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#1 2005-09-11 01:20:48

lkomarci
Member
Registered: 2005-08-24
Posts: 23

Integrals: substitution method

hi everyone, plz don't laugh at me, i'm a math noob, i'm currently learning integrals and i got a problem. i dont understand how to solve this exercise:

∫ 4th sqrt(3x-5)³ dx    (it's not 4 x, it's 4th root)


       1.  now...i usually take this under the brackets and derivate it:
3x - 5=t
3dx=dt /:3
dx=dt/3

the solution should be:   I = 4 X 4th sqrt (3x-5)^7  / 21 + C


my math-english is not that good so.. this 4th sqrt should mean fourth square root smile

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#2 2005-09-11 01:55:42

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

If you are learning integration you are less n00b than you think...: )

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#3 2005-09-11 02:10:45

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

∫ [ (3x-5)^(1/4) ]^3 dx = ∫ (3x-5)^(3/4) dx
z=3x-4
dz/dx=3
dz=3.dx

∫ (3x-5)^(3/4) dx = (3/3) ∫ (3x-5)^(3/4) dx = (1/3) ∫ 3 (3x-5)^(3/4) dx = (1/3) ∫ (3x-5)^(3/4) 3.dx =
= (1/3) ∫ z^(3/4) dz = (1/3) [ z^(3/4+1) ] / (3/4+1) +C = (1/3)(4/7) [ z^(7/4) ] + C = (4/21) (3x-5)^(7/4) + C
and that's the answer.

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#4 2005-09-11 08:03:18

lkomarci
Member
Registered: 2005-08-24
Posts: 23

Re: Integrals: substitution method

hey thanx for your detailed reply kylekatarn.
i dunno why i didn't think of writing that root as en exponent. i do that a lot smile

thanx again. i'll write back if i got some more problems smile

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#5 2005-09-11 11:49:32

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

ok! I'm always here for integration problems : )
Calculus IS my favorite subject!

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#6 2005-09-12 04:33:28

lkomarci
Member
Registered: 2005-08-24
Posts: 23

Re: Integrals: substitution method

hey kyle...i got another one for you. i want u to solve this one, because i think that the solution in the book is not correct.  the book was written by some of my teachers, they probably wrote the darn book in one night. so many mistakes.

ok here it goes:

∫ 3dx / 4(2-5x)^4/5 =


the solution given in the book goes like this:

I=-3(2-5x)^1/5  /  4 + C

please... smile

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#7 2005-09-13 08:42:44

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

The solution is correct.
I'm writting the explanation rigth now

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#8 2005-09-13 08:53:57

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

let z = 2-5x

dz
—— = -5 
dx

dz = -5.dx


⌠    3dx       3 ⌠    dx        3 1 ⌠   5dx        3 ⌠ dz    
│——————————— = -.│————————— = - —.—.│————————— =- ——.│————
│        4/5   4 │      4/5     4 5 │      4/5    20 │ 4/5
⌡ 4(2-5x)        ⌡(2-5x)            ⌡(2-5x)          ⌡z

           3
let α = - ——  and β = 4/5 to make things more clear.
          20
                                ┌      ┐
                                │ -β+1 │
 ⌠ dz    ⌠ 1         ⌠ -β       │z     │         α   -β+1
α│——— = α│——— dz = α ⌡z   dz = α│——————│ + C = ————.z     + C
 │  β    │  β                   │ -β+1 │       -β+1
 ⌡ z     ⌡ z                    └      ┘

-β+1 = 1/5

  α      3
———— = - —
-β+1     4


putting everything together we get:

  3      1/5
- —(2-5x)    + C
  4

and that's it : )

Last edited by kylekatarn (2005-09-13 08:55:56)

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#9 2005-09-13 09:21:18

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,529

Re: Integrals: substitution method

Excellent!


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#10 2005-09-13 09:49:35

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

MathsIsFun wrote:

Excellent!

thanks:)

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#11 2005-09-13 21:20:17

lkomarci
Member
Registered: 2005-08-24
Posts: 23

Re: Integrals: substitution method

Excellent indeed
and once more...the day is saved by kylekatarn. THANK YOU smile

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#12 2005-09-14 03:21:22

kylekatarn
Member
Registered: 2005-07-24
Posts: 445

Re: Integrals: substitution method

you're welcome :)

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#13 2006-01-09 05:55:59

angela
Guest

Re: Integrals: substitution method

2x-4y=16
3x+6y=-12



help

#14 2006-01-09 07:01:50

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Integrals: substitution method

For future reference, if you want to ask for help, then create a new topic rather than posting on someone else's.

Anyway, for these simultaneous equations you need to scale them up so that they have a common term, then combine them to give a simple linear equation.

2x-4y=16 --> (multiply by 3) 6x-12y=48
3x+6y=-12 --> (multiply by 2) 6x+12y=-24

Add these equations together: 6x-12y+6x+12y=48-24
12x=24
x=2

Knowing this, you can easily work out the other value.

2*2-4y=16
-4y=12
y=-3

Check: 3*2+6y=-12
6y=-18
y=-3

It works! So your final answer is x=2, y=-3


Why did the vector cross the road?
It wanted to be normal.

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