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AmTran_Bus

  • one year ago

Is there a way to get wolfram to show area of region? Trying to figure out how to use it.

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  1. phi
    • one year ago
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    Does not look promising... is this an integral problem?

  2. AmTran_Bus
    • one year ago
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    Well, yes and no. Here is all I know.

  3. AmTran_Bus
    • one year ago
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    Along with choices Area = 14.79 Area = 14.77 Area = 14.78 Area = 14.98

  4. phi
    • one year ago
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    then do \[ \int_{-4}^\infty e^{-\frac{x}{2}} \ dx\] if you know how do you know calculus?

  5. AmTran_Bus
    • one year ago
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    For real? Is that it? I was so overcomplicating this. I can do that.

  6. AmTran_Bus
    • one year ago
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    2e^2

  7. phi
    • one year ago
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    let u = -x/2 du = -dx/2 so dx = -2 du u goes from -(-4)/2 = 2 to u = -infinity/2 = -infinity thus \[ - 2 \int_2^{-\infty} e^u \ du \] or if we negate the integral we can swap the order of the integrand \[ 2 \int_{-\infty}^2 e^u \ du = 2 e^u \bigg|_{-\infty}^2= 2e^2 - 0= 2 e^2\]

  8. AmTran_Bus
    • one year ago
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    Phi, you are a lifesaver. Thanks a million.

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