Krysten
Reputation
399
Top tag
Next privilege 500 Rep.
Access review queues
 Mar 7 comment Find the area of the surface obtained by rotating the curve about the x axis After doing that I get: the integral from 0 to 1 of [(-2/pi) sqr(1 + u^2pi) du]. Would I have to do a trig substitution? Mar 7 asked Find the area of the surface obtained by rotating the curve about the x axis Mar 3 accepted Determining a closed-form solution for the following sum Mar 3 comment Determining a closed-form solution for the following sum thanks! that cleared it up Mar 3 comment Determining a closed-form solution for the following sum but when evaluating the first summation, it does not need to be multiplied by (n+1) because you are plugging in for i directly? Mar 3 comment Determining a closed-form solution for the following sum in the second summation, are you adding n+1 because the index starts at 0 and not 1? Mar 3 revised Determining a closed-form solution for the following sum added 1 characters in body Mar 3 comment Determining a closed-form solution for the following sum the second one. Mar 3 asked Determining a closed-form solution for the following sum Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ great. thanks for your help! Feb 27 accepted Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ Thanks. I finally understand, but shouldn't the evaluated integral be: -1/(10x^3) + x^5/6? Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ how did you get the second qauntity exactly? Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ is there supposed to be a sqr root? Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ @Tavares, yes it is Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ ok now i see. so then the integral would be sqr[(25x^8)/36 + 1/2 + 9/(100x^8)] Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ @Tavares where does the 1/2 come from? Feb 27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ It is 1/(10x^3) Feb 27 asked Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ Feb 25 comment Number Theory - Proof of divisibility by $3$ ahh i see now. makes perfect sense. thanks