Application to Improper Integrals

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SUMMARY

The discussion focuses on calculating the total number of infections from a virus outbreak modeled by the function y=200xe^-0.5x. The integral of this function from 0 to infinity is evaluated using integration by parts, leading to the conclusion that the total number of infections is 800. The improper integral type I is correctly applied, and the boundary terms are shown to evaluate to zero, confirming the result of 800 infections.

PREREQUISITES
  • Understanding of improper integrals, specifically type I.
  • Proficiency in integration by parts.
  • Familiarity with exponential decay functions.
  • Basic knowledge of limits and their evaluation at infinity.
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  • Study the properties of improper integrals in calculus.
  • Practice integration by parts with various functions.
  • Explore the behavior of exponential functions as limits approach infinity.
  • Learn about applications of integrals in modeling real-world phenomena, such as population dynamics.
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Students and professionals in mathematics, particularly those studying calculus and differential equations, as well as researchers modeling infectious disease outbreaks.

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Suppose that the rate that people are getting infected in an outbreak of a virus is given by y=200xe^-0.5x. How many people in total will get infected from this outbreak?

So i know I'm doing it right but i keep getting a strange number…

so i set up an integral of that function from 0 to infinity. I simplified it by integration by parts and i applied the improper integral type I since it has the infinity in its bound. and i got that the entire population will get infected or infinity…
∫_0^infinity▒〖200xe^(-0.5x) 〗

Is there something I'm doing wrong?
Thanks :)
 
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Integrating by parts,

$$ 200 \int_{0}^{\infty} xe^{-x/2} \ dx = 200x(-2e^{-x/2}) \Big|^{\infty}_{0} - 200 \int^{\infty}_{0} (-2 e^{-x/2}) \ dx$$

As $x$ goes to infinity, $e^{-x/2}$ goes to zero much faster than $x$ goes to infinity.

So the boundary term evaluates to zero at both the upper and lower limits.

And we have

$$200 \int_{0}^{\infty} xe^{-x/2} \ dx = 400 \int_{0}^{\infty} e^{-x/2} \ dx = - 800 e^{-x/2} \Big|^{\infty}_{0} = -800(0-1) = 800$$
 
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