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Calculus Help!! If f′(x)=5xe^(x^2) and f(0)=0, then f(x)=
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Use substitution u = x^2 du = 2x dx
punch what you have into the shape of U'.e^U which integrates into e^U with f'(x) = (5/2).2x.e^(x^2) giving f(x) = (5/2).e^(x^2) + c sub initial value to get at c.
that's what i did too....i put 2.5 and it is not correct!!
what did you put for the constant --> 2.5e^0 +c = 0
-2.5 is not correct either
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should i put 0?is the answer 0?
no -2.5 should be correct ??
it is incorrect for -2.5!!
http://www.wolframalpha.com/input/?i=y%27+%3D+5x*e^%28x^2%29%2C+y%280%29+%3D0 no it is correct
but it shows incorrect for it.stupid online assignment!!thanks anyways:)
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