Recent content by sbcdave

So true. Like Gauss' law. Which Newton discovered, in principle, 100 years before Gauss, and someone else probably discovered before Newton. Thank for the replies.

Does anyone know who coined the term "Improper" Integral, and why they chose the word improper? I couldn't find anything on the etymology of it via searching here or searching google, duckduckgo, and yahoo. Thanks for any info or direction to info.

Is there a mod than can move this to a forum where I may get some replies?

Forgot to say f prime was squared in the integrals I showed, but oddly the numbers work out the same either way. i.e. 1 squared and 0 squared, then add the one and square root. Also forgot to ask what dx represents here. I know this is probably beneath most of you, but I didn't see a better...

I'm feeling pretty stupid since no one even thought this deserved a reply...haha I'm giving it a bump just in case and have a few other questions. 1. How do I solve Cos(x) = y for x? I can see cos(x)-y=0, or cos(x)/y=1 but I'm lost from there. 2. I was looking into the procedure for...

Hi. I have been trying to find a way to create a function that I can use to solve sin(x) in my head. I'm familiar with the taylor series but a few weeks ago thought that f(x)= -(x^2)+1 might superimpose well over a sin wave and would be an easy way to approximate. e.g. from x= +1 to 0 would...

Thanks for the replies. Very interesting subject to me. Sorry for reviving the dead thread, hopefully some others were entertained by it.

I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a...

Thanks Roger But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote...

I'm new to calculus, but my impression is that an integral should be a function that represents the area under the curve of the function being integrated, which f(x)=ln(x) does not do for f(x)=1/x In the graph of 1/x you can see that from 0 toward infinity the area under the curve would come...