)^2+2t+2\Delta t+4-3t^2-2t-4$$
$$=6t\Delta t+3(\Delta t)^2+2\Delta t$$
$$=\Delta t(6t+3\Delta t+ 2)$$

Then divide by Δt, so:

$$v(t)=\lim_{\Delta t \to 0} (6t+3\Delta t+ 2)$$

For:

$$\Delta t \to 0$$

Then:

$$v(t)=6t+2$$

'On paper' we can work out:

$$v(t)=\frac{dx(t)}{dt}$$

... for any x(t) but you can imagine what kind of a kerfuffle that becomes for complicated functions x(t)!

And for that reason, aga, we simply apply The Rules! :)

[Edited on 27-3-2016 by blogfast25]

aga - 27-3-2016 at 09:21

I see !

So with a limit,

f seriously> 0 where v(caught) > limit

where f is the Speeding Fine and v is the Speed ;)

'Limit' is certainly no theory ...

[Edited on 27-3-2016 by aga]

blogfast25 - 27-3-2016 at 09:54

Quote: Originally posted by aga  


'Limit' is certainly no theory ...



Yours isn't, that much is clear! :D

"Dad, are we nearly there yet??"
"Integrals coming up soon, son!"

Take a deep breath...;)

aga - 27-3-2016 at 13:51

Quote: Originally posted by blogfast25  

$$q=\frac{\pi \Delta T}{\frac{1}{2k}\ln \frac{D}{d}+\frac{1}{hD}}$$

Determine for which value of D, in function of k and h, q is minimised.

Hint:

Set:

$$q=\frac{\pi \Delta T}{u}$$

Then see if:

$$\frac{1}{u}$$

... has an optimum.

Do WHAT ?

Erm, so we want q'=0 then ?

OK. so
$$q=\frac{\pi \Delta T}{u}$$
$$q'=\frac{(\pi \Delta T)'(u) - (\pi \Delta T))(u)'}{(u)^2}$$
pi and deltaT are given as constants, so
$$q'=\frac{-\pi \Delta T}{(u)^2}$$
Eh ? (u)<sup>2</sup> can't be any number to make it end up as zero.

Briefly toyed with
$$q'=-\pi \Delta T(u)^{-2}$$
... fainted for a moment then gave up.

Is it a trick question, did i just get it wrong, or sometimes there is no optimum ?

blogfast25 - 27-3-2016 at 14:25

$$q'=\frac{(\pi \Delta T)'(u) - (\pi \Delta T))(u)'}{(u)^2}$$

... is correct but should have become:

$$q'=-\pi \Delta T \frac{u'}{u^2}$$

And then:

$$u'=(\frac{1}{2k}\ln \frac{D}{d}+\frac{1}{hD})'$$

$$u'=\frac{1}{2kD}-\frac{1}{hD^2}=\frac{hD^2-2kD}{2khD^3}=\frac{hD-2k}{2khD^2}$$

$$\frac{dq}{dD}=-\frac{\pi \Delta T}{u^2} \frac{(hD-2k)}{2khD^2}$$

$$\frac{dq}{dD}=-\frac{\pi \Delta T}{2u^2khD^2} (hD-2k)$$

The factor:

$$-\frac{\pi \Delta T}{2u^2khD^2}$$

... never goes through zero, so that if there is optimum:

$$\frac{dq}{dD}=0$$

That can only mean:

$$hD-2k=0$$

So optimum is reached at:

$$D=\frac{2k}{h}$$
Quote: Originally posted by aga  

Is it a trick question, did i just get it wrong, or sometimes there is no optimum ?


Loads of functions don't have optima but this isn't one of them.


[Edited on 27-3-2016 by blogfast25]

yobbo II - 28-3-2016 at 04:08


The v(t) = dx(t)/dt


is quite confusing when you throw it out. A new name is being put on dx/dt (called v) You then have to announce that v is a function of t and then announce that x is also a function of t , hence the two (t)'s

Is that a sensible way of putting it?

Yob

blogfast25 - 28-3-2016 at 05:16

$$v(t)=\frac{dx(t)}{dt}$$

and:

$$v=\frac{dx}{dt}$$

... are synonymous expressions that can be used interchangeably. The choice depends on context.

Some texts will even use:

$$x'(t)$$

... for speed (velocity).


[Edited on 28-3-2016 by blogfast25]

Integration (“anti-derivation”):

blogfast25 - 28-3-2016 at 09:37

Woelen pointed out higher up and correctly that derivation is an operator that acts on a function f(x):

$$\frac{\mathrm d}{\mathrm d x} f(x)=f'(x)$$
There exist an anti-operator to derivation, called integration (rarely called anti-derivation), which achieves the opposite when acting on a function f(x).

If F(x) (note capital F, not f!) is the integral of f(x), then:

$$\frac{\mathrm d}{\mathrm d x} F(x)=f(x)$$
Let's just give it a try, on:

$$f(x)=x^2$$

So we're looking for a function F(x), so that when we take the derivative of it, that derivative gives us x<sup>2</sup>. Let's try:

$$\frac13 x^3$$

Take the derivative:
$$\Big(\frac13 x^3\Big)'= \frac13 3 x^2=x^2$$
Bingo! So:

$$F(x)=\frac13 x^3$$
... is the integral of x<sup>2</sup>.

Notation:

If f(x) is a function and we call y its integral, then we can write as above:

$$\frac{\mathrm d}{\mathrm d x} y=f(x)$$

And:

$$dy=f(x)dx$$

The operator "integration" (or "to integrate" or "take the integral of") is symbolised by:

$$\int$$
Applying it to both sides we get:

$$\int dy=y=\int f(x)dx$$

Applied to our first example, if y is the integral of x<sup>2</sup>, then:

$$y=\int x^2dx= \frac13 x^3$$

Next: The Indefinite Integral.

aga - 28-3-2016 at 10:38

er, "whimper" seems appropriate.

I'm going to hide in the shed and boil ethanol over an open flame, pretending that i know what i'm doing so nothing bad happens ...

... prefereably not dis-integration.

aga - 28-3-2016 at 12:32

That done, i see now with my remaining good eye that Integration is basically the opposite of a Derivative.

Using the accelleration example, knowing the accelleration at time t, the integral of accelleration a(t) would tell you the speed v(t) ?

Are there anti-rules to go with integration, same as rules for derivatives ?

Edit:

Oh ! They're the same rules backwards ?

[Edited on 28-3-2016 by aga]

blogfast25 - 28-3-2016 at 13:15

Quote: Originally posted by aga  
That done, i see now with my remaining good eye that Integration is basically the opposite of a Derivative.

Using the accelleration example, knowing the accelleration at time t, the integral of accelleration a(t) would tell you the speed v(t) ?

Are there anti-rules to go with integration, same as rules for derivatives ?

Edit:

Oh ! They're the same rules backwards ?



Yes, integration will get us from a to v and from v to x. Simply put:

$$v=\int a dt$$

and:

$$x=\int v dt$$

The rules of integration are kind of the 'anti-rules' of derivation and that's very handy as we shall soon see!
<hr>
Indefinite integrals:

There is, when we wrote:
$$y=\int x^2dx= \frac13 x^3$$
... a little snake in the grass. Although it's true that:

$$\Big(\frac13 x^3\Big)'= \frac13 3 x^2=x^2$$

... there's another function that satisfies that condition also, because:

$$\Big(\frac13 x^3+C\Big)'= \frac13 3 x^2+0=x^2$$

... is also true, if C is a constant. We call it the integration constant.

So we really have to write:

$$y=\int x^2dx= \frac13 x^3+C$$

This is not limited to x<sup>2</sup>, in fact it's universal.

For the general case we can write:

$$y=\int f(x)dx=F(x)+C$$

... provided that:

$$\frac{\mathrm d}{\mathrm d x} F(x)=f(x)$$
Then:
$$F(x) + C$$

... is called the indefinite integral because at this point we do not know the value of C yet. Later we'll find out its value or how to eliminate C.

Next up, simple integrals and the rules of integration. For now, just a few tasters (a is a constant):

$$\int dx=x+C$$
$$\int adx=a\int dx=ax+C$$
$$\int x^n dx= \frac{1}{n+1}x^{n+1}+C$$
$$\int ax^n dx= a\int x^n dx=\frac{a}{n+1}x^{n+1}+C$$

[Edited on 28-3-2016 by blogfast25]

aga - 28-3-2016 at 13:36

I, at least, am exremely grateful for the Effort you, blogfast25, have put into these Scientifically Educational threads, and the Time you have taken to do so, despite seemingly little interest from the majority of the ScienceMadness community.

Why that is the case amongst supposed Scientists is a mystery, as maths is pretty much the Key to understanding the subject.

I noticed that with my own children that their attention span is quite a lot shorter than mine was at their age, so perhaps that's the reason for the younger members' distinterest.

The older members may have studied this already and are scared of getting it wrong.

Either way, when the Fear of getting it Wrong stops you doing something, that's the point when you Stop Learning.

It is quivalent to the point where you think you know it all.

aga - 28-3-2016 at 13:45

Quote: Originally posted by blogfast25  
We call it the integration constant.

So we really have to write:

$$y=\int x^2dx= \frac13 x^3+C$$

That rings a dim and distant bell.

It sounds like 'bqoik' instead of 'Ding'.

