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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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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

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Other answers:

Oops the area (-3 to 6)∫ f(z) dz=3 not 8 sorry my bad
doesn't change anything but the value
so the area under f from 3to 6 would be 3
what about the negative sign in front of the integral? Does it affect 3 at all?
ohh 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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