The value of $$2m_{1} + 3n_{1} + m_{1}n_{1}$$ is ___ .

We first analyse the function $$f_1$$(x)$$ = \displaystyle \int_{0}^{x}\prod_{j=1}^{11}(t-j)\,dt\,,\;x \gt 0$$.

Denote $$P(x)=\displaystyle\prod_{j=1}^{11}(x-j)$$. Because the integrand of $$f_1$$ is $$P(t)$$, the first derivative is simply

$$f_1'(x)=P(x)=\prod_{j=1}^{11}(x-j)\,.$$

The critical points of $$f_1$$ are therefore the real zeros of $$P(x)$$, namely

$$x=1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9,\,10,\,11.\tag{A}$$

To decide whether each point in (A) is a point of local minimum or maximum, we study the sign of $$P(x)$$ in successive intervals. Note that $$P(x)$$ is a product of 11 linear factors with leading coefficient $$+1$$, so its sign alternates every time we cross a simple root.

• For $$0\lt x\lt1$$ all eleven factors $$(x-j)$$ are negative, hence $$P(x)\lt0$$.
• Crossing $$x=1$$ changes the sign once (odd multiplicity), so for $$1\lt x\lt2$$ we have $$P(x)\gt0$$.
• Crossing $$x=2$$ changes the sign again, so for $$2\lt x\lt3$$, $$P(x)\lt0$$, and so on.

Thus the sign pattern of $$f_1'(x)=P(x)$$ is

$$(-)\;(0)\;(+) \;(0)\;(-)\;(0)\;(+)\;(0)\;(-)\;(0)\;(+)\;(0)\;(-)\,,$$

corresponding to the successive open intervals
$$(0,1),\;1,\;(1,2),\;2,\;(2,3),\;3,\ldots,(10,11),\;11,(11,\infty).$$

Whenever $$f_1'(x)$$ changes from negative to positive we get a local minimum; when it changes from positive to negative we get a local maximum. Reading the pattern, we find

• Local minima at the odd integers: $$1,3,5,7,9,11$$.
• Local maxima at the even integers: $$2,4,6,8,10$$.

Hence

$$m_{1}=6,\qquad n_{1}=5.$$

Finally, the required expression is

$$2m_{1}+3n_{1}+m_{1}n_{1}=2(6)+3(5)+6\cdot5=12+15+30=57.$$

The value therefore lies in the interval $$56.90-57.10$$ requested in the answer range.