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Topic review (newest first)

George,Y
2006-03-30 13:55:36

e0=1   (1)
if you are confused, recommand you this websiteif you accept 1, we can go on to discuss the upper limit
Limit<x->infinity>e(t - 1/b)x = ?
it depends

t-1/b>0, after altering:
(ex)^(t-1/b) the base diverges, and the power also diverges, for t-1/b>0 and  the power growes bigger and bigger as the base does so. no answer.

t-1/b=0,  e0=1 the limit is 1

t-1/b<0, after altering
(e-x)^(1/b-t)
The base is always positive and grows smaller and smaller when x goes larger and larger. 1/b-t>0, thus the power goes smaller as the base goes smaller ( 0.0010.36 < 0.010.36 ) the base's limit is zero, and the power's limit is also zero. the limit is 0.

you can do the rest yourself

RickyGeorge
2006-03-30 00:39:41

thanks Man I will consider your advice.
its just that i dont have confidence in myself when solving maths problems. so i need to check with you mathmaticians.
for example,
(1/B)*e^(tx)*e^(-x/b) = (1/b)*e^(x(t - 1/b)) So our integral is simply:

(1/B)*S[e^x(t - 1/b)] dx =

(1/B) * (1/(t - 1/b)) * e^(x(t - 1/b)) = (1/(bt - 1)) * e^ ((t - 1/b)x)

this is a definite integral with an upper limit of infinity and a lower limit of zero. but i am not really sure of the answer after we evaluate this integral.

can anybody help me with that?

George,Y
2006-03-29 20:20:34

gnitsuk ,you are a better teacher than me, i should say. cool

moreover, i want to say a few words to RickyGeorge

You don't need to be confused by the arguement by the two parts of your id

Actually, mathematicians have carefully chosen a defination to avoid the "approaching vs being" controversy.

They define an indefinite integral as the limit that a definate integral approaches. so the same symbol may mean different things with regard to whether an infinity or an undefined point is involved.You can consult to your caculus book and check out if these two types of integrals are  in an advanced chapter.

whether being or approaching is not that important if you accept a consept that math itself is a theory, which is an approximition. As a philosopher once put it "You cannot find two idendical leaves" --1+1=2 is nonetheless, an approximation, too

gnitsuk
2006-03-29 00:09:55

You have to evaluate the function at the upper limit and subtract its value at the lower limit.

So, we have -(x + B)e^(-x/B)

At the lower limit of 0 this is equal to -B  (As e^0 = 1)

At the upper limit this is sequal to 0 as e^(-infinity) = 1/e^(infinity) = 0  (one over big number is small number)

So value of this integral between limits is 0 - -B = B (this is the magnitude of the area of the curve between x = 0 and x = infinity) this region is below the x-axis.

RickGeorge
2006-03-28 18:02:07

All i know is that the upper limit of the integral is positive infinity and the lower limit is zero.

i think we need to evaluate the integral after integrating it. but i am not really sure about how to do it.

mikau
2006-03-28 13:52:32

You may need to specify whether they approach zero from the right or left. :-)

RickGeorge
2006-03-28 12:18:19

Thanks a lot !!
I just have one "last" question.
Actually those integrals are definite with an upper limit of infinity and lower limit of 0.
Can you please tell me what is the result when we evaluate these functions.


Thanks again

gnitsuk
2006-03-28 02:11:47

I'll try:

Our original integral is:

∫(x/B * e^(-x/B)) dx

Let's take the 1/B that is premultiplying the x outside the integral as it is a constant factor, so we now have:

(1/B) ∫(x * e^(-x/B)) dx  Call this Eqn (1). Now we use the formula and so we have 'uv - ∫v * (du/dx)' where:

u = x
v = -B*e^(-x/B)
du/dx = 1.

This gives (for our integral):

-Bxe^(-x/B) - ∫-B*e^(-x/B)

Let's take constant B out of integral giving:

-Bxe^(-x/B) + B∫*e^(-x/B)

Let's do the integration by sight, this leave us with:

-Bxe^(-x/B) - B^2 * e^(-x/B)

Finally -Be^(-x/B) is a common factor of the two terms so lets take it out as a common factor:

-Be^(-x/B) * [x + B]

Remember we have a factor of 1/B multiplying out integral from Eqn (1) so our answer is:

-Be^(-x/B)*[x + B] =

-(x + B)e^(-x/B)

RickyGeorge
2006-03-28 01:29:38

Thanks a lot..
Could u please clarify the last bit of the first function because i got confused with it.

So S(x/B * e^(-x/B)) dx = 1/B * S(x * e^(-x/B)) dx ( until this point i am ok, the rest got me confused ) = (1/B) * [-Bx*e^(-x/B) - B^2 * e^(-x/B)] =

-(x + B)e^(-x/B)

gnitsuk
2006-03-27 22:39:33

Function 2:

(1/B)*e^(tx)*e^(-x/b) = (1/b)*e^(x(t - 1/b)) So our integral is simply:

(1/B)*S[e^x(t - 1/b)] dx =

(1/B) * (1/(t - 1/b)) * e^(x(t - 1/b)) = (1/(bt - 1)) * e^ ((t - 1/b)x)

Mitch.

gnitsuk
2006-03-27 22:26:40

Function 1:

Use integration by parts formula:

S(u * dv/dx)dx = uv - S(v * du/dx) dx

Let u = x
Let dv/dx = e^(-x/B)

Then du/dx = 1
and v = -B*e^(-x/B)  [By sight]

So S(x/B * e^(-x/B)) dx = 1/B * S(x * e^(-x/B)) dx = (1/B) * [-Bx*e^(-x/B) - B^2 * e^(-x/B)] =

-(x + B)e^(-x/B)

Mitch

RickyGeorge
2006-03-27 21:37:08

Hello guyzz...
Plzzz help me in integrating these functions

1.     ∫ x/β e^(-x/β)  dx  ( i believe that these function can be integrated using integration by parts but i don't know how to apply these technique to this function )



2.     ∫ (e^(tx))/β  e^(-x/β)  dx  ( I believe that this function should be simplified before integrating it but still am not sure about how to do it)

I hope that someone helps me..

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