I have got to solve this definite integral, though I have no idea in which direction to go. Another exercise I had is to solve something similar that looks like: $$\int_0^1((1-x)^{11}x^2)dx$$ so for this integral I could substitute $t=1-x$ and I get a much easier version of the integral. The same method I couldn't implement on the given integral $\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx$ as it just doesn't work. The only thing I could notice is: $$\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx=\int_0^1((1-x)^8x^8(x^3-(1-x)^3))dx$$ yet I'm stuck at here and I have no clue how to solve the integral. (sure there is the simplest solution just to open up the polynomial - but it's naive and foolish...)


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    $\begingroup$ Starting from the given expression, let $t=1-x$. Do the substitution and look. $\endgroup$ – André Nicolas Jun 6 '16 at 17:51
  • $\begingroup$ Try plotting the function in, say, WolframAlpha. What do you notice? $\endgroup$ – Semiclassical Jun 6 '16 at 17:52
  • $\begingroup$ @AndréNicolas for $\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx$??? $\endgroup$ – Ami Gold Jun 6 '16 at 17:52
  • $\begingroup$ Yes, very nice stuff will happen. (Maybe after you do the substitution, change the dummy variable of integration to $x$.) $\endgroup$ – André Nicolas Jun 6 '16 at 17:52
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    $\begingroup$ As to odd functions, that would be another way, let $y=x-1/2$. works out well, a little more typing because of the fractions. $\endgroup$ – André Nicolas Jun 6 '16 at 18:05


We recognize that the integral of interest is simply $B(12,9)-B(9,12)$, where $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$

is the Beta Function. Then, exploiting the property of Beta Function, $B(x,y)=B(y,x)$, we immediately find that the result is $0$.

I thought it would be instructive to present a way forward for those unfamiliar with the Beta Function. To that end, we proceed.

The result of the integral of interest can be generalized as follows. Let $I(x,y)$ be the integral given by

$$I(x,y)=\int_0^1 t^x (1-t)^y\,dt$$

Now, enforcing the substitution $t \to 1-t$ we find that

$$\begin{align} I(x,y)&=\int_0^1 t^x \,(1-t)^y\,dt\\\\ &=\int_1^0 (1-t)^x\,t^y\,(-1)\,dt\\\\ &=\int_0^1 (1-t)^x\,t^y\,dt\\\\ &=I(y,x)\\\\ \end{align}$$

Therefore, $I(x,y)=I(y,x)$ or $I(x,y)-I(y,x)=0$.


The symmetry of the expressions suggests shifting to center the integral around $0$. Substituting $u=x-1/2$, you have


The integral is of an odd function over $[-a,a]$, and so its value is $0$.


Since people are giving alternate approaches to the $t = 1 - x$ substitution, here's another one:

Integrate by parts in ${\displaystyle \int_0^1 (1 - x)^8x^{11}\,dx}$, integrating the $(1 - x)^8$ and differentiating the $x^{11}$. The endpoint terms are zero, so we get $$\int_0^1 (1 - x)^8x^{11}\,dx = {11 \over 9} \int_0^1 (1 - x)^9x^{10}\,dx$$ Doing the same thing two more times yields $$\int_0^1 (1 - x)^8x^{11}\,dx = {11 *10 * 9 \over 9 * 10 * 11} \int_0^1 (1 - x)^{11}x^8\,dx$$ This can be rearranged as $$\int_0^1 \big((1 - x)^8x^{11} - (1 - x)^{11}x^8\big)\,dx = 0$$


There exist more than one approach to your problem, this answer uses a more "brute force"-esque approach. Seeing different approaches can possibly help one have a greater understanding of what's going on here. $$\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx$$ Going to calculate the bounds after treating this as a definite integral. $$=\int{2x^{19}}dx-\int{19x^{18}}dx+\int{83x^{17}}dx-\int{221x^{16}}dx+\int{400x^{15}dx-\int{518x^{14}}dx+\int{490x^{13}}dx-\int{338x^{12}}dx+\int{166x^{11}}dx-\int{55x^{10}}dx+\int{11x^{9}}dx-\int{x^{8}}dx}$$ $$=\dfrac{x^{20}}{10}-x^{19}+\dfrac{83x^{18}}{18}-13x^{17}+25x^{16}-\dfrac{518x^{15}}{15}+35x^{14}-26x^{13}+\dfrac{83x^{12}}{6}-5x^{11}+\dfrac{11x^{10}}{10}-\dfrac{x^{9}}{9}+C$$ Now we calculate the bounds. $$\left(\dfrac{1^{20}}{10}-1^{19}+\dfrac{83(1)^{18}}{18}-13(1)^{17}+25(1)^{16}-\dfrac{518(1)^{15}}{15}+35(1)^{14}-26(1)^{13}+\dfrac{83(1)^{12}}{6}-5(1)^{11}+\dfrac{11(1)^{10}}{10}-\dfrac{1^{9}}{9}\right)-\left(\dfrac{0^{20}}{10}-0^{19}+\dfrac{83(0)^{18}}{18}-13(0)^{17}+25(0)^{16}-\dfrac{518(0)^{15}}{15}+35(0)^{14}-26(0)^{13}+\dfrac{83(0)^{12}}{6}-5(0)^{11}+\dfrac{11(0)^{10}}{10}-\dfrac{0^{9}}{9}\right)$$ $$=\left(\dfrac{1}{10}-1+\dfrac{83}{18}-13+25-\dfrac{518}{15}+35-261+\dfrac{83}{6}-5+\dfrac{11}{10}-\dfrac{1}{9}\right)-0$$ $$=0$$

  • $\begingroup$ That is the "brute force" approach. Exploiting the symmetry of the Beta function provided a one line answer. ;-)) $\endgroup$ – Mark Viola Jun 6 '16 at 18:22
  • $\begingroup$ Yes, showing the OP that there are multiple approaches to this. :) $\endgroup$ – Colbi Jun 6 '16 at 18:26
  • $\begingroup$ That is a good thing to do. You might consider stating that sentiment up front and state "We can proceed using a "brute force" approach," or "I thought it might be instructive to present a "brute force" approach." -Mark $\endgroup$ – Mark Viola Jun 6 '16 at 18:32
  • $\begingroup$ @Colbi Sir, I have honor for you for your bravery! $\endgroup$ – Ami Gold Jun 6 '16 at 18:51
  • $\begingroup$ @Ami Thanks, and it's no problem :) $\endgroup$ – Colbi Jun 6 '16 at 18:57

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