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...
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...
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...
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 similiar 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