
#1
Nov2404, 08:56 PM

P: 3

i've been trying to prove this question, but so far, there is no successs...
http://members.shaw.ca/brian103/theerrorfunction.jpg 



#2
Nov2404, 10:39 PM

Sci Advisor
HW Helper
P: 3,149

Does this help?
[tex]\int_a^b = \int_a^0+ \int_0^b[/tex] [tex]\frac {d}{dx} \int_0^x f(x') dx' = f(x)[/tex] 



#3
Nov2404, 10:40 PM

HW Helper
Thanks
P: 9,818

[tex]\int _0^bf(t)dt=\int_0^af(t)dt+\int_a^bf(t)dt [/tex] b. Remember that the differential quotient of definite integral with respect to the upper bound is the integrand itself. [tex] \big[\int _0^xf(t)dt\big]' = f(x)[/tex] ehild 


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