  Centrale des maths - centraledesmaths.uregina.ca  Dilemmes & doutes    « D & D »    Sujet: antiderivative   nouvelle recherche

9 articles trouvés pour ce sujet.    Page1/1            Distance as a function of acceleration 2013-07-10 Tom pose la question :If you start at a stoplight and your acceleration is 16t - t^2, how far have you gone after 8 seconds?Penny Nom lui répond.     Integral 1/(25-x^2)^3/2 2012-02-22 John pose la question :Integral 1/(25-x^2)^3/2Harley Weston lui répond.     An antiderivative of the square root of (8t + 3) 2011-04-19 Caitlyn pose la question :I know how to take an antiderivative. But this one's stumping me. I need it to finish a problem. What's the antiderivative of the square root of (8t + 3) ~Caitlyn=Penny Nom lui répond.     The integral of (x^2*exp(x)/(exp(x)-1)^2 2010-08-09 sujoy pose la question :please find this integral for me int(x^2*exp(x)/(exp(x)-1)^2Robert Dawson lui répond.     An antiderivative problem 2009-08-13 Indrajit pose la question :∫4e^x + 6e^-x/(9e^x + 4e^-x)dx = Ax + Bloge(9e2x - 4) + C then A=?......B=?.....C=? plz solve it...."^" stands for "to the power of"....Harley Weston lui répond.     A definite integral 2009-02-09 Mathata pose la question :Evaluate: integral from 0 to 1, x^2 e^x^3dxHarley Weston lui répond.     Antiderivative of 1/(x(1 - x)) 2008-10-22 Matt pose la question :derivative of dx/(x(1-x)) From what I've seen I should break apart the equation as such derivative of dx/x - dx/(1-x) and then get the 2 corresponding log functions. If that is correct why does this factoring work, if that is incorrect what is the proper way to find the derivative.Harley Weston lui répond.     f(x+y) = f(x) + f(y) + 2xy 2007-11-01 Marcia pose la question :For all real numbers x and y, let f be a function such that f(x+y) = f(x) + f(y) + 2xy and such that the limit as h -> 0 of f(h) / h = 7, find: f(0), use the definition of the derivative to find f'(x), and find f(x).Penny Nom lui répond.     Velocity and acceleration 2005-10-27 Candace pose la question :When taking the integral of the position function, you get the velocity function, and the same for velocity to acceleration. So when you do each of these, you get a function. But when you integrate on a graph, you get an area under a curve. The area is un units squared- where do the units go when you make it an equation? How can a function be an area?Harley Weston lui répond.      Page1/1    Centrale des maths reçoit une aide financière de l’Université de Regina et de The Pacific Institute for the Mathematical Sciences.    Qui sommes-nous :: carte du site :: our english site