Recent content by BlackMamba

Opps I also left out antisymmetric. I don't understand how if aRb and bRa then a=b. I could use a little guidance on this one too. Thanks.

Hello, It's been a while since I've had to determine whether a relation is reflexive, symmetric or transitive so I would appreciate a bit of guidance. I now that being reflexive means the relation xRx for all x and being symmetric implies xRy and yRx for all x, y and transitive implies that if...

Thanks FrogPad and thanks again to Astronuc. The help was greatly appreciated!

Thank you I understand how you got that. u=\sqrt{x} then, u^2 = x However, I'm getting lost as to how I'm supposed to substitute that into the original equation. I think the part that is throwing me is the d(u^2) = 2udu This is as far as I'm getting: \int_{0}^{4} {e^\sqrt{u^2}} du...

Hello, I have this integral that is stumping me. Once I get the basic start then I know I can finish the rest myself it's just getting started that's stumping me. Here is the problem: \int_{0}^{4} {e^\sqrt{x}} dx First this doesn't appear to me to be an improper integral. So I should just...

Thanks fire and radou. Fire ~ Your description is what my graph looks like so thanks again.

I still am unsure about this. I could use the help. Thanks. Actually I think I got it. But if you could look at confirm or deny what I have come up with I would appreciate it.

Hello again, Once I get this graphed correctly, I don't see any problems as to why I couldn't solve it. But for now I have a question about graphing this volume problem. The problem says to find the volume generated by rotating the region bounded by y=e^x, x=0, and y=\pi about the x-axis. I...

Oh my gosh...I was making that WAY harder then it needed to be. Thank you very much arildno.

Hello, I am having trouble getting started with this integral. \int sin2x ln(cos2x) dx My first thought goes to using integration by parts to solve this but I am still getting stuck. I also know that sin2x = 2sinxcosx but again I'm not seeing how that would help me so I'm back to...

Well that is what I was unsure of. I just double checked the math on my calculator, and I realized I made an error inputting the values. So that would explain the error. Thanks for the input. :smile:

Hello, I have a problem that I am having difficulties with. I'm told to find the general integral. So here is the problem: \int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {1 - sin^2 x} dx Here is my partial solution: \int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx}...

I think I'm getting confused. I mean I believe I understand what you are saying. I do recall learning that, but as I've proven before nothing really sticks with me. lol So if I keep all negative signs in the equations it would look something like: F(\pi) = -cos\pi + C F(-\pi) = -cos(-\pi) + C...

Uhmm not really: Let me try to break it down further. F(\pi) = - cos\pi + C F(-\pi) = cos\pi + C I'm assuming for F(-\pi) that since cos is already negative and that's paired with a negative \pi then that combo would be positive. I'm guessing that is not correct.

Actually I have a similar problem that I could use a bit of help on as well. I have to find the integral using the same method. Here's the problem: \int_{-\pi}^{\pi} sinx dx So basically it boils down to: = -cos\pi + C - (cos\pi + C) = -cos\pi - cos\pi = 2