Recent content by mcelgiraffe

"It is very important that we remember the definition for the unit step function. You just need to remember when t < 0, u(t) = 0 and it's 1 when t is greater or equal to 0." Thanks again for the additional clarification. I know this was like hammering a nail into concrete. I can see my...

I think I am catching on, but at t=4: wouldn't u(4) = 0, u(4-4) = u(0) = 1, & u(4-5) = u(-1) = 1? Since f(t) = 0 when 4≤t<5. I think it was just a matter of < vs ≤, either way we get the same answer. 0-1+1=0? vs 1-1+0=0. I appreciate you walking me through this. It finally cleared...

It rises again at t=5 so it would be u(t-5). I'm not sure why I changed u(t) to 1, part of my mis-understanding. So now we would have: f(t) = u(t) + u(t-4) - u(t-5) ? The signs are confusing me. If we are falling from 1 to 0 at t=4, it seems to me that it would be u(t) - u(t-4) and similarly...

It drops to zero at t=4, so I think it would be expressed as u(t-4). Does this lead to: f(t) = 1 - u(t-4)? This is what's getting confused in my head, but I will wait to see if my response is correct before I elaborate.

OK, the graph is attached (hope it shows up). In case it doesn't, the graph is a horizontal line at f(t) =1 from 0 to 4, drops to 0 from 4 to 5, and back up to 1 for t>5.

Homework Statement The problem is given as a piece-wise function f(t)={1, 0≤t<4; 0, 4≤t<5; 1, t≥5}. I am asked to write the function in terms of unit step functions and then find the Laplace Transform of the function. I can do the Laplace if I could ever get the unit step function...

I am not sure how I made it that difficult except that I have been staring at this way too long today. Dicks method was much easier and greatly appreciated. Thanks to both of you. James

Hi, I am trying to work a problem that seems to have me stumped. ∫x/√(x+1) dx I have tried to look at it as a right triangle with: hypotenouse = √(x+1) sideA = 1 sideB = √x So I have: cot^2 ∅=x, dx=-2cot∅csc^2 ∅ d∅ csc∅=√(x+1) Working through the problem I have -2∫(cot^2...

Thank You both for your response. My instructor told me the same thing but I had not accepted it yet. I did work it out both ways from 0 to pi and got the same answers. There is still something that I am uneasy about. I will spend some more time with it and see if I can get my question more...

Hi everyone, I have worked the following integral two different ways and received two different answers. ∫sin^3 xcosx dx+ ∫cosx dx The first method = ∫sin^3 xcosx dx+ ∫cosx dx Let u = sinx du/dx = cosx dx = 1/cosx du = ∫u^3 cosx*(1/cosx) du + ∫cosx dx = ∫u^3 du + ∫cosx dx =...