Memory may be very defective on this, given the decades.

Edit:

Yes ! Constants, whatever their value when 'derived' they equal 0, so in reverse they're there, but unknown.

OK.

[Edited on 28-3-2016 by aga]

[Edited on 28-3-2016 by aga]

blogfast25 - 28-3-2016 at 13:47

@aga:

Thank me when you've solved your first Real World differential equation. ;)

As regards the interest in this thread, its page views/posts ratio is fairly normal, so I expect a lot of 'lurkers' and that's fine with me. Of course it would have been nice to get a few more active participants but it is what it is. Among aspiring scientists the interest in theory is often surprisingly low. 'They'll learn!' is my attitude to that. ;)

aga - 28-3-2016 at 13:50

Does it Blend ? is my answer to that.

Edit:

no need for the @aga thing,.just us here.

I expect yobbo(x) was just visiting with some googled garbage and is highly unlikely to actually attempt the questions you have posed.

Maybe a future Probability Calculation lesson would be good, so we could calculate stuff like that.

[Edited on 28-3-2016 by aga]

blogfast25 - 28-3-2016 at 14:18

Quote: Originally posted by aga  


Maybe a future Probability Calculation lesson would be good, so we could calculate stuff like that.



Ahhh... probability! One of the hardest, most counter-intuitive and most misunderstood concepts in mathematics. :D

yobbo II - 28-3-2016 at 15:40



$$ \int dx=x+C $$

is that the same thing as

$$ \int 1 dx=x+C $$



@ aga Name is x, yobbo x. Licenced to differentiate!

[Edited on 28-3-2016 by yobbo II]

blogfast25 - 28-3-2016 at 16:24

Quote: Originally posted by yobbo II  


$$ \int dx=x+C $$

is that the same thing as

$$ \int 1 dx=x+C $$



Yes, it is.

So what is:

$$\int 0 dx=?$$

yobbo II - 28-3-2016 at 17:54


C


Do you see that aga, it's C

blogfast25 - 28-3-2016 at 18:08

Quote: Originally posted by yobbo II  

C



Correct. Of course that doesn't mean the integral of nothing is not nothing because C could be nothing . But it's possible that the integral of the zero function is some number.

[Edited on 29-3-2016 by blogfast25]

aga - 28-3-2016 at 22:28

Quote: Originally posted by yobbo II  
Do you see that aga, it's C

Thanks for joining in yobbo-eye-eye !

Not sure i do entirely C clearly.

Surely
$$\int n dx=x^n+C$$
The special case of n=0 appears to have a (possibly) important intermediate :-
$$\int 0 dx=x^0+C = 1+C$$
rolling that 1 into the Constant gets the lone C.

Whilst doing that is entirely valid, it warps the integration constant by 1.

[Edited on 29-3-2016 by aga]

woelen - 28-3-2016 at 23:09

Aga, here you make a mistake. What is the derivative of xn? It is not n, but nxn-1. The integeral of n equals nx + C and not xn + C.

---------------------------------------------------------------------------------------------

The problem of integration of 0 leading to a non-zero value is a very subtle one. Only formally, it can be equal to a non-zero constant.

I'm quite sure blogfast25 will introduce the concept of definite integrals, which compute the surface of a function between its graph and the x-axis. In such cases, the constant disappears and definite values are generated. In such cases, the integral of 0 always is zero, except for the special case when integration occurs from a constant to infinity. Specially defined 0-functions can yield non-zero values in such cases (in fact, the inverse of a dirac-function). A dirac function is a function which is zero everywhere, except for one single value, where it is infinite with the property that the surface area of the single spike, which has zero thickness and infinite height, equals 1.

For now, just forget about the 0-integral being non-zero if you are a starter in this subject ;)

[Edited on 29-3-16 by woelen]

blogfast25 - 29-3-2016 at 05:35

@aga:

You need to distinguish sharply between n as an exponent and n as a coefficient.
<hr>
Basic Rules of integration (indefinite integrals):

Remember that the indefinite integrals of most simple functions can be found here (scroll down):

http://www.ambrsoft.com/Equations/Derivation/Derivation.htm

1. Constant rule (with a a constant):
$$\int af(x)dx=a\int f(x)dx$$
2. Power rules:
$$\int x^n dx=\frac{x^{n+1}}{n+1}+C$$
$$\int e^xdx=e^x+C$$
3. Sum/difference rule:
$$\int[f(x) \pm g(x)]dx=\int f(x)dx \pm \int g(x)dx$$
4. Chain Rule:

While the above three rules are somewhat self-evident, the chain rule of integration requires a bit of elaboration.

When taking derivatives we saw that the following chain rule applies:

If y is a function of u, with u a function of x, then:

$$y'(x)=\frac{dy}{dx}=y'(u) \times u'$$
Also:
$$dy=y'(u)u'dx$$
Since as:

$$u'=\frac{du}{dx}$$
$$u'dx=du$$

So:

$$dy=y'(u)du$$

Integrate both sides:
$$\int dy=y=\int y'(u)du$$
(Note that the chain rule for differentiation is sometimes also written in the following form: )

$$\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$$

If all this sounds like abstract gobbledygook, a few simple examples should soon clear things up. Applying the chain rule to integrals is also known as ‘integration by substitution’.

1st example:

$$y=\int \sqrt{x-5}dx$$

Set:

$$u=x-5$$

Then:

$$du=(x-5)'dx=(1-0)dx=dx$$

Now substitute:

$$y=\int \sqrt u du=\frac{1}{1+1/2}u^{1+1/2}=\frac23u^\frac32$$

Substitute back:

$$y=\frac23 (x-5)^{\frac32}+C$$

2nd example:
$$y=\int \frac{x}{x^2+6}dx$$
Set:

$$u=x^2+6$$

$$du=(x^2+6)'dx=(2x+0)dx=2xdx$$
So:

$$xdx=\frac12 du$$
Now substitute:

$$y=\int \Big(\frac12\Big) \frac{du}{u}=\frac12 \int \frac{du}{u}$$
$$y=\frac12 \ln u$$

Substitute back:

$$y=\frac12 \ln(x^2+6)+C$$
3rd example:

$$y=\int \sin(7x)dx$$
Set:

$$u=7x$$

So:

$$du=(7x)'dx$$

$$du=7dx$$

So:

$$dx=\frac17 du$$

Now substitute:

$$y=\int \frac{\sin u du}{7}=\frac17 \int \sin u du=-\frac17 \cos u$$
Substitute back:
$$y=-\frac17 \cos (7x) + C$$

After ‘questions and answers’ on this section, a limited number of exercises will be put up.

[Edited on 29-3-2016 by blogfast25]

aga - 29-3-2016 at 05:40

Quote: Originally posted by woelen  
Aga, here you make a mistake. What is the derivative of xn? It is not n, but nxn-1. The integeral of n equals nx + C and not xn + C

Doh ! Obvious error now you point it out.
Quote: Originally posted by woelen  
For now, just forget about the 0-integral being non-zero if you are a starter in this subject ;)

Wise words which i shall follow.

Cheers woelen

blogfast25 - 29-3-2016 at 06:17

@aga:

Ok. Have a close read of my last post and see if all is clear.

aga - 29-3-2016 at 12:36

Sorry, did not see that.

Must have cross posted O Praetor.

OK. Read it. Looks like some seriously heavy stuff.

(usually means that temporary interloper(s) will duck out seeing as there's heavy lifting to be done ...)

Er, the +/- thing is presumably because x^2 could be with x=-2, or +2, so the derived 4 could be from +2 or -2.

I like the 'set the hard part = one letter' bit, then substitute back later.

the 'dx' thing is confusing, mostly because i have not properly studied your earlier explanation of the various notations.

3rd example looses me when you set u = 7x and suddenly the sin() function evaporates.

Apart from that, all ready to spectacularly fail the next question !

blogfast25 - 29-3-2016 at 13:02

Quote: Originally posted by aga  

Er, the +/- thing is presumably because x^2 could be with x=-2, or +2, so the derived 4 could be from +2 or -2.

the 'dx' thing is confusing, mostly because i have not properly studied your earlier explanation of the various notations.

3rd example looses me when you set u = 7x and suddenly the sin() function evaporates.

+/-:

$$\int[f(x) + g(x)]dx=\int f(x)dx + \int g(x)dx$$
or:
$$\int[f(x) - g(x)]dx=\int f(x)dx - \int g(x)dx$$

Does that clarify it?

Not really sure what you mean by the 'dx' thingy. Say you make a substitution, e.g.:
$$u=x^2-9$$
Then:

$$u'=\frac{du}{dx}=2x$$
So:

$$du=2xdx$$
And:
$$xdx=\frac12 du$$

Not sure about the sin thing:

$$\frac17 \int \sin u du=-\frac17 \cos u$$
... because the integration turns the sine into a minus cosine (as per the rules: see the table I provided).



[Edited on 29-3-2016 by blogfast25]

aga - 29-3-2016 at 13:24

The +/- thing is clear now. Thankyou.

