AmTran_Bus
  • AmTran_Bus
Is there a way to get wolfram to show area of region? Trying to figure out how to use it.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AmTran_Bus
  • AmTran_Bus
http://www.wolframalpha.com/input/?i=%28area+of%29+x+greater+than+or+equal+to+-4%2C+0+less+than+or+equal+to+y+less+than+or+equal+to+e%5E%28-x%2F2%29
phi
  • phi
Does not look promising... is this an integral problem?
AmTran_Bus
  • AmTran_Bus
Well, yes and no. Here is all I know.

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AmTran_Bus
  • AmTran_Bus
Along with choices Area = 14.79 Area = 14.77 Area = 14.78 Area = 14.98
phi
  • phi
then do \[ \int_{-4}^\infty e^{-\frac{x}{2}} \ dx\] if you know how do you know calculus?
AmTran_Bus
  • AmTran_Bus
For real? Is that it? I was so overcomplicating this. I can do that.
AmTran_Bus
  • AmTran_Bus
2e^2
phi
  • phi
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 \[ 2 \int_{-\infty}^2 e^u \ du = 2 e^u \bigg|_{-\infty}^2= 2e^2 - 0= 2 e^2\]
AmTran_Bus
  • AmTran_Bus
Phi, you are a lifesaver. Thanks a million.

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