A integral problem

coolwind · 2006-04-13 16:05:28

∫(t+1)e^(-jwkt)dt

Thanks...

Prakash Panneer · 2006-06-01 08:56:02

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)}

This is our required answer.

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

John E. Franklin · 2006-06-01 11:11:28

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) ?

George,Y · 2006-06-03 21:22:56

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

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

Prakash Panneer · 2006-06-04 01:09:22

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

I made a mistake.

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)

Math Is Fun Forum

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

A integral problem

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

Re: A integral problem

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

Re: A integral problem

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

Re: A integral problem

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

Re: A integral problem

Board footer