Approximating integral of x^4 e^-x

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The forum discussion focuses on approximating the integral of the function \( \int_0^{r_0} r^4 e^{-r/a} dr \) where \( r_0 \ll a \). The user attempts to use a Taylor series expansion of the exponential function, specifically \( \int_0^{r_0} r^4 (1 - \frac{r}{a}) dr \), but encounters increasing errors as \( r_0/a \) decreases. The discrepancy between the approximated and exact values, particularly when \( r_0/a \) is around \( 10^{-5} \), raises concerns about potential numerical errors in the computation. The exact integral is expressed in the form \( \int_0^1 x^m e^{-bx} dx \), with the exact solution given as \( \frac{m!}{b^{m+1}} \left[ 1 - e^{-b} \sum_{r=0}^{m} \frac{b^r}{r!} \right] \).

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I'm trying to approximate an integral of the form:

[tex]\int_0^{r_0} r^4 e^{-r/a} dr[/tex]

where r0<<a. I figured I could write out the first few terms of the expansion of the exponential and integrate that, since the extra terms would quickly become negligible:

[tex]\int_0^{r_0} r^4 (1-\frac{r}{a}) dr[/tex]

However, when I calculate this answer and compare it to the exact answer from mathematica, the error gets bigger the smaller r0/a gets. I can't figure out why. Specifically, with a few extra terms with the same form, but different powers of r, and r0/a of the order 10^-5, my approximated integral comes out to be about 10^-10 while the exact value is about 0.02. Could this be numerical error? I told it to algebraically come up with the ratio of the approximation and the exact integral, and there were 1/r^5 and 1/r^6 terms, but I can't figure out where they're coming from. Even as I add more terms to the taylor approximation, the answer barely changes.
 
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express the integral in the form
[tex]\int^{1} _{0} x^{m} e^{-bx} dx[/tex] the exact answer is

[tex]\frac{m!}{b^{m+1}} \left[ 1 - e^{-b} \sum^{m} _{r = 0} \frac{b^{r}}{r!} \right][/tex]
 
Have you considered that, after you've approximated the integrand, you made a mistake in your numerical computation of the integral?
 

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