The 'meaning of d' i will probably understand better having used it a bit more.
Likely that will be the source of several errors, and soon.

sin() evaporation was just reading without brain engaged - obvious now, re-reading your text (i read it as if it were the full equation, and it wasn't).

blogfast25 - 29-3-2016 at 13:44

Ok, I'll just post these now, just to keep you awake at night! ;)

Exercises:

1.
$$y=\int (3x^5-2\sqrt{x})dx$$
2.
$$y=\int 5\cos dx$$
3. (b is a constant)
$$y=\int (bx^2-4\sin x)dx$$
4. (alpha is a constant)
$$y=\int(2x-\alpha)^3dx$$
Hint: chain rule: set:
$$u=2x-\alpha$$
5.
$$y=\int x \cos(x^2-3)dx$$
Hint: chain rule: set:
$$u=x^2-3$$
6.
$$y=\int e^{2x-7}dx$$
Hint: chain rule.

7.
$$y=\int x^6\ln |x^7|dx$$
Hints:

Set:
$$u=x^7$$
and:
$$\int \ln|u|=u \ln u -u$$
8.
$$y=\int \cos(3x)[\sin(3x)]^2dx$$
Hint: set:
$$u=\sin(3x)$$
9.
$$y=\int x\sqrt{x+1}dx$$
Hint: set:
$$x=u-1$$


[Edited on 29-3-2016 by blogfast25]

[Edited on 29-3-2016 by blogfast25]

aga - 29-3-2016 at 13:48

Oh bugger.

Erm, pretty please do the easiest one as an example ?

(clever trick: just 8 to do then)

Edit:

Hey ! WTF does this mean ?
$$\ln|u|$$
Parallel lines (pipe character) are new, to me, at least in maths.

Please be explainings the meanings.

[Edited on 29-3-2016 by aga]

blogfast25 - 29-3-2016 at 13:57

Sorry:

$$\ln|u|$$

$$|u|$$

... means the modulus of u. A modulus always returns a positive number, e.g.:

$$|3|=3$$
and:
$$|-3|=3$$

The modulus is used here because the ln of a negative number is not allowed in mathematics.

[Edited on 29-3-2016 by blogfast25]

aga - 29-3-2016 at 14:00

OK. Cheers.

Kinda got it.

|n| = abs(n) as in the absolute value with no sign of a sign ...

blogfast25 - 29-3-2016 at 14:02

Quote: Originally posted by aga  
OK. Cheers.

Kinda got it.

|n| = abs(n) as in the absolute value with no sign of a sign ...


Yep. 'Absolute value' is another term for 'modulus'.

aga - 30-3-2016 at 11:48

OK. Hands up.

My legs have fallen off at the first hurdle.

With 'Example 1' here http://www.sciencemadness.org/talk/viewthread.php?tid=65532&...

i don't follow how you go from
$$y=\int \sqrt{x-5}dx$$
to
$$du=(x-5)'dx=(1-0)dx=dx$$
by setting u = x - 5 .

Specifically the du = (u)'dx part.

Edit:

I mean, obviously you're processing the du thing in the second step, just that i can't see how that works later with the
$$\sqrt udu$$
part


[Edited on 30-3-2016 by aga]

blogfast25 - 30-3-2016 at 13:06

Quote: Originally posted by aga  
OK. Hands up.

My legs have fallen off at the first hurdle.



$$y=\int \sqrt{x-5}dx$$
If instead the problem was:
$$y=\int \sqrt{x}dx$$
... then we could integrate it directly. But we can't because of that x - 5 bit. So we'll try and 'get rid of it' by a substitution.

Call:
$$u=x-5$$
Then:
$$u'=(x-5)'=1$$

and with:

$$u'=\frac{du}{dx} \implies du=dx$$
Now, in:
$$y=\int \sqrt{x-5}dx$$
Insert:
$$x-5=u$$
$$dx=du$$
So we get:
$$y=\int \sqrt{u}du$$
This can be integrated by power rule:
$$y=\frac23 u^{\frac32}$$
So:
$$y=\frac23 (x-5)^{\frac32}+C$$

I don't really see how much clearer to make it. Maybe more examples?

aga - 30-3-2016 at 13:21

Appologies for not getting it immediately.

Definitely i need to backtrack and go over this some more, particularly the 'd' notation.

It's not your explanations at fault, just my understanding thereof, which will take some time to resolve (probably more questions too).

Basically i 'must try harder', and will do.

Edit:

so dx = (x)' ?

[Edited on 30-3-2016 by aga]

blogfast25 - 30-3-2016 at 13:37

Quote: Originally posted by aga  
Appologies for not getting it immediately.

Definitely i need to backtrack and go over this some more, particularly the 'd' notation.

It's not your explanations at fault, just my understanding thereof, which will take some time to resolve (probably more questions too).

Basically i 'must try harder', and will do.


This isn't easy stuff. The whole thread is probably worth several months of A/GCSE level math.

So spend a little time on it (and off it) and get back when things have cleared up a bit? :)

[Edited on 30-3-2016 by blogfast25]

aga - 30-3-2016 at 13:44

GCSE ?

I recall integration, trigonometry and quadratic equations from my 'O' level course aged 16.

Sadly didn't progress further than that - money was a more immediate need than qualifications at the time.

Sadder still is that i can't remember all that i learned then !

blogfast25 - 30-3-2016 at 16:53

I don’t know if this will help a lot with aga’s “d woes” but here’s an attempt to enlighten.

Look at the following figure: a smooth and continuous function y=f(x) is shown with the tangent line (red) in the point (x,y):

Differential aga.png - 7kB

The inset shows the situation in (x,y) hugely magnified. Because of the limit taking:

$$\Delta x \to 0$$

... dx and dy are infinitesimals, very small and about as close to zero as can be imagined. But they are real numbers and we showed above that:
$$\frac{dy}{dx}=y'$$
So:
$$dy=y’dx$$
But also for example:
$$dx=\frac{dy}{y'}$$

So, dy and dx, although infinitesimally small, are ‘mathematical objects’ that can be manipulated like any other. And that is essentially what we do when we use the substitution method.

Another example of substitution:

$$y=\int \frac{x^2}{(x^3+3)^4}dx$$
Call:
$$u=x^3+3$$
So:
$$u'=\frac{du}{dx}=3x^2$$
So:
$$du=3x^2dx$$
Also:
$$x^2dx=\frac13 du$$
Now substitute these expressions into y:
$$y=\int \Big(\frac13\Big) \frac{du}{u^4}$$
Apply the Constant Rule and rework a bit:
$$y=\frac13 \int u^{-4}du=-\frac19 u^{-4+1}=-\frac19 u^{-3}$$
So:
$$y=-\frac19 (x^3+3)^{-3}+C$$

[Edited on 31-3-2016 by blogfast25]

aga - 1-4-2016 at 12:32

Amazingly that little inset of the dx / dy actually is mega helpful !

It is the difference in x and the differnce in y when those differences are almost not there.

A (brief) explanation of that there Limit Theory could be helpful here.

Specifically the notation of x -> 0 i vaguely remember meaning something like 'tending towards zero'.

blogfast25 - 1-4-2016 at 14:14

Quote: Originally posted by aga  
Amazingly that little inset of the dx / dy actually is mega helpful !

It is the difference in x and the differnce in y when those differences are almost not there.

A (brief) explanation of that there Limit Theory could be helpful here.

Specifically the notation of x -> 0 i vaguely remember meaning something like 'tending towards zero'.


Ok, upwards and onwards.

So I'm trying to show what:

$$y'(x)=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}= \lim_{\Delta x \to 0}\frac{y(x+\Delta x)-y(x)}{\Delta x}=\frac{dy}{dx}$$
... really means. Look at the picture:

Derivative limit.png - 10kB

The ACTUAL gradient ( = 1st derivative) of f(x) is the gradient of the red tangent line in (x,y). y = f(x) is the blue line.

Now say we draw a chord between (x,y) and point 1. We would then calculate Δy and Δx and the ratio Δy/Δx would give us a crude approximation of the gradient.

Now if we draw another chord between (x,y) and point 2 then notice that Δy and Δx would both decrease and the ratio Δy/Δx would be slightly better approximation of the actual gradient.

Even smaller become Δy and Δx for a chord between (x,y) and point 3 and the ratio Δy/Δx would be yet closer to the true gradient.

It follows that if we keep reducing the size of the interval Δx, the estimate of the real gradient becomes more and more accurate and that is the meaning of:

$$\Delta x \to 0$$

Remember that for a simple function like y = x2:
$$y'(x)=\lim_{\Delta x \to 0}\frac{y(x+\Delta x)-y(x)}{\Delta x}$$
$$y'(x)=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}$$
$$y'(x)=\lim_{\Delta x \to 0}\frac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x}$$

$$y'(x)=\lim_{\Delta x \to 0}\frac{2x\Delta x+(\Delta x)^2}{\Delta x}$$
$$y'(x)=\lim_{\Delta x \to 0}\frac{\Delta x(2x +\Delta x)}{\Delta x}$$
$$y'(x)=\lim_{\Delta x \to 0} (2x +\Delta x)$$
$$y'(x)=2x$$



Finally, the required incantation: 'gremlins, gremlins, go away, don't come back another day' :D



[Edited on 2-4-2016 by blogfast25]

Integration by parts:

blogfast25 - 4-4-2016 at 05:31

In addition to the rules above there’s another popular integration technique that bears mentioning, called integration by parts.

