Is there a way to get wolfram to show area of region? Trying to figure out how to use it.
Does not look promising... is this an integral problem?
Well, yes and no. Here is all I know.
Along with choices Area = 14.79 Area = 14.77 Area = 14.78 Area = 14.98
then do \[ \int_{-4}^\infty e^{-\frac{x}{2}} \ dx\] if you know how do you know calculus?
For real? Is that it? I was so overcomplicating this. I can do that.
2e^2
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 \[ 2 \int_{-\infty}^2 e^u \ du = 2 e^u \bigg|_{-\infty}^2= 2e^2 - 0= 2 e^2\]
Phi, you are a lifesaver. Thanks a million.
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