It came up when finding a constant such that the integral is equal to 1 and thus behaves like a pdf. I used the parts method but have made an error, just curious how others might approach the problem.
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7$\begingroup$ I'd probably either use integration by parts directly or take $u=1-x$ so that I could expand a binomial to the third power instead of the sixth power. $\endgroup$– IanDec 31, 2014 at 21:00
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2$\begingroup$ As a side note this is a Beta Distribution with $cf(x),\alpha=4,\beta=7$ where $c=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}$. $\endgroup$– ChrisDec 31, 2014 at 21:05
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2$\begingroup$ See beta function. $\endgroup$– LucianDec 31, 2014 at 21:16
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6 Answers
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The beta function is the best idea.
$$\beta(a, b) = \int_{0}^{1} x^{a-1}(1-x)^{b-1} dx$$
For here $I$, let $a = 4, b = 7$
Using:
$$\beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
$$\beta(4, 7) = \frac{\Gamma(4)\Gamma(7)}{\Gamma(11)} = \frac{3!6!}{10!}$$
$$= \frac{3\cdot2\cdot1}{10\cdot9\cdot8\cdot7} = \frac{1}{840}$$
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$\begingroup$ Nice, was just working on this answer after thinking about it :) $\endgroup$– ChrisDec 31, 2014 at 21:22
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This way might be faster, but it ultimately depends on your personal preference.
Let $u = 1-x$. Then the integral becomes:
$$\int (u-1)^3u^6 \ \text{d}u$$
And that expansion is much easier to work with than the original!
answered Dec 31, 2014 at 21:01
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$\begingroup$ Be careful with the sign of the definite integral if you do this method though. $\endgroup$– GFauxPasDec 31, 2014 at 21:07
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$\begingroup$ Indeed @GFauxPas. We picked up a negative since $du = -dx$, which I incorporated into the binomial. $\endgroup$ Dec 31, 2014 at 21:10
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$\begingroup$ But also, the transformation will have the bounds be from $1$ to $0$ rather than from $0$ to $1$. $\endgroup$– GFauxPasDec 31, 2014 at 21:16
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1$\begingroup$ Oh sure. I usually do the integral as is, then at the end sub back in whatever $u$ was and evaluate at the original bounds. Just a weird habit of mine. $\endgroup$ Dec 31, 2014 at 21:19
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$\begingroup$ I also tend to sub back and evaluate the original bounds. $\endgroup$– ChrisDec 31, 2014 at 21:31
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By symmetry,
\begin{align} \int_0^1 x^3(1-x)^6\,\mathrm{d}x &= -\int_1^0x^6(1-x)^3\mathrm{d}x\\ &=-\int_0^1 (x-1)^3x^6\,\mathrm{d}x \\ &=-\int_0^1 x^9-3x^8+3x^7-x^6 \,\mathrm{d}x\\ &=-\left(\frac{1}{10}-\frac{1}{3}+\frac{3}{8}-\frac{1}{7}\right)+0\\\\ &=-\frac{84-280+315-120}{840}\\ &=\boxed{\displaystyle\frac{1}{840}} \end{align}
answered Dec 31, 2014 at 21:10
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$\begingroup$ Check your signs; you're integrating something positive. Also this is the solution I just posted... $\endgroup$ Dec 31, 2014 at 21:11
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1$\begingroup$ @pre-kidney Thanks. I didn't see yours when I posted mine (look at the differences in time - I don't think anyone can type that fast). Also it's the solution Kaj posted 15 minutes ago :P $\endgroup$ Dec 31, 2014 at 21:14
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Along the lines of Kaj, I think the fastest is to just expand: $$ \int x^3(1-x)^6 = \int (1-u)^3u^6=\int u^6-3u^7+3u^8-u^9=\frac{1}{7}-\frac{3}{8}+\frac{3}{9}-\frac{1}{10}. $$
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If we realize the integral is part of the Beta Distribution: $$\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},0<x<1$$ We find $\alpha=4,\beta=7$. We also know the integral of any pdf equals $1$ and thus this our coefficient must be the inverse of the result.
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using that $$x^3(1-x)^6=x^9-6 x^8+15 x^7-20 x^6+15 x^5-6 x^4+x^3$$ we can calculate the integral without the beta function, we get $$\int_{0}^{1} x^9-6 x^8+15 x^7-20 x^6+15 x^5-6 x^4+x^3dx=\frac{1}{840}$$
answered Dec 31, 2014 at 22:02