A slightly different way of writing the derivation product rule, with u and v both functions of x is:

$$d(uv)=udv+vdu$$
Integrating both sides we get:

$$\int d(uv)=uv=\int udv+\int vdu$$
Or:
$$\int udv=uv-\int vdu$$

This format is useful to integrate certain types of integrals. A classic example is:
$$\int xe^xdx$$
Set:
$$dv=e^xdx$$
Integrate to get v (ignore the integration constant for now):
$$\int dv=v=\int e^xdx=e^x$$
Then set:
$$u=x$$
So:
$$du=dx$$
Now plug it all in:
$$\int xe^xdx=xe^x-\int e^xdx=xe^x-e^x+C=(x-1)e^x+C$$
Another example:
$$\int x \cos(ax)dx$$
Set:
$$dv=\cos(ax)dx$$
Integrate:
$$v=\frac1a \sin(ax)$$
Then:
$$u=x$$
$$du=dx$$
$$\int x\cos(ax)dx=\frac{x}{a} \sin(ax)-\int \frac1a \sin(ax)dx=\frac{x}{a} \sin(ax)-\frac1a \int \frac1a \sin(ax)dx$$
$$\int x\cos(ax)dx=\frac{x}{a} \sin(ax)+\frac{1}{a^2}\cos(ax)+C$$
And with C just a constant, we can rework slightly to:
$$\int x\cos(ax)dx= x\sin(ax)+\frac{1}{a}\cos(ax)+C_1$$

aga - 4-4-2016 at 12:24

So, if it's just 'applying the rules', with immense trepidation, the Exercises ...

1. (cribbed from teach)
$$y=\int (3x^5-2\sqrt{x})dx$$
Sum/difference rule
$$y=\int 3x^5dx - \int 2 \sqrt{x}dx$$
Constants rule
$$y=3 \int x^5dx - 2 \int \sqrt{x}dx$$
Power rule, left hand side
$$y=3 \frac {x^6}{6}+C - 2 \int x^{\frac 12}dx+C$$
Roll the constants into one, Power rule Right hand side
$$y=\frac {3x^6}{6} - 2 \big ( \frac {x^{\frac 32}}{\frac 32} \big)+C$$
Simplify fraction left hand side, flip fractional divisor right hand side
$$y=\frac {x^6}{2} - \frac 43 x^{\frac 32}+C$$


2.
$$y=\int 5\cos dx$$
$$y=5 \int \cos dx$$
$$y=5 \sin x + C$$

3.
$$y=\int (bx^2-4\sin x)dx$$
Sum/difference rule
$$y=\int bx^2dx - \int 4 \sin(x)dx$$
Constants jobbie
$$y=b \int x^2dx - 4 \int \sin(x)dx$$
Power rule, roll constants together
$$y=b \frac{x^3}{3} + 4 \cos(x) + C$$
Simplify
$$y=\frac{bx^3}{3} + 4 \cos(x) + C$$

4.
$$y=\int(2x-\alpha)^3dx$$
set $$u = 2x-\alpha$$
$$y=\int u^3dx$$
Power rule
$$y= \frac {u^4}{4}+C$$
Return to the 'real meaning of u'
$$y= \frac {(2x -\alpha)^4}{4}+C$$

5.
$$y=\int x \cos(x^2-3)dx$$
set $$u = 2x +3$$
$$y=\int x \cos(u)dx$$
$$y=x\int \cos(u)dx$$
$$y=-x \sin(u)$$
$$y=-x \sin(x^2-3)$$

Here, and in exercise 6, the chain rule is simply not understood yet. so i got to go back and see how that all works again.

blogfast25 - 4-4-2016 at 13:46

@aga:

1, 2 and 3 are correct. The problems start with the substitutions.

Let's see if the following can make you see the problem.

Another, more intuitive way of understanding integration by substitution:

Let’s take a simple example:
$$y=\int \cos x dx$$
The part:
$$\cos x$$
... is called the integrand ("cosine of x"), the part:
$$dx$$
... is called the differential ("the differential of x"). Note that in both cases we have "... of x". In that case we can simply integrate directly:
$$y=\sin x +C$$
Now what happens if we change the integrand to:
$$\cos (5x)$$
So:
$$y=\int \cos (5x)dx$$
Now the integrand is "cosine of 5x" and the differential "the differential of x". Note: "... of 5x" and "... of x". They don't 'match': this cannot be integrated directly. To try and resolve this problem, we carry out a substitution:
$$u=5x$$
So:
$$du=(5x)'dx=5dx$$
So:
$$dx=\frac15 du$$
Substitute all this into the integral:
$$y=\int \cos(u) \frac15 du=\int \frac15 \cos (u) du=\frac15 \int \cos (u) du$$
Now the integrand is "cosine of u" and the differential "differential of u", so now we can integrate directly:
$$y=\frac15 \sin (u)$$
Substitute back and add integration constant:
$$y=\frac15 \sin (5x)+C$$
Second example:
$$y=\int (4x+5)^{\frac13}dx$$
Again, the integrand ("cubic root of 3x+5") does not 'match' the differential ("differential of x").

Substitute:
$$u=4x+5$$
$$du=(4x+5)'dx=(4+0)dx=4dx$$
$$dx=\frac14 du$$
So:
$$y=\int u^{\frac13} \frac14 du=\frac14 \int u^{\frac13}du$$
Now the integrand and differential 'match' and the integral can be integrated directly:
$$y=\frac14 \times \frac34 u^{\frac43}$$
$$y=\frac{3}{16}(4x+5)^{\frac43}+C$$


[Edited on 4-4-2016 by blogfast25]

aga - 4-4-2016 at 14:33

erm, so the bits before the dx have to be 'simple' as in no operators like +,-/sin,tan etc or you got to substitute those bits before integrating ?

To be honest, it's all rather more difficult with Integration than it was with derivatives.

The concepts are still not 100% clear, although the way to perform the computations are.

The dx/dy are still hanging about causing mischief gremlin-wise.

Perhaps i'm at the limit of my brain capacity.

blogfast25 - 4-4-2016 at 15:08

@aga:

Look at it like this.

We saw that:

$$\int \cos(5x)=\frac15 \sin(5x)+C$$
Integration is the anti-operator of derivation, so the derivative of:
$$\frac15 \sin(5x)+C$$
... should give us the starting function cos(5x). So:
$$(\frac15 \sin(5x)+C)'=\frac15 (\sin(5x))'+0$$
$$=\frac15 \cos(5x)(5x)'=\frac15 \times 5 \cos(5x)$$
$$=\cos(5x)$$
... which proves the integration was correct.

But had we not put the factor:
$$\frac15$$
... upfront, the result would not have been correct. W/o the substitution we would have missed that factor.

Same for:
$$y=\frac{3}{16}(4x+5)^{\frac43}+C$$
Derive:
$$y'=\frac{3}{16}\big[(4x+5)^{\frac43}\big]+0$$
$$=\frac{3}{16} \times \frac43 (4x+5)^{\frac13}(4x+5)'$$
$$=\frac14 (4x+5)^{\frac13} (4+0)$$
$$y'=(4x+5)^{\frac13}$$


[Edited on 4-4-2016 by blogfast25]

aga - 5-4-2016 at 09:30

The bit that i'm still failing to understand is the dx/dy thing, so a step backwards is required.

e.g.
$$y=\int (x-1)dx$$
... is still gobeldygook. Does it mean that
$$\frac {dy}{dx} = (x-1)$$
therefore
$$dy = (x-1)dx$$
and if
$$\int dy = y$$
meaning that if
$$\int dy = \int (x-1)dx$$
and if int(dy) actually does equal y ...
$$y=\int (x-1)dx$$
?

Applogies again : this must be like pulling teeth from a titanium robo-donkey.

blogfast25 - 5-4-2016 at 12:00

@aga:

All that is correct.

And yes:

$$\int dy=y, \int dx=x, \int du=u, \int dv =v,\:\text{etc, etc}$$

Because integration is the opposite of differentiation, it 'cancels' the differentiation.

Another segment to follow tonite...

[Edited on 5-4-2016 by blogfast25]

aga - 5-4-2016 at 12:45

Phew ! Great.

One last bit and i think i'll get it :

$$\int dy = \int (x-1)dx$$
$$y = \int (x-1)dx$$
so far it seems that the entire 'dx' part just disappears : would that happen anyway if it was worked out fully instead of applying the handy rules ?

This bit is being stubborn too
$$u = x^2-9$$
this bit works fine !
$$u' = 2x$$
$$u'=\frac{du}{dx}=2x$$
why is dx = 1 ?
Quote:
Another segment to follow tonite...

Eeek ! Not got this one properly sorted yet !

edit:

in the case of
$$y = \int (x-1)dx$$
i would have to use
$$u = (x-1)$$
then
$$y = \int udu$$
then substitute back yes ?

[Edited on 5-4-2016 by aga]

blogfast25 - 5-4-2016 at 13:12

Quote: Originally posted by aga  
Phew ! Great.

One last bit and i think i'll get it :

$$\int dy = \int (x-1)dx$$
$$y = \int (x-1)dx$$
so far it seems that the entire 'dx' part just disappears : would that happen anyway if it was worked out fully instead of applying the handy rules ?

This bit is being stubborn too
$$u = x^2-9$$
this bit works fine !
$$u' = 2x$$
$$u'=\frac{du}{dx}=2x$$
why is dx = 1 ?

Quote:
Another segment to follow tonite...

