9 articles trouvés pour ce sujet.
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Distance as a function of acceleration |
2013-07-10 |
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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. |
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Integral 1/(25-x^2)^3/2 |
2012-02-22 |
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John pose la question :
Integral 1/(25-x^2)^3/2
Harley Weston lui répond. |
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An antiderivative of the square root of (8t + 3) |
2011-04-19 |
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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. |
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The integral of (x^2*exp(x)/(exp(x)-1)^2 |
2010-08-09 |
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sujoy pose la question :
please find this integral for me
int(x^2*exp(x)/(exp(x)-1)^2
Robert Dawson lui répond. |
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An antiderivative problem |
2009-08-13 |
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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. |
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A definite integral |
2009-02-09 |
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Mathata pose la question :
Evaluate: integral from 0 to 1, x^2 e^x^3dx
Harley Weston lui répond. |
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Antiderivative of 1/(x(1 - x)) |
2008-10-22 |
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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. |
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f(x+y) = f(x) + f(y) + 2xy |
2007-11-01 |
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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. |
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Velocity and acceleration |
2005-10-27 |
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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. |
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