How much multiplying will you do?

$\begin{aligned} I & = \int_0^1 x^{2016}(1-x)^{2017} dx \\ & = \color{#3D99F6}{B(2017,2018) \quad \quad \small B(m,n) \text{ is beta function.}} \\ & = \color{#3D99F6}{\frac {\Gamma(2017) \Gamma(2018)}{\Gamma(4035)} \quad \quad \small \Gamma(x) \text{ is gamma function.}} \\ & = \frac {2016!2017!}{4034!} \\ & = \frac 1{\frac{2017\cdot 4034!}{2017!2017!}} \\ & = \frac 1{2017{4034 \choose 2017}} = \frac ab \end{aligned}$

Both $a$ and $b$ are integers and $a=\boxed{1}$ .

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References

• Beta function
• Gamma function

The function $x^{2016}(1-x)^{2017}$ can be integrated by parts to remove the $x$ term like such:

$\displaystyle\int_{0}^{1}x^{2016}(1-x)^{2017} = \dfrac{-x^{2016}(1-x)^{2018}}{2018}-\dfrac{2016x^{2015}(1-x)^{2019}}{2018*2019}-...$

$-\dfrac{2016!(1-x)^{4033}}{\frac{4033!}{2017!}}$

The only term we care about is the last term since when plugging in the limits into $x$ , it is the only term that does not go to zero.

Plugging the limits in results in $\displaystyle\int_{0}^{1}x^{2016}(1-x)^{2017} = \dfrac{2016!*2017!}{4033!} = \dfrac{1}{\binom{4033}{2016}}$ a binomial coefficient, as pointed out to me by @Kyle Coughlin .

The binomial coefficient returns an integer, $b$ , and so $\Large\boxed{a = 1}$