Eeek ! Not got this one properly sorted yet !

edit:

in the case of
$$y = \int (x-1)dx$$
i would have to use
$$u = (x-1)$$
then
$$y = \int udu$$
then substitute back yes ?

[Edited on 5-4-2016 by aga]


Yes, integration makes the differential dx disappear, because of the same reason: integration is the anti-operator of differentiation. It does that when you apply the handy rules.

See:

$$y=\int (x-1)dx$$

Here we can apply the sum rule:

$$y=\int xdx -\int 1dx=\frac12 x^2-x+C$$

But it can also be solved by substitution:

$$u=x-1$$
$$u'=x$$
$$du=dx$$
$$y=\int udu=\frac12 u^2$$
$$=\frac12 (x-1)^2+C$$
$$=\frac12 (x^2-2x+1)+C$$
$$y=\frac12 x^2-x+\frac12 +C$$
Which is of course the same solution (C is a constant: adding 0.5 to is doesn't change that)
<hr>
But:
$$u'=\frac{du}{dx}=2x$$

... in no way does it follow that 'dx is 1': it isn't.

Will hold my horses for now... :)

[Edited on 5-4-2016 by blogfast25]

aga - 5-4-2016 at 13:34

Quote:
... in no way does it follow that 'dx is 1'.

Will hold my horses for now... :)

$$u = x^2-9$$
$$u' = 2x$$
$$u'=\frac{du}{dx}=2x$$
$$du \not = u'$$

I see my 'tard-like error there. dx could only be 1 if u' were actually 2x, which could not happen unless x were zero, in which case the mind boggles with 0-> 1 differences.

Thanks for resisting the 'kick-in-the-knackers-then-yank' titanium tooth extraction method : it's cyber carrots tomorrow.

[Edited on 5-4-2016 by aga]

yobbo II - 5-4-2016 at 13:40



Just to be sure what problem 1) above is saying is:

$$ F(x) = 3x^5-2\sqrt{x} $$

Find y

Correct?

And the integral of dx = x ?
It's hard to grasp. How is the intergal of an infinitesimal (a constant) = x



[Edited on 5-4-2016 by yobbo II]

aga - 5-4-2016 at 13:41

$$y=\int xdx -\int 1dx=\frac12 x^2-x+C$$
Oh ! I See now !

I treat the 'dx' term as a whole unit, and no matter where it ends up, in however many places in intermediate fiddling about, it buggers off after applying the rules !

Edit:

That's Awesome. Must have been some seriously dedicated people working out those rules and then checking them !

[Edited on 5-4-2016 by aga]

blogfast25 - 5-4-2016 at 13:42

We'll let it sink in for now and continue bright and bushy tailed on the morrow. :cool:

blogfast25 - 5-4-2016 at 13:48

Quote: Originally posted by yobbo II  


Just to be sure what problem 1) above is saying is:

$$ F(x) = 3x^5-2\sqrt{x} $$

Find y

Correct?

And the integral of dx = x ?
It's hard to grasp. How is the intergal of an infinitesimal (a constant) = x



[Edited on 5-4-2016 by yobbo II]


Ex. 1. was:

Find y, if:

$$y=\int(3x^5-2\sqrt{x})dx$$

Here we reserve the notation F(x) for:

$$y=\int(3x^5-2\sqrt{x})dx=F(x)+C$$

Find F(x) and you've found y.

blogfast25 - 5-4-2016 at 13:55

Quote: Originally posted by aga  
$$y=\int xdx -\int 1dx=\frac12 x^2-x+C$$
Oh ! I See now !

I treat the 'dx' term as a whole unit, and no matter where it ends up in intermediate fiddling about, it buggers off after applying the rules !


Yes.

For example, a general expression:

$$y=\int[f(x)+g(x)+u(x)+z(x)]dx$$

With the sum rule becomes:

$$y=\int f(x)dx+\int g(x)dx+\int u(x)dx+\int z(x)dx$$

Then (perhaps after more fiddling the individual integrals), the dx will disappear on actual integration. That 'disappearing act' of the differential will also become clearer in future, I promise.

Darkstar - 5-4-2016 at 21:11

Quote: Originally posted by aga  
GCSE ?

I recall integration, trigonometry and quadratic equations from my 'O' level course aged 16.

Sadly didn't progress further than that - money was a more immediate need than qualifications at the time.


Which is why we at B&D University strongly believe in educating first and billing later. Just don't mistake our unusual generosity for weakness and try to take advantage of us. Our financial department has been known to occasionally contract out ex-mob enforcers to help "persuade" the more reluctant former students to pay up when the conventional third-party collection agencies fail. So whether you pay your tuition upfront or decide to wait until after you've completed your degree, just know that, one way or another, you're going to pay.

Speaking of which, our records indicate that you currently owe B&D University for introductory courses in quantum mechanics, organic chemistry and calculus. So consider this a friendly reminder from one of your former professors. I would hate for my favorite pupil to end up permanently confined to a wheelchair after a visit from Tony "The Widow Maker" Gambino.

aga - 6-4-2016 at 04:38

'Gambino' was Tony ? We mistook him for a prawn and ate him.

The dorm Rules state that all moveable objects must be clearly and accurately labelled to avoid confusion such as this.

Exercise 6:

$$y=\int e^{2x-7}dx$$
isn't e the 'teflon number' that is the same when derived, so probably the same when integrated
$$y=e^{2x-7}+C$$


[Edited on 6-4-2016 by aga]

woelen - 6-4-2016 at 04:51

Checking your answer is easy. Take the derivative of your answer and you must get the same function as the one given as exercise. If we do that, then you see the following:

First we use the sum-rule:
d(e2x-7 + C)/dx = d(e2x-7)/dx + dC/dx = d(e2x-7)/dx + 0.

Now we use the chain rule:

let u = 2x-7

d(eu)/dx = d(eu)/du * du/dx = eu * d(2x-7)/dx = eu * 2 (here * stands for multiplication).

Substituting back and rearranging, the derivative is: 2e2x-7

You see, this is not the original function, given in the exercise, so something went wrong. Now try to find the correct answer and point out what went wrong.

blogfast25 - 6-4-2016 at 05:06

Quote: Originally posted by aga  

$$y=\int e^{2x-7}dx$$
isn't e the 'teflon number' that is the same when derived, so probably the same when integrated
$$y=e^{2x-7}+C$$

As woelen showed, check your proposed solution by deriving it. You must find the original function. This is not the case here, as the derivative of your solution is:
$$y'=2e^{2x-7}$$
The error occurred because once again the integrand and differential didn't match.

Try the substitution:

$$u=2x-7$$

Had the integral been of the form:

$$y=\int e^{something}dsomething$$

... then direct integration would have been possible.

But:

$$y=\int e^{somethingelse}dsomething$$

... cannot be integrated directly.
<hr>
Products are commutative:

Here’s something we all know but less experienced practitioners of algebra/calculus may sometimes overlook:
$$a .b=b.a$$
But also:
$$a.b.c=c.a.b=b.c.a=c.b.a=b.a.c$$
Products are commutative: the result of the product is invariant to the sequence in which the multiplications are carried out.

Sometimes this can help ‘see’ how an integral can be solved because for example:
$$\int x\sqrt{x^2+1}dx=\int \sqrt{x^2+1}xdx$$
Or even:$$=\int dx x\sqrt{x^2+1}$$
(the latter notation, with the differential directly right of the integration symbol, is increasingly popular in modern textbooks)

We can do this because:
$$ x\sqrt{x^2+1}dx$$
... is just a product, even though it’s preceded by an integration symbol.

But look at this notation:
$$\int \sqrt{x^2+1}xdx$$
This might jig the mind a bit because isn’t that last bit?
$$xdx$$
... the differential of:
$$x^2+1$$

Well, not exactly but close:
$$(x^2+1)’=2x$$
$$d(x^2+1)=2xdx$$
So really:
$$xdx=\frac12 d(x^2+1)$$
Substituting we get:
$$\int \sqrt{x^2+1}\frac12d(x^2+1)$$
$$=\frac12 \int\sqrt{x^2+1}d(x^2+1)$$
This can be integrated directly, because integrand and differential ‘match’:
$$=\frac12 \times \frac23 (x^2+1)^{\frac32}+C$$
$$=\frac13 (x^2+1)^{\frac32}+C$$
This is basically a variant on the substitution method.


[Edited on 6-4-2016 by blogfast25]

aga - 6-4-2016 at 10:28

Oh dear. I think i should give up .....

... imagining that i 'get it' regarding to the chain rule, certainly how to reverse it.

It would probably be easier to 'get' if the functions in the examples were simpler : instead of
$$\int \frac {sin(5x) cos(x^3) \sqrt [x^{\frac 12}] {x^2-2x+1}}{\sqrt {popeye}} dx$$
something like
$$\int (x-1)x^2dx$$

I'll go back to page 3 or 4 and start again with the derivatives chain rule.

woelen - 6-4-2016 at 10:53

I myself learned integrating by regarding it as the reverse of taking a derivative. Knowing the rules for taking derivatives it frequently is easy to integrate. Combine this with the rule that f(u)du can be integrated directly, but f(u)dv, with u not equal to v cannot be integrated directly.

