1. Find the area of triangle $PQR$

The vertices are $P(0,0)$, $Q(1,1)$ and $R(2,0)$.

Here, base $PR$ lies on the $x$-axis from $x=0$ to $x=2$, so $PR = 2.$ The height from $Q(1,1)$ to the $x$-axis is $1.$ Therefore, $\text{Area of } \triangle PQR = \frac{1}{2}\times 2 \times 1 = 1.$

2. Identify the region taken away by farmer $F_2$

The side $PQ$ joins $(0,0)$ and $(1,1)$, so its equation is $y=x, \quad 0\le x\le 1.$

The curve is $y=x^n, \quad n>1.$

For $0<x<1$ and $n>1$, we have $x^n < x.$ So the region between $y=x$ and $y=x^n$ from $x=0$ to $x=1$ is the part taken away.

Hence its area is $A = \int_0^1 (x - x^n)\,dx.$

3. Use the given percentage condition

The taken area is exactly $30\%$ of the triangle's area.

Since area of the triangle is $1$, $A = 0.3 = \frac{3}{10}.$

So, $\int_0^1 (x-x^n)\,dx = \frac{3}{10}.$

4. Evaluate the integral

$\int_0^1 x\,dx = \frac{1}{2}, \qquad \int_0^1 x^n\,dx = \frac{1}{n+1}.$

Thus, $\frac{1}{2} - \frac{1}{n+1} = \frac{3}{10}.$

5. Solve for $n$

$\frac{1}{2} - \frac{3}{10} = \frac{1}{n+1}$ $\frac{5-3}{10} = \frac{1}{n+1}$ $\frac{2}{10} = \frac{1}{n+1}$ $\frac{1}{5} = \frac{1}{n+1}$

Therefore, $n+1=5 \implies n=4.$

6. Compare with stored answer

Derived answer is $4$, which matches the stored correct answer.