integration by substitution (1 Viewer)

[izzy_xo]

hey guys

for the question:

using the substitution u^2= x + 2

find the integral of (x-2)/square root of (x+2)

to solve this would you just square root the u^2 substitution
so u= square root of (x+2) and just sub that in?

many thanks

 

[shaon0]

hey guys

for the question:

using the substitution u^2= x + 2

find the integral of (x-2)/square root of (x+2)

to solve this would you just square root the u^2 substitution
so u= square root of (x+2) and just sub that in?

many thanks

S (x-2)/sqrt(x+2) dx
= S (u^2-4)/sqrt(u^2) 2udu
= 2 S (u^2-4) du
= 2[(1/3)u^3-4u]+C
= 2sqrt(x+2).[(1/3)(x+2)-4]+C

Yes, is the answer to your question where applicable

 

[Ayatollah]

let u=s.r(x+2)
then find du/dx
then rearrange to get du=''''''dx
then sub that into the original (it fits nicely)
now you have integral of 2(x-2) with respect to 'u'
now sub in x=(u^2)-2 (just a rearrangement from original subsitution)
then its just simply intergral with respect to 'u'
then sub back in u=s.r(x+2)

 

[marmsie]

I am going to go against popular opinion and say that it is not entirely correct.

Firstly the mistake, if

u^2 = x + 2

then

u = sqrt(x + 2) AND u = -sqrt(x + 2)

e.g. if x = 7, then u^2 = 9 but then u could be either +3 or -3.

So unless you are going to take both cases and do the integration for each (which you are more then free to do so), then I suggest you wait until the end.

To do the substition using u^2:

u^2 = x + 2

x = u^2 -2

dx / du = 2u

dx = 2u du

subbing into the integral gives:

[(x - 2) / sqrt(x + 2)] dx

[(u^2 - 4) / sqrt(u^2)] 2u du

2u[(u^2 - 4) / u] du

2u^2 - 8 du

integrating:

2/3 u^3 - 8u + C

2/3 u (u^2 - 12) + C

subbing the x's back into the equation: (here we will have to take the sqrt of u^2 to sub it in for the u out the front)

+/- 2/3 sqrt(x + 2) (x + 2 - 12)

which gives the answer to be:

+/- 2/3 (x -10) sqrt(x + 2)

which with the exception of the +/- is the same answer that shaon0 calculated.

 

[cyl123]

um... try differentiating both answers you got... clearly only the positive case is valid

 

[marmsie]

Good point, well made.

Clearly, differentiating my answers showed that only the +ve solution holds true (also looking over it again, I realise I dropped my + C).

But I can not see any reason why if u^2 = x + 2, then u only equals +sqrt(x+2)

After all, if you were to graph u^2 = x + 2 then you have a complete side ways parabola with its vertex at (0 , -2).

So unless there is a reason why we can ignore the -ve value of u (and I am beginning to think there might be, but apart from the final answer not working I can not see it), then to be mathematically correct you would have to give both +ve and -ve answers but then state that only one of your answers actually fits. It would be like an absolute value question where you must test your answers because sometimes one or more values do not actually fit when you sub them back in.

Also, as an aside. If the question did not state which substitution to use, then I would have just let u = +sqrt(x+2) and that would avoid this situation completely.

 

[yugi]

hey guys

for the question:

using the substitution u^2= x + 2

find the integral of (x-2)/square root of (x+2)

to solve this would you just square root the u^2 substitution
so u= square root of (x+2) and just sub that in?

many thanks

hey, sorry no idea how to use latex and um might wanna check this but i got:

u^2=x+2
so differentiating both sides at once gives you:
2udu=dx
from the given substitution we can deduce that
x=u^2-2

so ...
integral of x-2/root (x+2)dx
= integral of u^2-2-2/root of u^2 multiplied by 2udu (substituting for dx)
= integral of u^2-2/u multiplied by 2udu
= 2 integral of u^2-2 du
= 2 [u^3/3 -2u] + C

 

[YashC3]

Is the substitution of u squared still part of the 3uint syllabus. I know a lot of 3u text books still have u squared, but Im pretty sure it got moved to 4unit? Anyone know anything about that ?

 

[Trebla]

[(x - 2) / sqrt(x + 2)] dx

[(u^2 - 4) / sqrt(u^2)] 2u du

2u[(u^2 - 4) / u] du

2u^2 - 8 du

integrating:

2/3 u^3 - 8u + C

2/3 u (u^2 - 12) + C

subbing the x's back into the equation: (here we will have to take the sqrt of u^2 to sub it in for the u out the front)

+/- 2/3 sqrt(x + 2) (x + 2 - 12)

which gives the answer to be:

+/- 2/3 (x -10) sqrt(x + 2)

which with the exception of the +/- is the same answer that shaon0 calculated.

Using your argument of the positive and negative case, the expression in bold above would technically be the absolute value of u. You're correct in pointing out there must be a positive and negative case but there is a small flaw in the working you showed. This is what it should look like:



Notice that both positive or negative case give the same answer. To avoid confusion most people opt for the positive case, though technically one should write u > 0.

 

[stephoe]

help? evaluate:

 

[Pwnage101]

Complete the square in the denominator first.

 

[stephoe]

thanks pwnage101
I got another question as well..


i got upto..



.. I dont know how to simplify from that? Yes I know my algebra and technique is useless

 

[shaon0]

thanks pwnage101
I got another question as well..


i got upto..



.. I dont know how to simplify from that? Yes I know my algebra and technique is useless

Yeah, this is how i used to do it during my HSC.
just multiply the 3 into the square root. So; dy/dx=4/sqrt(9-16x^2)

 

[Drongoski]

I know this question has been flogged to death.

Without formal substitution, using the drongoski approach:



Edit:
I'm sorry; I did not notice question was originally posted in February.

 

[Lukybear]

Is the drogonski method ok for 3unit?

 

[Trebla]

lol @ "drogonski method"

 

[adomad]

Is the drogonski method ok for 3unit?

its a 4-unit method but u can use it in the 3U paper

 

[Drongoski]

Is the drogonski method ok for 3unit?

Most teachers have never seen it done this way or are uncomfortable with it because they are not familiar with it, when shown, perhaps the first time. They think there is something fishy about it. There is nothing illegitimate about it except that most teachers would rather you did it the official (conventional) way, just to be safe.

 

[Trebla]

It's pretty much substitution anyway but without calling it u = ... formally. Either way, the integral with the indices (x+2)^0.5 etc can be done wrt x rather than wrt (x + 2) anyway. It's just a standard reverse chain rule problem.