Unfortunately, in general, integration is much more difficult than differentiation and there are infinitely many more functions which cannot be integrated analytically than ones that can be integrated. E.g. taking the integral of esin(x) is not possible, at least not in terms of functions like sin(x), cos(x) or ex or combinations of those. Even something like e-x² cannot be integrated and requires introduction of new special functions like erf(x). Such functions are introduced, simply by definition. Another example is the so-called li(x) function. It is defined by

dli(x)/dx = 1/ln(x)

For such non-integrable functions only approximations can be given, beit numerically, or trucated series. As a high school boy I was really fascinated by this concept that a seemingly easy operation like taking a derivative cannot be reversed always analytically.

blogfast25 - 6-4-2016 at 11:19

Quote: Originally posted by aga  
Oh dear. I think i should give up .....

... imagining that i 'get it' regarding to the chain rule, certainly how to reverse it.

It would probably be easier to 'get' if the functions in the examples were simpler : instead of
$$\int \frac {sin(5x) cos(x^3) \sqrt [x^{\frac 12}] {x^2-2x+1}}{\sqrt {popeye}} dx$$
something like
$$\int (x-1)x^2dx$$

I'll go back to page 3 or 4 and start again with the derivatives chain rule.


It's a deal! :cool:

Compute:

$$\int (x-1)x^2dx$$

aga - 6-4-2016 at 12:27

Quote: Originally posted by blogfast25  
Compute:
$$\int (x-1)x^2dx$$

I will when i feel sure i actually DO understand the chain rule for derivatives properly, especially knowing which bit is a function of a function of x, and how to identify those bits in the thing we're deriving, then identify what those results look like when integrating.

If it's any consolation, i watched a few khan acedemy videos on this, and they were no clearer, seemingly randomly picking a chunk to be u then saying that all the other bits were therefore du, which equals dx, obviously ...

Thanks for the input woelen. It is clear that i'm not yet good enough with the derviative chain rule to be able to apply it in reverse.

I might get fascinated at the forward/reverse discrepancy when i can actually properly apply the rules !

blogfast25 - 6-4-2016 at 13:08

Quote: Originally posted by aga  
Quote: Originally posted by blogfast25  
Compute:
$$\int (x-1)x^2dx$$

I will when i feel sure i actually DO understand the chain rule for derivatives properly, especially knowing which bit is a function of a function of x, and how to identify those bits in the thing we're deriving, then identify what those results look like when integrating.



Sure but this one doesn't even need substitution:

$$(x-1)x^2=x^3-x^2$$:)

aga - 6-4-2016 at 13:45

$$y = \int (x-1)x^2dx$$
$$(x-1)x^2dx$$
$$=(x-1)dx^3$$
$$=dx^4-dx^3$$
$$=dx(x^3-x^2)$$
$$=(x^3-x^2)dx$$

$$y = \int (x^3-x^2)dx$$
Sum/diff rule
$$y = \int x^3dx - \int x^2dx$$
$$y = \frac {x^4}{4} - \frac {x^3}{3}+C$$
$$12y = 3x^4 - 4x^3+C$$
$$4y = x^4 - 1\frac 13 x^3+C$$

Those bits seem fine, yet the Chain Rule, forward and reverse are <strike>mis</strike> not understood.

Edit: (there have been many !)

For those wondering why i did not apply any operations to the constant, C, well, it's an unknown constant, so if it's 10,000 times bigger or smaller, or +/- 0.001, it does not matter yet, seeing as we do not know what it is anyway.

At this stage it is just 'C', no matter what you do to it (apart from trying to include it in a function of x or y).

[Edited on 6-4-2016 by aga]

blogfast25 - 6-4-2016 at 15:05

Do by all means revisit the chain rule in derivation: it's the key to understanding the chain rule/substitution in integration.
Quote: Originally posted by aga  
$$y = \int (x-1)x^2dx$$
$$(x-1)x^2dx$$
$$=(x-1)dx^3$$
$$=dx^4-dx^3$$
$$=dx(x^3-x^2)$$
$$=(x^3-x^2)dx$$

$$y = \int (x^3-x^2)dx$$
Sum/diff rule
$$y = \int x^3dx - \int x^2dx$$
$$y = \frac {x^4}{4} - \frac {x^3}{3}$$
$$12y = 3x^4 - 4x^3$$
$$4y = x^4 - 1\frac 13 x^3$$

Those bits seem fine, yet the Chain Rule, forward and reverse are <strike>mis</strike> not understood.



Getting from line 1 to line 6 had already been done because as I wrote:

$$y=\int (x-1)x^2dx=\int (x^3-x^2)dx$$

Then:

$$y=\int x^3dx-\int x^2dx=\frac14 x^4-\frac13 x^3+C$$
So you forgot the integration constant there, the rest of your reworking wasn't necessary.
Quote: Originally posted by aga  


At this stage it is just 'C', no matter what you do to it (apart from trying to include it in a function of x or y).



That's correct but it's easiest to tag on C when all integrations and back-substitutions are fully done and dusted.

<hr>
Tomorrow I really want you do do some simple exercises, according the following template.

If f is a function of ax+b (and a and b are constants) find the integral:
$$y=\int f(ax+b)dx$$
... where f is a simple function like:
$$(ax+b)^n, \cos(ax+b), e^{ax+b}, \ln(ax+b),\:\text{etc, etc}$$
Using substitution we can show that, if we call:

$$u=ax+b$$
Then:
$$dx=\frac1a du$$
So:
$$\int \frac1a f(u)du=\frac1a \int f(u)du=\frac1a F(u)$$
And:
$$y=\frac1a F(ax+b)+C$$

Example:

$$y=\int (7-5x)^{-5}dx$$

Solution:
$$y=-\frac15 \times -\frac14 (7-5x)^{-4}+C$$
$$y=\frac{1}{20}(7-5x)^{-4}+C$$

Exercises:

1.

$$y=\int \cos(3x)dx$$

2.

$$y=\int e^{-4x}dx$$

3.

$$y=\int \frac{1}{(x+10)^2}dx$$

4.

$$y=\int \sin(6-7x)dx$$

5.

$$y=\int \ln(4x+3)dx$$

[Edited on 7-4-2016 by blogfast25]

aga - 7-4-2016 at 11:58

Unfortunately today was quarter-end day, so had to do 3 months of accounts where the derivative is Tax payment, which is integral to avoiding jail.

The morrow shalt thou see efforts most calculatious.

aga - 8-4-2016 at 11:09

Somehow that explanation seems a lot clearer. Let's see.

So if
$$y=\int f(nx+a)dx$$
and
$$y=\frac 1n \int f(nx+a)$$
If i understand that rightly, Exercise 1 ...
$$y=\int \cos(3x)dx$$
$$y=\frac 13 \int \cos(3x)$$
$$y=\frac 13 \sin(3x) + C$$

Exercise 2
$$y=\int e^{-4x}dx$$
dunno yet - gimme a sec

Exercise 3
$$y=\int \frac{1}{(x+10)^2}dx$$
erm...

Exercise 4
$$y=\int \sin(6-7x)dx$$
$$y=-\frac 17 \int \sin(6-7x)dx$$
$$y=-\frac 17 \big ( -\cos(6-7x) \big ) + C$$
$$y=\frac {\cos(6-7x)}{7} + C$$


Exercise 5
$$y=\int \ln(4x+3)dx$$
$$y=\frac 14 \int \ln(4x+3)dx$$
$$y=\frac {(4x+3)ln(4x+3)-(4x+3)}{4} + C$$

blogfast25 - 8-4-2016 at 13:34

@aga:

Thanks for these answers.

1, 4 and 5 all have the correct final answer.

BUT. For example:

$$y=-\frac 17 \int \sin(6-7x)dx$$

... is not correct and should have been:

$$y=-\frac 17 \int \sin(6-7x)d(6-7x)$$

Because:

$$d(6-7x)=0-7dx=-7dx$$

That's where that factor:

$$-\frac 17$$

... comes from.

So for the more general case:

$$\int f(ax+b)dx=\frac 1a \int f(ax+b)d(ax+b)$$

Then integrate.
<hr>
I suggest we draw a temporary line under the substitution/chain rule and move on to newer pastures. I will revisit integration by substitution/chain rule again a little later.

Deal? :)

[Edited on 8-4-2016 by blogfast25]

aga - 8-4-2016 at 13:41

Stuff it down the the back of the cupboard and pretend it's all OK ?

That would be unethical, maybe cause the deaths of many integrants, and is downright wrong ...

... yet probably profitable.

OK. Deal !

blogfast25 - 8-4-2016 at 15:57

Trust me, we'll get back to it. It's hard to do simple integrals without it.

But we'll give it a rest for now.

Tomorrow: how to determine C and compute the determined integral.

<hr>
The solution for 2 and 3 are:

Ex.2.
$$y=\int e^{-4x}dx$$
Note that:
$$d(-4x)=-4dx$$
$$\implies dx=-\frac14 d(-4x)$$
Insert into integral:
$$y=\int e^{-4x} \Big[-\frac14\Big]d(-4x)$$
$$=-\frac14 \int e^{-4x} d(-4x)$$
$$=-\frac14 e^{-4x}+C$$
Ex.3:
$$y=\int \frac{1}{(x+10)^2}dx$$
$$=\int (x+10)^{-2}dx$$
Note that:
$$d(x+10)=dx+0=dx$$
Insert into the integral:
$$y=\int (x+10)^{-2}d(x+10)$$
$$=-(x+10)^{-1}+C$$
$$y=-\frac{1}{x+10}+C$$

[Edited on 9-4-2016 by blogfast25]

The Definite Integral:

blogfast25 - 9-4-2016 at 11:15

Two methods are available, here’s the first one.

