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karisos
Group Title
Suppose that f is continuous and that (3to3)∫f(z)dz=0 and (3to6)∫f(z)dz=8
Find (3to6) ∫4f(z) dz.
NOTE: (By 3 to 3 I mean that the 3 is below the integral and the 3 is up or above the integral)
 one year ago
 one year ago
karisos Group Title
Suppose that f is continuous and that (3to3)∫f(z)dz=0 and (3to6)∫f(z)dz=8 Find (3to6) ∫4f(z) dz. NOTE: (By 3 to 3 I mean that the 3 is below the integral and the 3 is up or above the integral)
 one year ago
 one year ago

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Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
think of them as areas... the area under f from 3..3 is zero; the area under f from 3..6 is eight; so the area under f from 3..6 must be 8
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
dw:1354143287330:dw
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
dw:1354143374332:dw
 one year ago

karisos Group TitleBest ResponseYou've already chosen the best response.0
Oops the area (3 to 6)∫ f(z) dz=3 not 8 sorry my bad
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
doesn't change anything but the value
 one year ago

karisos Group TitleBest ResponseYou've already chosen the best response.0
so the area under f from 3to 6 would be 3
 one year ago

karisos Group TitleBest ResponseYou've already chosen the best response.0
what about the negative sign in front of the integral? Does it affect 3 at all?
 one year ago

karisos Group TitleBest ResponseYou've already chosen the best response.0
ohh never mind...i get it. I have to plug in the 3 into the integral and the answer would be 12
 one year ago
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