|
seriously you dont have to look at my blogs since it contains 80% homework.
there is something wrong in it, but idk where i did wrong. the result comes out as 3/5x^(5/3)+C
i understand that the dx needs to get taken out so we get the reciprocal of it, then replace f(x) into u for substitution.
im just doing some basic ones to understand it then doing the more harder ones:
the last part, do i plug back the original in and then do the antiderivative of it which would make it: u^4/4 - 2u^2/2 + 7u dx?
the problem i am having at the moment i think, is once i get the du out(if it IS du), i dont know when to do the antiderivative, and if i do it, lets say the original function is x^2, does it go x^3/3 or it stays as x^2 while the rest of the function gets their powers +1. or it gets antiderivatives twice? yes ive opened the book but it skips around a bit as that it doesnt go step by step which my brain needs.
thanks.
EDIT - dx = du, i tend to mistype that.
|
thedeadhaji
39489 Posts
ya i've definitely been going "wtf another intregral blog by raithed" haha
|
that was definite integral this is indefinite which has a + constant, the indefinite, the only issue with that is solving this, after i get THIS, then i just plug it in. its not hard its just that the graph threw me off a little,
|
You didn't need to use u substitution for the former, just write the radical in fraction form and do power rule.
I don't quite follow that you tried to do in the second one, what is the original problem? Int u^3 - 2u + 7?
EDIT - lol, Haji, I am sure most react the same way when you posted a blog like daily sometime the last couple of weeks. :p
|
= integral x^2/3
= 1/(2/3+1) * x^(2/3+1) + c
= 3/5 * x^(5/3) + c
|
For the following, I'll use the INT() to designate integral. For example, the first line of the first example will be INT( (x^2)^(1/3), x). The x after the comma is the dx.
Your substitution for the first one doesn't work. It'd give you this INT( u^(1/3) * (1/2x), u) which is just ugly. Instead, rewrite it like this: INT( u^(2/3), u) which should be easy to integrate (2/3 + 1 = 5/3, etc.).
For the second one, you don't need to substitute stuff. Just use linearity and separate it. INT( u^3 - 2u + 7, u) = INT( u^3, u) + INT( 2u, u) + INT( 7, u). This should be easy.
ETA: By the way, when you do the second step of the substitution, it's easier to write it in terms of du/dx. For example, when you say u = x^2, then write du/dx = 2x which gives you dx = du/2x. Then substitute that in for the dx.
|
On June 07 2008 10:23 Ecael wrote:
You didn't need to use u substitution for the former, just write the radical in fraction form and do power rule.
I don't quite follow that you tried to do in the second one, what is the original problem? Int u^3 - 2u + 7?
EDIT - lol, Haji, I am sure most react the same way when you posted a blog like daily sometime the last couple of weeks. :p
yeah to the second one. i didnt know that i needed to use the substitution or not, i do have a hard time looking at it.
On June 07 2008 10:24 fight_or_flight wrote:
= integral x^2/3
= 1/(2/3+1) * x^(2/3+1) + c
= 3/5 * x^(5/3) + c
uh oh, algebra issue i didnt see i guess. (x^2)^(1/3) = x^(2/3)? but then i dont understand how:
1/(2/3+1) * x^(2/3+1) + c
how come its 1/(5/3)?
On June 07 2008 10:26 Zelc wrote:
For the following, I'll use the INT() to designate integral. For example, the first line of the first example will be INT( (x^2)^(1/3), x). The x after the comma is the dx.
Your substitution for the first one doesn't work. It'd give you this INT( u^(1/3) * (1/2x), u) which is just ugly. Instead, rewrite it like this: INT( u^(2/3), u) which should be easy to integrate (2/3 + 1 = 5/3, etc.).
For the second one, you don't need to substitute stuff. Just use linearity and separate it. INT( u^3 - 2u + 7, u) = INT( u^3, u) + INT( 2u, u) + INT( 7, u). This should be easy.
ETA: By the way, when you do the second step of the substitution, it's easier to write it in terms of du/dx. For example, when you say u = x^2, then write du/dx = 2x which gives you dx = du/2x. Then substitute that in for the dx.
INT( u^3 - 2u + 7, u) = INT( u^3, u) + INT( 2u, u) + INT( 7, u).
u^3, u <-- what does this mean? if its the INT, then, antiderivative?
|
remember that for u subs, you have to have whatever du is in part of the original.
for example:
the u is x^3+3, du is 3x^2 dx, which is right there in the equation.
So you can't do a u-sub, since there is no 2x in the equation.
"1/(2/3+1) * x^(2/3+1) + c
how come its 1/(5/3)? "
1/(2/3+1) is 1/(5/3) which is 3/5
"
u^3, u <-- what does this mean? if its the INT, then, antiderivative?"
i think the u means du. He's just splitting the integrals into different sums (http://en.wikipedia.org/wiki/Sum_rule_in_integration)
if you ever have trouble with u-subs, just pretend it's a system of equations and you're just substitution. using my first example, u = x^3+3, du = 3x^2 dx, dx=du/3x^2, so wherever you see a dx, you replace it with a du/3x^2
|
OH. so when i see something like that, it basically LOOKS the same as if looking at a chain rule, correct?
INT (1 / 2x^3) dx
INT (1)(2x^(-3))dx
INT 2x^(-3)dx <-- can i do this now?
then we take the derivative out, so should i look at it as 2x = 2 = 1/2 ?
(1/2) INT .... uh oh, whats the next step or am i even going into the right direction?
|
yeah, a u-sub is just a chain rule in reverse.
i dunno what you wrote, so i just put the answer. i hope you see why this happens
|
okay going back to the first problem i listed:
from fof:
= integral x^2/3 understood
= 1/(2/3+1) * x^(2/3+1) + c i dont know why its 1 over (2/3 +1), can you try to explain? i get the x^(2/3+1) [anti?]