1. Determining C with a boundary condition (aka an ‘initial condition’):

The indefinite integral is given by:

$$y=\int f(x)dx=F(x)+C$$
(provided: )
$$\frac{\mathrm d}{\mathrm d x} F(x)=f(x)$$

Since as C is any constant, y is not fully defined. So to fully define y we need to determine the integration constant.

One way of doing this is to know a single data point (x,y) = (a,b) of the function y(x).

For example, if we knew that:

$$y(a)=b$$
So:
$$y(a)=b=F(a)+C$$
Then:
$$C=b-F(a)$$
And:
$$y(x)=F(x)+b-F(a)$$
Example:
$$y=\frac13 x^3+C$$

Let's say we know that:
$$y(3)=10$$
So:

$$10=\frac13 3^3+C=9+C$$
Or:
$$C=1$$
So:
$$y=\frac13 x^3+1$$
Another example:

Compute the definite integral y:
$$y=\int (x-1)^3dx$$
With boundary condition:
$$y(1)=5$$
Solution. First, compute the indefinite integral:

Use substitution:
$$u=x-1$$
$$du=dx$$
$$\int u^3du=\frac14 u^4$$
Substitute back and add C:
$$y=\frac14 (x-1)^4+C$$
Now determine C from the boundary condition:
$$5=\frac14 (1-1)^4+C=0+C=C$$
So the definite integral is:
$$y=\frac14(x-1)^4+5$$

aga - 9-4-2016 at 15:07

Bugger.

If i could do the previous bit, this bit looks a lot easier.

Perhaps a break from pure maths and back to synthesising stuff for a while ?

Over a kilo of sodium acetate is itching to become GAA thence Zinc acetate, so it can become nice clean white crystals (same as the 150g advance party is after recrystallisation from ethanol) thence to acetic anhydride via vac distillation.

There's also this bottle o'vodka itching to become TCA, which would come in handy for catalysing some of this bottle of Turps into Turpineol ...

[Edited on 9-4-2016 by aga]

blogfast25 - 9-4-2016 at 15:32

Quote: Originally posted by aga  
Bugger.

If i could do the previous bit, this bit looks a lot easier.

Perhaps a break from pure maths and back to synthesising stuff for a while ?



The next few episodes are probably easier to grasp than much of what went before.

But if I build in a break now, chances are we'll never resume again... :D

So let's bite the bullet!

Tomorrow: definite integrals by integrating between two boundaries. :cool:

[Edited on 9-4-2016 by blogfast25]

aga - 9-4-2016 at 15:36

Being Chain-Rule-Disabled is a bit Hampering, and we all know what happens to Hampers.

Perhaps i should just get a Maths Disabled badge and carry on regardless ?

(Free parking sounds good - i can certainly calculate that one !)

blogfast25 - 9-4-2016 at 15:54

Quote: Originally posted by aga  


Perhaps i should just get a Maths Disabled badge and carry on regardless ?

(Free parking sounds good - i can certainly calculate that one !)


We'll address the u-gremlins again later, then we'll see what badge fits.

The purpose of this mini-tour of calculus is mainly to understand its place in science, less to churn out integration-proficient under-graduates.

Trust me, you haven't done so bad so far. :)

And u-gremlins will soon be destroyed by means of computer assisted integration, coming to you soon, HERE!


[Edited on 10-4-2016 by blogfast25]

2. Eliminating C with two boundary conditions:

blogfast25 - 10-4-2016 at 07:22


If F(x) + C is the integral y:
$$y=\int f(x)dx=F(x)+C$$
Now suppose we want to calculate:
$$\Delta y=y(b)-y(a)$$
Of course:
$$y(b)=F(b)+C$$
and:
$$y(a)=F(a)+C$$
Then:
$$\Delta y=y(b)-y(a)=F(b)+C-F(a)-C$$
C drops out, so:
$$\Delta y=y(b)-y(a)=F(b)-F(a)$$
Notation:
$$\Delta y=\int_a^b f(x)dx=\Big[F(x)\Big]_a^b=F(b)-F(a)$$
We say that the function f(x) is being integrated between the boundaries x=a and x=b.

Applied example:

An object moves along a straight line with the following speed - time function:
$$v(t)=6t+2$$
Problem: What distance Δx does the object travel between t = 2 and t = 5?
We know that:
$$v(t)=\frac{dx}{dt}$$
So:
$$dx=v(t)dt$$
$$dx=(6t+2)dt$$
$$\Delta x=\int_2^5(6t+2)dt$$
The indeterminate integral of (6t+2)dt is:
$$ \int(6t+2)dx=6\times \frac12 t^2+2t+C=3t^2+2t+C$$
So:
$$\Delta x=\Big[3t^2+2t\Big]_2^5$$
$$\Delta x=3 \times 5^2+2 \times 5 – [3 \times 2^2 + 2 \times 2]$$
$$\Delta x= 69$$
Incidentally: we’ve just solved a differential equation (albeit a really simple one)!

Darkstar - 10-4-2016 at 11:35

Quote: Originally posted by aga  
Bugger.

If i could do the previous bit, this bit looks a lot easier.

Perhaps a break from pure maths and back to synthesising stuff for a while?


If you ever want to go back to learning organic chemistry, I'm down for picking up where we left off in blog's QM/QC thread.

blogfast25 - 10-4-2016 at 13:20

Quote: Originally posted by Darkstar  

If you ever want to go back to learning organic chemistry, I'm down for picking up where we left off in blog's QM/QC thread.


Me too. I've got a follow-up post ready on organo-metallics applied to Grignard reactions. But I've held back to avoid 'information overflow'... ;)

[Edited on 10-4-2016 by blogfast25]

aga - 10-4-2016 at 13:49

Yay ! Lab time ! Woohoo !
Quote: Originally posted by Darkstar  
If you ever want to go back to learning organic chemistry,

Never stopped Sir, was just in Detention for quite a long time ...

[Edited on 10-4-2016 by aga]

blogfast25 - 10-4-2016 at 14:09

@aga:

Any problems/questions/comments on definite integrals so far?

aga - 10-4-2016 at 14:24

Quote: Originally posted by blogfast25  
Any problems/questions/comments on definite integrals so far?

(to self: Shit ! might get into Detention again ! Quick ! Think !)

No Sir.

They look fantastically clear and amazing.

I will do my homework on indefinite integrals, master that, then apply them to the Definite integrals with no difficulties at all.

(to self: add in some Sweetener or he'll Test you and you might miss the practical)

With all of the information you have kindly provided, plus all the Extra information you graciously added due to my innate dim-ness, it should all become clear in a few days with some hard work and diligence.

(to self: he's hesistating. Get out Now !)

I must go to the bathroom Sir.

Thank you again for your patient efforts and i cannot express my gratitude enough.

(self: Exeunt stage left, running)

[Edited on 10-4-2016 by aga]

blogfast25 - 10-4-2016 at 15:02

Ok.

Tomorrow: two-boundary definite integrals as surface areas.

Then I'm going to bring software assisted integration forward. I was going to keep that till the end but it seems a little silly to hold back on a method nearly everybody uses nowadays.

And then we can finally talk about differential equations proper.

Surface area under a function's curve:

blogfast25 - 11-4-2016 at 05:26

It can be shown that the expression:
$$\int_a^b f(x)dx=\Big[F(x)\Big]_a^b$$
... mathematically corresponds to the green hatched area in the figure below:
Integration surface.png - 5kB
So that calculation yields the surface area contained between the curve of f(x), the x-axis and the boundaries x=a and x=b.

This can be understood somewhat intuitively as follows. Imagine a very thin slice of surface area with width dx and height y(x). Its surface area is:
$$dA=f(x)dx$$

The sum total of all these slices between a and b would the actual hatched surface area A. This is where the integration symbol:
$$\int$$
... comes from: integration between boundaries is an extended summation and the integration symbol an elongated ‘S’.

Worked example:

Determine the hatched area A for:
$$y=\sin x$$
Between:
$$x=0, x=\pi$$
Sine area.png - 3kB
$$A=\int_0^{\pi} \sin x dx$$
$$A=\Big[-\cos x\Big]_0^{\pi}$$
$$A=[- \cos \pi -(-\cos 0)]=(-(-1)-0)=1$$
Worked physics example: Isothermal compression of an Ideal Gas, work needed.

If a volume V0 of ideal gas at pressure p0 is kept at constant temperature then according to Boyle’s Law for any other pressure p and volume V the following relationship holds:
$$p_0V_0=pV$$
Indicator work.png - 5kB

We can also write accordingly:
$$p(V)=p_0\frac{V_0}{V}$$
Thermodynamics tells us that to carry out an infinitesimally small compression dV, an infinitesimal amount of mechanical work dW has to be expended, according to:
$$dW=-p(V)dV$$
(The negative sign is needed because dV < 0 but dW > 0)

It can be shown that for a compression from V0 to V (at constant temperature) the total amount W corresponds to the red hatched area in the graph above, so:
$$W=\int_{V_0}^V (-p(V))dV$$
$$W=\int_{V_0}^V (-p_0\frac{V_0}{V})dV$$
$$W=-p_0V_0\int_{V_0}^V \frac{1}{V}dV$$
$$W=-p_0V_0\big[\ln V\big]_{V_0}^V$$
$$W=-p_0V_0(\ln V-\ln V_0)$$
$$W=-p_0V_0\ln \frac{V}{V_0}$$
Finally:
$$W=p_0V_0\ln \frac{V_0}{V}$$
<hr>
Simple Exercise:

Determine the value of the blue hatched area:

surface area exercise.png - 3kB

Where:
$$y=5-2x$$
with boundaries:
$$x=0, x=2$$

[Edited on 11-4-2016 by blogfast25]

aga - 12-4-2016 at 12:31

If expulsion does get discussed again, may i remind the teaching staff of the modulus of those negatives i still have in safe keeping from the 1992 school cultural visit to Paris.

