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anonymous
 3 years ago
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)
anonymous
 3 years ago
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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0think 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354143287330:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1354143374332:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oops the area (3 to 6)∫ f(z) dz=3 not 8 sorry my bad

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0doesn't change anything but the value

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the area under f from 3to 6 would be 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what about the negative sign in front of the integral? Does it affect 3 at all?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohh never mind...i get it. I have to plug in the 3 into the integral and the answer would be 12
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