Evaluation of Local Extrema Expression for First Given Integral Function

To find the value of $2m_1 + 3n_1 + m_1 n_1$, we first need to determine the number of points of local minima ($m_1$) and local maxima ($n_1$) for the function $f_1(x)$ in the interval $(0, \infty)$.

The function $f_1(x)$ is given by: $f_1(x) = \int_0^x \prod_{j=1}^{21} (t - j)^j dt, \quad x > 0$

Using the Newton-Leibniz formula, we differentiate $f_1(x)$ with respect to $x$: $f_1'(x) = \prod_{j=1}^{21} (x - j)^j = (x - 1)^1 (x - 2)^2 (x - 3)^3 \cdots (x - 21)^{21}$

The critical points of $f_1(x)$ in $(0, \infty)$ are $x = 1, 2, 3, \dots, 21$.

To determine the local extrema at these critical points, we examine the sign change of $f_1'(x)$ around each point:

1. Even powers ($j = 2, 4, 6, \dots, 20$): For even $j$, the factor $(x - j)^j \ge 0$ for all $x$. Therefore, $f_1'(x)$ does not change sign as $x$ crosses an even integer $j$. Hence, these 10 points are neither local minima nor local maxima.

2. Odd powers ($j = 1, 3, 5, \dots, 21$): For odd $j$, the factor $(x - j)^j$ changes sign from negative to positive as $x$ increases through $j$. Thus, $f_1'(x)$ changes sign at every odd critical point.

Now, let's analyze the sign of $f_1'(x)$ across the intervals defined by the odd critical points:

• For $x > 21$, all factors $(x - j)^j > 0$, so $f_1'(x) > 0$.
• Moving from right to left through the odd points:
  • At $x = 21$: $f_1'(x)$ changes sign from $-$ to $+$ as $x$ increases through 21 $\implies$ Local Minimum.
  • At $x = 19$: $f_1'(x)$ changes sign from $+$ to $-$ as $x$ increases through 19 $\implies$ Local Maximum.
  • At $x = 17$: Local Minimum
  • At $x = 15$: Local Maximum
  • At $x = 13$: Local Minimum
  • At $x = 11$: Local Maximum
  • At $x = 9$: Local Minimum
  • At $x = 7$: Local Maximum
  • At $x = 5$: Local Minimum
  • At $x = 3$: Local Maximum
  • At $x = 1$: Local Minimum

Grouping the extrema:

• Points of local minima: $x \in \{1, 5, 9, 13, 17, 21\}$ $\implies m_1 = 6$

• Points of local maxima: $x \in \{3, 7, 11, 15, 19\}$ $\implies n_1 = 5$

Now, we evaluate the required expression: $2m_1 + 3n_1 + m_1 n_1 = 2(6) + 3(5) + (6)(5)$ $= 12 + 15 + 30 = 57$