You know, the ones of yourselves with 'Monsieur Fouetter la Derriere' at madame Fifi's place on the Rue de La Huchette ...

Speaking of that Blue Hatched Area clearly visible in the photos :-

$$y=5-2x$$
$$x=0, x=2$$
$$BHA = \int_0^2 (5-2x)dx$$
sum/diff rule
$$BHA = \Big[\int 5dx - \int 2xdx \Big]_0^2$$
$$BHA = \Big[(5x + C) - (\frac {2x^2}{2} + C) \Big]_0^2$$
C cancels out, 2's cancel out
$$BHA = \Big[5x - x^2 \Big]_0^2$$
$$BHA = (5(2) - 2^2) - (5(0) - 0^2)$$
$$BHA = (10 - 4) - (0 - 0)$$
$$BHA = 6$$

... although the total area in the photos seems much larger ;)

[Edited on 12-4-2016 by aga]

The Volatile Chemist - 12-4-2016 at 13:45

Looking good Blogfast! Fun to see you teaching stuff I've just been learning this past year. Looking forward to seeing you go over Rotational volumes in a week or so :)

aga - 12-4-2016 at 13:55

Hmm. i forgot. you're not in the 'Fifi Affair' photos.

Erm, how about the Epcot Centre 2012 photos ?

Yes. I have those too.

The Volatile Chemist - 12-4-2016 at 14:29

Quote: Originally posted by aga  
Hmm. i forgot. you're not in the 'Fifi Affair' photos.

Erm, how about the Epcot Centre 2012 photos ?

Yes. I have those too.

Uh, drunk...? I don't get it...

Your answers, upon quick review, are correct, but I didn't check your basic math skills at the end :)

blogfast25 - 12-4-2016 at 14:49

Quote: Originally posted by aga  

$$y=5-2x$$
$$x=0, x=2$$
$$BHA = \int_0^2 (5-2x)dx$$
sum/diff rule
$$BHA = \Big[\int 5dx - \int 2xdx \Big]_0^2$$
$$BHA = \Big[(5x + C) - (\frac {2x^2}{2} + C) \Big]_0^2$$
C cancels out, 2's cancel out
$$BHA = \Big[5x - x^2 \Big]_0^2$$
$$BHA = (5(2) - 2^2) - (5(0) - 0^2)$$
$$BHA = (10 - 4) - (0 - 0)$$
$$BHA = 6$$



... is 100 % correct. :cool::cool:

You can also verify it by geometrical means. The area is the sum of a rectangle and a right angled triangle, both with width = 2.

The height of the rectangle is 1 (because y=5 - 2 . 2 = 1), so its area is 2 . 1 = 2.

The height of the triangle is 5 - 1 = 4, so its area is (4 . 2)/2 = 4.

The sum is 2 + 4 = 6.

[Edited on 12-4-2016 by blogfast25]

blogfast25 - 12-4-2016 at 14:52

Quote: Originally posted by The Volatile Chemist  
Looking good Blogfast! Fun to see you teaching stuff I've just been learning this past year. Looking forward to seeing you go over Rotational volumes in a week or so :)


I wasn't going to do Rotational volumes but I'll gladly do it by special request. It's a good example of how to set up a differential equation. :)

The Volatile Chemist - 12-4-2016 at 15:03

Haha, I'll leave it to aga/others to request it, as I've already been taught it and have three calculus textbooks at my house right now, two of which I own. I just particularly like them because they're a nice simple example of the practical use of the integral.

aga - 12-4-2016 at 15:15

Excellent result from an Excellent lecture.
(to sneaky self: attempts to insert the word Excellent into teacher's brain for writing in results)

So, where next ?
(from self: simple diversion at the point of indecision)

Oh ! It's QM/OC time.
(from self: give direction)

I do believe that's the bell.
(all: the bell rang)

blogfast25 - 12-4-2016 at 15:58

Quote: Originally posted by aga  

So, where next ?
(from self: simple diversion at the point of indecision)

Oh ! It's QM/OC time.
(from self: give direction)



The next bit you'll really like: how to compute integrals using software, on the tinkerwebs, no downloads. Hands free! :cool:

OC/QM to resume very shortly...

[Edited on 13-4-2016 by blogfast25]

Integration with Wolfram Alpha’s DSolve:

blogfast25 - 13-4-2016 at 12:14

The ‘computational knowledge engine’ Wolfram Alpha (WA) offers a large number of free, online mathematical tools, one of which is called DSolve.

As the name implies, DSolve is a tool for solving differential equations (DEs). Since as the simplest of DE takes on the general shape:
$$y’(x)=f(x)$$
Because:
$$\frac{dy(x)}{dx}=f(x)$$
Thus:
$$y(x)=\int dy(x)=\int f(x)dx$$
This type of DE is known as a ‘DE with separation of variables’ because all terms containing y are on one side of the equation and all terms containing x on the other side of the equation. So we can use DSolve’s capability of solving that type of DE to compute integrals.

1. Indefinite integrals:

To have DSolve compute an indefinite integral go to WA and type the following command in the box:

DSolve[y’(x)==f(x)]

Where f(x) is a mathematical expression that is interpretable by WA. WA does not require (nor accept) LaTex but is generally very good at correctly interpreting simply formatted mathematical expressions. Let’s try:
$$x\cos(3x^2-1)$$

Input:

DSolve[y’(x)== xcos(3x^2-1)]

This generates the following page

First and foremost, check if WA has interpreted your input correctly by checking the ‘input’ section. If the interpretation was incorrect you’ll have to make the expression clearer (more explicit), often by adding brackets or other forms of mathematical ‘punctuation’. Note that DSolve will also handle expressions that contain constants like a or b.

Once input is correct, scroll down for the computed integral. DSolve uses ‘c<sub>1</sub>’ as the symbol for the integration constant.

2. Definite integrals:

If the problem is a stated as:
$$y’(x)=f(x), f(a)=b$$
Where f(a)=b is the boundary (or initial) condition.

The required WA input is:

DSolve[y’(x)==f(x),y(a)==b]

Example:

$$y’=x\sqrt{x+1}, y(0)=0$$
WA input:

DSolve[y’(x)== xsqrt(x+1),y(0)==0]

Result page

Note that WA now puts the input between {} brackets because it’s a set of two equations.

In either case of definite or indefinite integrals the computed integral may superficially differ from a manual solution. That’s due to the fact that most algebraic expressions can be written differently while returning the same function values.

Can DSolve compute any integral? Not by a long shot! As woelen pointed out, many, the majority even, of integrals do not have an analytical solution. In plain English that means a solution cannot be obtained by means of a finite number of steps.

This problem was recognised from early on in calculus’ long history and algorithms that offer numerical solutions have been developed ever since (the earliest recorded method is probably Euler’s Method.

WA offers its own suite for numerical integration called NDSolve and depending on interest I might spend some time on it.

[Edited on 13-4-2016 by blogfast25]

aga - 13-4-2016 at 12:26

Erm, i like the Human method best.

The computers only know what we tell 'em (mostly) so MUCH better for humans to be able to do the maths instead of relying on mechanical devices.

Also, the Human programmer can make mistakes, causing the computer to give the Wrong answers (which always look Right in nicely formatted text/graphics).

One could cite millions of pre-release tests in defence, but then so could Microsoft.

blogfast25 - 13-4-2016 at 13:42

Quote: Originally posted by aga  
Erm, i like the Human method best.

The computers only know what we tell 'em (mostly) so MUCH better for humans to be able to do the maths instead of relying on mechanical devices.

Also, the Human programmer can make mistakes, causing the computer to give the Wrong answers (which always look Right in nicely formatted text/graphics).


That's a bit of an idealistic position, aga. This is an extremely basic course.

Now imagine Big Calculus: problems that involve systems of simultaneous DEs, non-separable, perhaps second order (or higher) and non-linear. You'd need all the help you could get, trust me!

Of course it's to be preferred to solve 'by hand' what you can (can I take this as your commitment to battle your own u-gremlins? :D) but that's not always possible.

Re. MS, we all like to bitch about them but would you really want to do without Excel (or similar)?

[Edited on 13-4-2016 by blogfast25]

aga - 13-4-2016 at 14:01

I have no problem with Computer-Assisted.

Big problem with Computer-Enabled.

As a programmer, i have seen what happens.
(they blame the programmer as they had no way to check the computer's result)

Edit:

u-gremlin source is still being tracked.
It Will be found and assimilated.
Resistance is Futile.

Excel was not the first, but still probably the best.

[Edited on 13-4-2016 by aga]
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