A farmer F1 has a land in the shape of a triangle with vertices at P(0, 0), Q(1, 1) and R(2, 0). Fro

Q: Wait, why is the area of the triangle equal to 1? I thought it was more complex.

It is a classic trap to overcomplicate! The triangle has vertices at $(0,0)$, $(1,1)$, and $(2,0)$. The base $PR$ lies on the x-axis from $x=0$ to $x=2$, so its length is $2$. The height is the y-coordinate of point $Q$, which is $1$. Using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, we get $\frac{1}{2} \times 2 \times 1 = 1$. It is that simple!

Q: Why is the integral $\int_{0}^{1} (x - x^n) dx$ and not $\int_{0}^{1} (x^n - x) dx$?

Great question! When calculating the area between two curves, we must subtract the lower curve from the upper curve to ensure the result is positive. For $x$ in the interval $(0, 1)$ and $n > 1$, $x^n$ is always less than $x$ (e.g., if $x=0.5$ and $n=2$, $0.5^2 = 0.25 < 0.5$). Thus, $y=x$ is the upper boundary and $y=x^n$ is the lower boundary.

Q: Is there any trick to the integration part?

Not at all! You just need the power rule for integration: $\int x^k dx = \frac{x^{k+1}}{k+1}$. Applying this to $x$ gives $\frac{x^2}{2}$ and to $x^n$ gives $\frac{x^{n+1}}{n+1}$. Just be careful with your limits of integration from $0$ to $1$.

Question

A farmer $F_1$ has a land in the shape of a triangle with vertices at $P(0, 0), Q(1, 1)$ and $R(2, 0)$. From this land, a neighbouring farmer $F_2$ takes away the region which lies between the side PQ and a curve of the form $y = x^n (n > 1)$. If the area of the region taken away by the farmer $F_2$ is exactly 30% of the area of $	riangle PQR$, then the value of $n$ is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Field

Imagine you are standing on the edge of this farmer's land. We have a triangular field defined by the vertices $P(0, 0)$, $Q(1, 1)$, and $R(2, 0)$.

The base of the triangle, $PR$, lies perfectly along the x-axis, stretching from $x=0$ to $x=2$, giving us a base length of $2$ units. The peak of the triangle is at $Q(1, 1)$, which means the height of our triangle is exactly $1$ unit.

Using the fundamental formula for the area of a triangle, $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$, we calculate:

$$	ext{Area} = \frac{1}{2} 	imes 2 	imes 1 = 1 	ext{ square unit}$$

This is our baseline.

### The Stolen Sliver

Now, enter the neighbor, farmer $F_2$. They are taking a region bounded by the line segment $PQ$ and a curve $y=x^n$.

First, let us define the line $PQ$. Since it connects $(0,0)$ and $(1,1)$, its equation is simply $y=x$.

The curve $y=x^n$ (where $n>1$) is a classic power function. Because $n>1$, for any $x$ between $0$ and $1$, $x^n$ will always be smaller than $x$. This means the curve bows downwards, creating a sliver of land trapped between the straight line $y=x$ and the curve $y=x^n$.

### The Calculus of Encroachment

To find the area of this stolen region, we turn to the power of definite integrals. We are looking for the area between two functions, which is defined as the integral of the upper function minus the lower function over the interval of interest.

Here, our interval is from $x=0$ to $x=1$. So, the area taken by farmer $F_2$ is:

$$	ext{Area}_{F_2} = \int_{0}^{1} (x - x^n) dx$$

Do not let the variable $n$ intimidate you. We treat it just like any other constant. Applying the power rule for integration, the integral of $x$ is $\frac{x^2}{2}$, and the integral of $x^n$ is $\frac{x^{n+1}}{n+1}$.

Evaluating this from $0$ to $1$:

$$\left[ \frac{x^2}{2} - \frac{x^{n+1}}{n+1} ight]_{0}^{1}$$

Substituting the upper limit $x=1$, we get $\frac{1}{2} - \frac{1}{n+1}$. Substituting the lower limit $x=0$ yields $0$, so our expression for the stolen area is simply $\frac{1}{2} - \frac{1}{n+1}$.

### The Final Solve

The problem gives us a crucial piece of information: this stolen area is exactly $30\%$ of the total area of $	riangle PQR$. Since the total area is $1$, the stolen area must be $0.3$.

Now, we set up our final algebraic equation:

$$\frac{1}{2} - \frac{1}{n+1} = 0.3$$

To solve for $n$, we isolate the term with $n$. Subtracting $0.3$ from $0.5$ (which is $\frac{1}{2}$), we get:

$$\frac{1}{n+1} = 0.5 - 0.3 = 0.2$$

Since $0.2$ is the same as $\frac{1}{5}$, we have:

$$\frac{1}{n+1} = \frac{1}{5}$$

By comparing the denominators, it is clear that $n+1 = 5$, which leads us directly to **$n=4$**. Through the elegance of calculus and a bit of algebraic persistence, we have determined the exact nature of the curve that defined the neighbor's encroachment.

Enter Numerical Value:

Visualized Solution

Plotting the Land

Area of $△ P QR$

The Boundaries: Line and Curve

The Stolen Region

Setting up the Integral

Integrating the Terms

Applying the Limits

The 30\% Condition

Substituting Known Values

Rearranging the Equation

Solving for $n$

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry of the Field

The Stolen Sliver

The Calculus of Encroachment

The Final Solve

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