problem integrals

tomislav91 · 2011-03-01 21:19:15

i have a few integrals for maths class...but i dont know how to solwe them..Please help me.

Last edited by tomislav91 (2011-03-01 21:22:05)

Bob · 2011-03-01 22:21:45

hi tomislav91

That's quite a list so I'll start with a hint for each.

Post again, if that's not enough.

(i) Write as sin squared, squared. Use the double cosine formula to get as cos2x (squared), then again to get in terms of cos4x.
I think that should do it.

(ii) Haven't tried it, but what I'd try is substitution u = cube root (1 + x)

(iii) tricky. I think substitute u = root x, then partial fractions might do it.

Next three all look like integration by parts.

(Last one) You have a function and its derivative so guess (x squared +9) to the power 3/2 and modify to correct any multiplier.

See if those hints help. If not, post again.

Bob

tomislav91 · 2011-03-01 22:39:23

can you write like I..because I am from Serbia,i dont understand all...example: CUBE ROOT and so on...if you be so kind..i would be thanksfull..

gAr · 2011-03-01 23:00:29

Hi tomislav91 and bob,

1)
I would do what bob says.

2)
Substitute 1+x=t, then you end up with this:

3)

Substitute 1+x^(1/6)=t, we get:

4)
Substitute x²-1=t
⇒ ∫ (1/2)*ln(t) dt

5) and 6)
Integration by parts

7)
Substitute x²+9=t; we would get

Have a nice day.

gAr · 2011-03-01 23:03:36

Hi tomislav91,

Please tell what you did not understand.

tomislav91 · 2011-03-02 06:43:05

@gar this is 7. That what you do is not same like mine..or i somewhere make mistake?

and can you help me...i dont know how to continue..2 example..

and

Bob · 2011-03-02 07:29:05

hi tomislav91

Q1) Your answer is OK up to this point.

You should cancel the 'x' not the '2'

Q2)

let 1 + x = t

Can you do it from here?

Bob

Last edited by Bob (2011-03-02 07:46:01)

tomislav91 · 2011-03-02 08:36:01

i do not undrestand how is sqrt(t)dt/2 = 2/3 *pow(t,3/2)/2

and please do it all q.2 (because i dont undrestand all what i have to do)

Last edited by tomislav91 (2011-03-02 08:39:13)

Bob · 2011-03-02 09:06:21

hi

i do not undrestand how is sqrt(t)dt/2 = 2/3 *pow(t,3/2)/2

Standard integration of a power = "raise the power by 1, then divide by the new power"

Q2. Divide each term by "t to the power 1/3".

This leads to the expression given by gAr in post #4

The integral is then just a list of different powers of t. Do each one.

Bob

Last edited by Bob (2011-03-02 09:09:41)

tomislav91 · 2011-03-02 09:10:28

i know that,but...pow(x,1/2)*dt/2= pow(t,3/2)/3/2*2 that is the same? isnt?

Last edited by tomislav91 (2011-03-02 09:15:48)

gAr · 2011-03-02 14:56:44

Hi tomislav91,

I hope you meant:

It is true.

tomislav91 · 2011-03-03 21:43:55

calculate the volume of the body caused by the rotation of the surface which is limited by curves: about the axis Ox

Bob · 2011-03-03 22:27:34

hi tomislav91

How did you get on with the other integrals ?

Bob

Bob · 2011-03-03 22:40:20

hi tomislav91,

I always start with a sketch of the curves.

so this is a circle, centre (-3,0) and radius 3.

The other is a standard quadratic curve.

See below for the plot from

http://www.mathsisfun.com/data/function-grapher.php

Now find where the curves cross to get the limits.

Can you do this step ?

The expressions

will enable you to do

which is the volume of one minus the volume of the other.

Integrate between the limits you found above.

Post back for more help, if needed. I'm going off-line now for 2 hours, but I'll check again then.

Bob

Last edited by Bob (2011-03-03 22:50:07)

tomislav91 · 2011-03-04 20:34:10

Calculate the surface figure that is limited lines

gAr · 2011-03-04 21:05:45

Hi tomislav91,

Did you understand and complete the previous problems?

For the above problem, Is it the surface area that you require?

tomislav91 · 2011-03-04 21:48:15

yes,we finished in the school...
Yes,the surface..and please the function grapher..:) i have written task in Monday, and I have the few tasks for solving..Sorry if I bother you.

gAr · 2011-03-04 22:58:29

Ok, no problem.

I'm not sure which exactly is the area you require.

Anyway, I'll make assumptions and give a solution.
From the figure, I assume you need the area of the parabola when revolved about x-axis, from x=0 to x=1.

Last edited by gAr (2011-03-04 22:59:12)

tomislav91 · 2011-03-04 23:10:54

wait wait..:D and this task should be done as well as the volume, only that f (x), and the volume of f (x) ^ 2 and another PI. If you can only intersection points to find the plot and through the integral.

gAr · 2011-03-04 23:23:15

Hi,

Do you mean you want to find the volume as well?
Do you have the answers to check?

Last edited by gAr (2011-03-04 23:31:07)

Math Is Fun Forum

#1 2011-03-01 21:19:15

problem integrals

#2 2011-03-01 22:21:45

Re: problem integrals

#3 2011-03-01 22:39:23

Re: problem integrals

#4 2011-03-01 23:00:29

Re: problem integrals

#5 2011-03-01 23:03:36

Re: problem integrals

#6 2011-03-02 06:43:05

Re: problem integrals

#7 2011-03-02 07:29:05

Re: problem integrals

#8 2011-03-02 08:36:01

Re: problem integrals

#9 2011-03-02 09:06:21

Re: problem integrals

#10 2011-03-02 09:10:28

Re: problem integrals

#11 2011-03-02 14:56:44

Re: problem integrals

#12 2011-03-03 21:43:55

Re: problem integrals

#13 2011-03-03 22:27:34

Re: problem integrals

#14 2011-03-03 22:40:20

Re: problem integrals

#15 2011-03-04 20:34:10

Re: problem integrals

#16 2011-03-04 21:05:45

Re: problem integrals

#17 2011-03-04 21:48:15

Re: problem integrals

#18 2011-03-04 22:58:29

Re: problem integrals

#19 2011-03-04 23:10:54

Re: problem integrals

#20 2011-03-04 23:23:15

Re: problem integrals

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