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

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

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

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

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

∫ [ (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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**lkomarci****Member**- Registered: 2005-08-24
- Posts: 23

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

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

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

ok! I'm always here for integration problems : )

Calculus IS my favorite subject!

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

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...

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

The solution is correct.

I'm writting the explanation rigth now

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

```
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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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,660

Excellent!

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

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

MathsIsFun wrote:

Excellent!

thanks:)

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

Excellent indeed

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

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

you're welcome :)

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**angela****Guest**

2x-4y=16

3x+6y=-12

help

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

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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