Calculate Exponent in Curve Bounding Fractional Area of Triangular Region

To find the value of $n$, we proceed step-by-step by calculating the total area of $\Delta PQR$ and the area of the region taken away by farmer $F_2$.

Step 1: Area of $\Delta PQR$

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

• The base of the triangle lies along the $x$-axis from $x = 0$ to $x = 2$, so the length of the base is: $b = 2 - 0 = 2$
• The height of the triangle is the $y$-coordinate of the vertex $Q(1,1)$, which is: $h = 1$

Thus, the total area of $\Delta PQR$ is: $\text{Area}(\Delta PQR) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$

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Step 2: Area of the Region Taken Away

The region taken away by farmer $F_2$ is bounded between the line segment $PQ$ and the curve $y = x^n$ for $n > 1$.

• The equation of the line segment $PQ$ connecting $P(0,0)$ and $Q(1,1)$ is given by: $y = x \quad \text{for } x \in [0, 1]$

• For $n > 1$ and $x \in (0, 1)$, we have $x > x^n$. Therefore, the line $y = x$ lies above the curve $y = x^n$ in the interval $x \in [0, 1]$.

The area of this bounded region is: $\text{Area}_{\text{taken}} = \int_{0}^{1} \left( x - x^n \right) dx$

Evaluating the integral: $\text{Area}_{\text{taken}} = \left[ \frac{x^2}{2} - \frac{x^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{n+1}$

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Step 3: Solve for $n$

We are given that the area taken away is equal to $30\%$ of the total area of $\Delta PQR$: $\text{Area}_{\text{taken}} = 30\% \times \text{Area}(\Delta PQR) = \frac{30}{100} \times 1 = \frac{3}{10}$

Equating the two expressions for $\text{Area}_{\text{taken}}$: $\frac{1}{2} - \frac{1}{n+1} = \frac{3}{10}$

Rearranging the equation: $\frac{1}{n+1} = \frac{1}{2} - \frac{3}{10} = \frac{5 - 3}{10} = \frac{2}{10} = \frac{1}{5}$

Taking reciprocals on both sides: $n + 1 = 5 \implies n = 4$

Since $n = 4 > 1$, it satisfies the given condition $n > 1$.

Final Answer: 4