| Pages:
1
..
5
6
7
8
9
..
19 |
aga
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
| 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
National Hazard
Posts: 709
Registered: 28-3-2016
Member Is Offline
Mood: No Mood
|
|
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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
$$y=\int xdx -\int 1dx=\frac12 x^2-x+C$$
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
We'll let it sink in for now and continue bright and bushy tailed on the morrow. 
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| 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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| 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
Hazard to Others
Posts: 279
Registered: 23-11-2014
Member Is Offline
Mood: Sleepy
|
|
| 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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
'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
Super Administrator
Posts: 7976
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| 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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
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
Super Administrator
Posts: 7976
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| 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!
Compute:
$$\int (x-1)x^2dx$$
|
|
|
aga
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by aga |
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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
$$y = \int (x-1)x^2dx$$
$$(x-1)x^2dx$$
$$=(x-1)dx^3$$
$$=dx^4-dx^3$$
$$=dx(x^3-x^2)$$
$$=(x^3-x^2)dx$$
$$y = \int (x^3-x^2)dx$$
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
@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
Forum Drunkard
Posts: 7030
Registered: 25-3-2014
Member Is Offline
|
|
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
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
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]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
The Definite Integral:
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:
$$y=\frac14(x-1)^4+5$$
|
|
|
| Pages:
1
..
5
6
7
8
9
..
19 |
![[*]](images/xpblue/default_icon.gif){kind=icon}
