Definite Integral

[Talon_Karrde]

I have a definite integral that needs to be evaluated. I can only use basic integration formulas like these to solve it.
The integral of du = u + C.
The integral of du/u = ln|u| + C.
The integral of [f(u) ± g(u)]du = The intergral of f(u)du ± the integral of g(u)du.
The integral of (u to power n)du = [u power of (n+1)]/[n+1].
And any other basic intergration formulas.
Someone told me that you were supposed to break it up into two different intergals.

Here is the problem. I neet to evaluate the definite integral from 1 - 3 of [4Px times (the square root of 1 - (x-2)²)]dx. Where P = pie (or 3.14)

[[Digital-X-Treme]]

I think i understand what you are trying to do. I am guessing you are trying to find the area underneath the function you specified between the points 1 and 3 on the x-axis.

f(x) = 4*pi*Sqr(1-(x-2)^2) 'Where pi is a constant.

Is this what you want to do? BTW, that is a shocking biatch of a function to attempt to integrate... ;)

[Guv]

I thought it was integral of the following.

4*Pi*X*sqr(1-((X-2)^2)

Neither the above nor the Digit-X interpetation is easy.

I scanned a table of integrals and did not see anything which would help. I did not study it carefully, so I might have missed something.

Where did you find the above integral?

The table I scanned had some entries like the following.

Integral( F(X) ) = G(X) + integral( H(X) )

In some of the examples, F( X ) = Sqr(A*X^2 + B*X + C)

The above shows some, but not much promise. Something like the above might be what some body was trying to tell you about.

Note that derivative of (1- (x-2)^2)^3/2) is something like the following.

(3/2)*sqr(1-(x-2)^2)*2*X*(X-2)

The above does not help, but fooling around with something to power of 3/2 might get you an answer.

[Talon_Karrde]

Digital-X-Treme,
That is right except that it has an "X" in it like this...

4*Pi*X*sqr(1-((X-2)^2)) (as Guv wrote) then integrating from 1 to 3 (as you wrote)

Do you know how to do this? I know you have to break it into two intervals.

[noble]

I did this really quick so I don't know if it's right.

Let I = 4*Pi*x*sq[1-(x-2)^2])

1) Int(I,1,3) :problem
2) Let u = x-2 so u+2 = x and du = dx
3) Int(4*Pi*(u+2)*sq(1-u^2),1,3) :substitute
4) 4*Pi*Int(u*sq(1-u^2),1,3) + 8*Pi*Int(sq(1-u^2),1,3) :seperate
5) Let v = u^2 so dv = 2u*du
6) 2*Pi*Int(sq(1-v),1,3) + 8*Pi*Int(sq(1-u^2),1,3) :substitute

The rest is simple integration techniques
the first integral is basic and the second is a bit more
complicated but there are integration tables for these.

Int(I, 1, 3) = (4*Pi/3)*[1-(x-2)^2]^3 + 4*Pi*[ (x-2)*sq[1-(x-2)^2]+asin(x-2) ]


Like i said, I did it quick, but it should be right

[noble]

btw, a step i skipped was replacing u^2 for v and then
replacing x-2 for u

[Talon_Karrde]

how I'm going to solve the first int is
let z = 1 - v,
dz = -dv,
dv = -dz,
I'm still working on the second one becuase
I am only allowed to use some basic
integral formulas. I worked on that intgral
for 1 1/2 hours, and you did it quick?????!!!!!

[parksie]

How about integration by parts?

John - if you're reading this...please can we have LaTeX?

[Talon_Karrde]

Intergration by what??

[parksie]

Way cool technique...

Okay, for example you have a function:

f(x) = x^2 * (x + 2)



If f(x) is a product of two simpler functions, you can use the integration by parts formula:

INT[u * (dv/dx)] = uv - INT[v * (du/dx)]

All you need to do is decide which part is going to be u, and which is v. And this is the cool part:

u = x^2 ==> (du/dx) = 2x

dv/dx = (x + 2) ==> v = x/2 + 2x

...which gives:

[i]your orig. expr.
INT[x^2 * (x + 2)] = (2x)(x/2 + 2x) - INT[(x/2 + 2x)(2x)]

Which simplifies to:

INT[f(x)] = 5x^2 - INT[5x^2]

= 5x^2 - (5/3)x^3 + C

[noble]

np, yes i did it quick :P

integration by parts is this simple formula.......
there are proofs for it on the internet if you want to
verify it


int(u*dv) = v*u - int(v*du)

[Talon_Karrde]

Thanks noble, parksie, [Digital-X-Treme], guv.

One problem with the integration by parts is that
I can't use it yet. (I am only allowed to use the
"basic" integration formulas)

without using other integration techniques I have
got it down to int(u²/(sqr(1-u²))) integrating
from 1 to 3 with respect to u (du). Where u = x - 2.

I know you can use one of the arcsine rules, but
does anyone know how to do it with basic integral rules?

[parksie]

Why aren't you allowed to use the efficient and much easier techniques?

Because let's face it...substitution is...well...crap :)

[Talon_Karrde]

It's for a math assignment and my professor said I can't.

[Magnetic-Mike]

What basic integral formulas are you allowed to use?

[Talon_Karrde]

I am allowed to use these intergals.

The integral of du = u + C.
The integral of du/u = ln|u| + C.
The integral of [f(u) ± g(u)]du = The intergral of f(u)du ± the integral of g(u)du.
The integral of (u^n)du = [u^(n+1)]/[n+1].
The integral of du/[sqr(a² - u²)] = arcsin (u/a) + C, where (a) is a constant.
The integral of du/(a² + u²) = (1/a)*arctan (u/a) + C, where (a) is a constant.
The integral of du/[u*sqr(u² - a²)] = (1/a)*arcsec (|u|/a) + C, where (a) is a contstant.
The integral of (a^u)du = 1/(ln (a))*a^u + C.
The integral of (e^u)du = e^u + C.
I can also move the constant outside the integral if I want to.

Can anybody help?

[Edwin_Drood_1870]

WAASSSSUPPP!!!
yo, wassup TK?????
what's with the 100.5??????

[Talon_Karrde]

Better than your 94.5, eddy man!

[Talon_Karrde]

Hey thanks parksie, noble, gov, [Digital-X-Treme], I got an answer from my professer.
He boke it up into to integrals like this.
int [4*pie*x*sqr(1-(x-2)^2)dx] from 1 to 3,
= -2*pie*int[ (-2x+4-4)*sqr(1-(x-2)^2)dx].
= -2*pie*int[ (-2x+4)*sqr(1-(x-2)^2)dx] +
(-2)*pie*int[ -4*sqr(1-(x-2)^2)dx].
You then let u = 1-(x-2)^2, and the rest is simple substution.
By the way Magnetic-Mike, I learn integration by parts in two weeks.
Bye.

[Magnetic-Mike]

I'm glad for you Talon_Karrde
keep on learning :)