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Application to Improper Integrals


New member
Nov 4, 2013
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 :)

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
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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