MHB Application to Improper Integrals

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The discussion revolves around calculating the total number of infections from a virus outbreak modeled by the function y=200xe^-0.5x. The improper integral from 0 to infinity is set up to find the total infections, but confusion arises regarding the evaluation leading to an infinite result. The integration by parts is correctly applied, and the boundary terms evaluate to zero, confirming that the integral converges. Ultimately, the calculation shows that the total number of infections is 800, resolving the initial concern about obtaining an unexpected infinite result. Proper application of improper integrals confirms the finite total infections.
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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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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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