= 3/5 * x^(5/3) + c
and for:
(1/2) INT x^(-3) dx <-- it was (1/2) INT 2x^(-3) dx and you took this out right? i know why but can you explain why its always that number to be taken out?
|
"= 1/(2/3+1) * x^(2/3+1) + c i dont know why its 1 over (2/3 +1), can you try to explain? i get the x^(2/3+1) [anti?]"
power rule of integration: http://en.wikibooks.org/wiki/Calculus/Integration#Basic_Properties_of_Indefinite_Integrals
"(1/2) INT x^(-3) dx <-- it was (1/2) INT 2x^(-3) dx and you took this out right? i know why but can you explain why its always that number to be taken out?"
eh? well you wrote 1/2 * x^(-3), so you just take the coefficient out. i don't see where you got the 2 in front of the x from
|
what i mean is:
INT (1) / (2x^3)dx a
INT (1)(2x^(-3))dx b
you basically saw it as (1/2) INT x^(-3) dx from a instead of the need to rewrite? because i looked at it again and you just took the (1/2) out of the x^(-3)
im trying to really recognize integrals like:
INT (x^2 + 1)^2(2x)
g(x) would 2x right?
f(g(x)) = (x^2 + 1)^2 right?
and as i recognized it, what are the steps into solving it, as the answer is (1/3)(x^2 + 1)^3 + C
ugh, sorry im being repetitive, i cant seem to grasp the concept and understand it. antiderivatives are so much easier, integrals are so hard.
|
uhh yeah i just took out the 1/2 lol.
but for the second one, yeah you're right.
so you solve it by doing a substitution. i'm bad at using f and g notation, so i'll just demonstrate.
in your example, you let u=x^2+1, right?
then you take the derivative of both sides, so you get du = 2x dx.
then you see that you have a 2x dx in the equation, and you already had du = 2x dx, so you substitute 2x dx with du.
and then you get the u-sub, just remember to switch back to x's when you're done.
and well, u-subs should always be your second line of defense after the power rule and like, given integrals like cosine and sine.
and if you can't do a u-sub you go on to more difficult ways of solving it.
|
if you are still wondering about the first integral. Here are the steps, sorry I had to catch a bus so I didn't have time to make it more clear.
On June 07 2008 11:52 Raithed wrote:
im trying to really recognize integrals like:
INT (x^2 + 1)^2(2x)
g(x) would 2x right?
f(g(x)) = (x^2 + 1)^2 right?
and as i recognized it, what are the steps into solving it, as the answer is (1/3)(x^2 + 1)^3 + C
ugh, sorry im being repetitive, i cant seem to grasp the concept and understand it. antiderivatives are so much easier, integrals are so hard.
same thing....
|
youtube is helping me out too. =] i THINK, im not 100% but i THINK i got it. but for people who are viewing this, what happens to the other sin? o_o i mean, d' of cosx is sinx, so i guess it went into the sinx^4 yes?
INT (cosx)(sinx)^3 dx
(1/4)sin^4 + C
|
thats a u substitution
u = sinx
du = cosx dx
therefore
INT sinx^3 (cosx dx) = INT u^3 du
|
On June 07 2008 10:34 Raithed wrote:Show nested quote +On June 07 2008 10:26 Zelc wrote:
For the following, I'll use the INT() to designate integral. For example, the first line of the first example will be INT( (x^2)^(1/3), x). The x after the comma is the dx.
Your substitution for the first one doesn't work. It'd give you this INT( u^(1/3) * (1/2x), u) which is just ugly. Instead, rewrite it like this: INT( u^(2/3), u) which should be easy to integrate (2/3 + 1 = 5/3, etc.).
For the second one, you don't need to substitute stuff. Just use linearity and separate it. INT( u^3 - 2u + 7, u) = INT( u^3, u) + INT( 2u, u) + INT( 7, u). This should be easy.
ETA: By the way, when you do the second step of the substitution, it's easier to write it in terms of du/dx. For example, when you say u = x^2, then write du/dx = 2x which gives you dx = du/2x. Then substitute that in for the dx. INT( u^3 - 2u + 7, u) = INT( u^3, u) + INT( 2u, u) + INT( 7, u).u^3, u <-- what does this mean? if its the INT, then, antiderivative?
The function before the comma is the stuff that's being integrated. The u after the comma indicates that u is the variable that is being integrated. If you have more than one variable in the function, this gets important  . In other words, that u after the comma specifies that there's a du, as opposed to say a dx (if it was a dx, it'd be an x after the comma). So the INT( u^3, u) would look like squigly u^3 du. Haven't you used the integral function on a TI calculator before?
By the way, I made a typo. That should be a - INT( 2u, u) in the second part, not a + INT( 2u, u).
Basically integrals are linear in that INT( a*f(x) + b*g(x), x) = a*INT( f(x), x) + b*INT( g(x), x), where a and b are constants.
|
|
|
|
|
|
EPT![[image loading]](http://i32.tinypic.com/255rj3t.jpg)
![[image loading]](http://i26.tinypic.com/348nhty.jpg)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21%5Cint%203x%5E2%28x%5E3%2B3%29%5E2dx.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21%5Cint%20%5Cfrac%7B1%7D%7B2x%5E3%7Ddx%3D%5Cfrac%7B1%7D%7B2%7D%5Cint%20x%5E%7B%28-3%29%7Ddx%3D%5Cfrac%7B1%7D%7B2%7Dx%5E%7B-2%7D%2A%5Cfrac%7B1%7D%7B%28-2%29%7D.gif)
![[image loading]](http://img248.imageshack.us/img248/8413/calc101zj2.png)
