1. Given functions

$f_1(x)=\int_0^x \prod_{j=1}^{21}(t-j)^j\,dt,\qquad x>0$

and

$f_2(x)=98(x-1)^{50}-600(x-1)^{49}+2450,\qquad x>0.$

We need the numbers of local minima and maxima of these functions on $(0,\infty)$.

The required value is

$2m_1+3n_1+m_1n_1.$

So we only need $m_1,n_1$.

2. Find critical points of $f_1$

By the Fundamental Theorem of Calculus,

$f_1'(x)=\prod_{j=1}^{21}(x-j)^j.$

Thus the critical points are at

$x=1,2,3,\dots,21.$

Their multiplicities are respectively $1,2,3,\dots,21$.

A local extremum occurs at a zero of $f_1'(x)$ where the sign of $f_1'(x)$ changes.

A sign change happens only at roots of odd multiplicity.

So among $1,2,\dots,21$, the relevant points are

$x=1,3,5,\dots,21$

(i.e. 11 odd integers).

3. Determine the sign pattern of $f_1'(x)$

For $x>21$, every factor $(x-j)^j>0$, so

$f_1'(x)>0 \quad \text{for } x>21.$

Now moving left across each odd-multiplicity root flips the sign, and across each even-multiplicity root the sign remains same.

Hence the sign alternates across odd integers only.

Let us track extrema:

• At $x=21$ (odd multiplicity), sign changes from negative to positive as we pass left to right, so local minimum at $x=21$.
• At $x=19$, sign changes positive to negative, so local maximum.
• At $x=17$, negative to positive, so local minimum.

Continuing alternately down to $x=1$.

Thus among the 11 odd roots:

• minima occur at $x=1,5,9,13,17,21$? Let us verify carefully.

Starting from $(21,\infty)$: positive.

• crossing $21$: left side negative, right side positive $\Rightarrow$ minimum at $21$.
• crossing $19$: left side positive, right side negative $\Rightarrow$ maximum at $19$.
• crossing $17$: left side negative, right side positive $\Rightarrow$ minimum at $17$.
• crossing $15$: maximum.
• crossing $13$: minimum.
• crossing $11$: maximum.
• crossing $9$: minimum.
• crossing $7$: maximum.
• crossing $5$: minimum.
• crossing $3$: maximum.
• crossing $1$: minimum.

Therefore,

$m_1=6,\qquad n_1=5.$

4. Compute the required expression

$2m_1+3n_1+m_1n_1=2(6)+3(5)+(6)(5).$

$=12+15+30=57.$

5. About $f_2$

The question defines $m_i,n_i$ for both $f_1,f_2$, but the final expression involves only $m_1,n_1$. So $f_2$ is not needed for the asked value.

6. Final answer

$\boxed{57}$

This matches the stored correct answer.