substitution? integration 0_o (1 Viewer)

[sukiyaki]

im really lost in this topic1.. i dunno i jsut dont get it
lets say we given a question like the one below where do we even start? 0_o

oh u = x^2 - 4

and how do we do it if we given the range of 2 to 0?

 

[McLake]

Let u = x^2 - 4
du = 2x dx
dx = du/2x

find new limits

u = (2)^2 - 4 = 0
u = (0)^2 - 4 = -4

substitue into inertrgation (I)

I = -2x/u * du/2x
I = -du/u

Now intergate to get
-[ln u] with limits 0 and -4

But 0 and 2 arn't valid limits ....

(0 causes a negative under the square root and 2 causes the intergration to go to infinity ...)

 

[Lazarus]

McLake is right... it's a hyperbola, and it's undefined for -2 <= x <= 2.

Were the bounds for x or for u? Because u is only undefined for u < -4.

 

[Halo]

So basically, what you want to do is convert the whole integral with "x" (or another pronumeral) to an integral with "u" (etc.). You are usually given what substitution is to take place, so you take that and work with it. Differentiate etc. until you get everything you need to completely replace the "x", and since you need to replace "dx" as well, you need to find out what the new limits are. You therefore put the limit values into the given equation to produce new values in terms of "du".

 

[sukiyaki]

Originally posted by Lazarus
McLake is right... it's a hyperbola, and it's undefined for -2 <= x <= 2.

Were the bounds for x or for u? Because u is only undefined for u < -4.

oh!! yeh !!i didnt notice that..

i just made up that range... outta me head.. bad range to fit the question thoz

anyways thansk your help

what if we had dx/(2x + 3)^3??

i try to do it but my answer turned wrong :/

 

[marsesbars]

Originally posted by sukiyaki

what if we had dx/(2x + 3)^3??

Let u = 2x+3
du = 2 dx
dx = du/2

Change the fraction into a power to make things easier:
I = (2x+3)^(-3) dx

Substitute u
I = u^(-3) du/2

Integrate and you get
u^(-2)
-----
-4
Maybe you forgot it was du/2 and didn't multiply by 1/2...

Plug back in u=2x+3, and there you are


Hm, I remember Coroneos claiming it's possible to do substitution integrations in your head! I didn't believe him, but by practicing I could at least do the substitution and the fudging of the expression in my head. I reckon if you keep practicing, integration is an easy topic (if tedious...)

 

[McLake]

Originally posted by marsesbars
Hm, I remember Coroneos claiming it's possible to do substitution integrations in your head! I didn't believe him, but by practicing I could at least do the substition and the fudging of the expression in my head. I reckon if you keep practicing, integration is an easy topic (if tedious...)

Sometimes you can intergrate equations that would normally require substitution by imagining what the intergral would look like (well, I can anyway )

 

[BlackJack]

Yes, it is quite possible. The markers derive the answer to get to the question in the first place; and since they have to think up of a logical one in the syllabus you can figure it out w/ practice.

 

[Weisy]

Originally posted by McLake


Sometimes you can intergrate equations that would normally require substitution by imagining what the intergral would look like (well, I can anyway )

easier, and much less to write. *remembers the endless t-substitutions questions*

 

[McLake]

Originally posted by Weisy
easier, and much less to write. *remembers the endless t-substitutions questions*

That reminds me: learn off by heart the derivitive of tan(x/2) [for use in "t" ratio intergration]

 

[sukiyaki]

Originally posted by McLake


That reminds me: learn off by heart the derivitive of tan(x/2) [for use in "t" ratio intergration]

er wHAT? wait we havin learnt dat yeh *shOO

 

[kini mini]

Originally posted by McLake


That reminds me: learn off by heart the derivitive of tan(x/2) [for use in "t" ratio intergration]

I disagree with that, not all t substitutions occur with t = tan(x/2). You might want in a tougher question to do substitute, say, for t = tan (3x/4). It's much safer, in my opinion, just to rederive the thing or at least know how to.