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karisos

  • 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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  1. Algebraic!
    • 3 years ago
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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

  2. Algebraic!
    • 3 years ago
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    |dw:1354143287330:dw|

  3. Algebraic!
    • 3 years ago
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    |dw:1354143374332:dw|

  4. karisos
    • 3 years ago
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    Oops the area (-3 to 6)∫ f(z) dz=3 not 8 sorry my bad

  5. Algebraic!
    • 3 years ago
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    doesn't change anything but the value

  6. karisos
    • 3 years ago
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    so the area under f from 3to 6 would be 3

  7. Algebraic!
    • 3 years ago
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    yes

  8. karisos
    • 3 years ago
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    what about the negative sign in front of the integral? Does it affect 3 at all?

  9. karisos
    • 3 years ago
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    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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