9 items are filed under this topic.
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    Distance as a function of acceleration | 
  2013-07-10 | 
  
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  From Tom: If you start at a stoplight and your acceleration is 16t - t^2, how far have you gone after 8 seconds? Answered by Penny Nom. | 
  
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    Integral 1/(25-x^2)^3/2 | 
  2012-02-22 | 
  
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  From John: Integral 1/(25-x^2)^3/2 Answered by Harley Weston. | 
  
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    An antiderivative of  the square root of (8t + 3) | 
  2011-04-19 | 
  
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  From Caitlyn: 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= Answered by Penny Nom. | 
  
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    The integral of (x^2*exp(x)/(exp(x)-1)^2 | 
  2010-08-09 | 
  
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  From sujoy: please find this integral for me
  
int(x^2*exp(x)/(exp(x)-1)^2 Answered by Robert Dawson. | 
  
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    An antiderivative problem | 
  2009-08-13 | 
  
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  From Indrajit: ∫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".... Answered by Harley Weston. | 
  
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    A definite integral | 
  2009-02-09 | 
  
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  From Mathata: Evaluate: integral from 0 to 1, x^2 e^x^3dx Answered by Harley Weston. | 
  
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    Antiderivative of 1/(x(1 - x)) | 
  2008-10-22 | 
  
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  From Matt: 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. Answered by Harley Weston. | 
  
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    f(x+y) = f(x) + f(y) + 2xy | 
  2007-11-01 | 
  
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  From Marcia: 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). Answered by Penny Nom. | 
  
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    Velocity and acceleration | 
  2005-10-27 | 
  
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  From Candace: 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? Answered by Harley Weston. | 
  
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