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## #1 2006-04-13 16:05:28

coolwind
Member
Registered: 2005-10-30
Posts: 30

### A integral problem

∫(t+1)e^(-jwkt)dt

Thanks...

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## #2 2006-06-01 08:56:02

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

### Re: A integral problem

You didn't give any information about j,k,and w. So, i took these are constants.

Problem: Integral ( t+1 ) e ^ ( -jwkt ) dt

To solve this problem, we have to  use "Integration by parts method,"

Take u = t+1 , dv = e^ (-jwkt)dt

du = dt  and v = e^(-jwkt)/(-jwk)       (Oops)

Integrantion by parts formula  Integral ( u dv) = uv- integral(vdu)

So, we will get,     Integral ( t+1 ) e ^ ( -jwkt ) dt = {(t+1)e^(-jwkt) /(-jwk)} - Integral ( -e^(-jwkt)/(jwk)) dt

= {-(t+1)e^(-jwkt)/(jwk)} - e^ (-jwkt)/ (jwk)^2.

= {e^(-jwkt)/(jwk)} {-(t+1)- (1/jkw)}

Last edited by Prakash Panneer (2006-06-04 01:05:40)

Letter, number, arts and science
of living kinds, both are the eyes.

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## #3 2006-06-01 11:11:28

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: A integral problem

I'm rusty on calculus.
Why in above is the integral of  dv = e^(-jwkt)dt
become:
v = (e^(-jwkt)) / (-jwkt)
Does this mean that if y = e^(kx) then y'=kx e^(kx) ?

igloo myrtilles fourmis

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## #4 2006-06-03 21:22:56

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: A integral problem

is there an "i' instead of a "j"?

Last edited by George,Y (2006-06-03 21:23:28)

X'(y-Xβ)=0

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## #5 2006-06-04 01:09:22

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

### Re: A integral problem

Oops....................

That means  dv = e^(kx) dx

Integrating with respect to x on both sides, we get,

v = e^(kx)/k.

If   y = e^(kx) then y' = k e^kx.

Last edited by Prakash Panneer (2006-06-04 01:12:21)

Letter, number, arts and science
of living kinds, both are the eyes.

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