Definite Integral

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)

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... ;)

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.

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.

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.

Code:

---

  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

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

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?????!!!!!

How about integration by parts?

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

Intergration by what??

Way cool technique...

Okay, for example you have a function:

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

[ignore the fact that it's simple]

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:

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

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

Code:

---

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

---

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?

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

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

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

What basic integral formulas are you allowed to use?

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?

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

Better than your 94.5, eddy man!

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.

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