Prove that the area bounded by the curve y = kx^n

Prove that the area bounded by the curve $y = kx^n$, the line $x = a$, and $y = 0$ equals $\frac{1}{n+1}$ times the area of the rectangle constructed over the coordinates of the point on the curve $y = kx^n$ with abscissa $a$.

We will prove the general relationship and verify it with specific examples.

Step 1: Identify the point on the curve

For the curve $y = kx^n$ at $x = a$:

\[y = ka^n\]

So the point on the curve is $(a, ka^n)$.

Step 2: Find the area of the rectangle

The rectangle has:

• Width: $a$ (from $x = 0$ to $x = a$)
• Height: $ka^n$ (from $y = 0$ to $y = ka^n$)

Area of rectangle:

\[A_{\text{rect}} = a \cdot ka^n = ka^{n+1}\]

Step 3: Find the area under the curve

The area bounded by $y = kx^n$, $x = a$, and $y = 0$ is:

\[A_{\text{curve}} = \int_0^a kx^n \, dx\]

Evaluating:

\[A_{\text{curve}} = k\int_0^a x^n \, dx = k\left[\frac{x^{n+1}}{n+1}\right]_0^a\]

\[= k \cdot \frac{a^{n+1}}{n+1} - 0 = \frac{ka^{n+1}}{n+1}\]

Step 4: Calculate the ratio

\[\frac{A_{\text{curve}}}{A_{\text{rect}}} = \frac{\frac{ka^{n+1}}{n+1}}{ka^{n+1}} = \frac{ka^{n+1}}{n+1} \cdot \frac{1}{ka^{n+1}} = \frac{1}{n+1}\]

Therefore:

\[A_{\text{curve}} = \frac{1}{n+1} \cdot A_{\text{rect}}\]

\[\boxed{A_{\text{curve}} = \frac{1}{n+1} \cdot A_{\text{rect}}}\]

Step 5: Verification with specific examples

Example 1: $n = 1$ (linear function)

Curve: $y = kx$. Point at $x = a$: $(a, ka)$. Rectangle area: $a \cdot ka = ka^2$.

Area under curve: $\int_0^a kx \, dx = k\left[\frac{x^2}{2}\right]_0^a = \frac{ka^2}{2}$

Ratio: $\frac{ka^2/2}{ka^2} = \frac{1}{2} = \frac{1}{1+1}$ ✓

Example 2: $n = 2$ (parabola)

Curve: $y = kx^2$. Point at $x = a$: $(a, ka^2)$. Rectangle area: $a \cdot ka^2 = ka^3$.

Area under curve: $\int_0^a kx^2 \, dx = k\left[\frac{x^3}{3}\right]_0^a = \frac{ka^3}{3}$

Ratio: $\frac{ka^3/3}{ka^3} = \frac{1}{3} = \frac{1}{2+1}$ ✓

Example 3: $n = 3$ (cubic)

Curve: $y = kx^3$. Point at $x = a$: $(a, ka^3)$. Rectangle area: $a \cdot ka^3 = ka^4$.

Area under curve: $\int_0^a kx^3 \, dx = k\left[\frac{x^4}{4}\right]_0^a = \frac{ka^4}{4}$

Ratio: $\frac{ka^4/4}{ka^4} = \frac{1}{4} = \frac{1}{3+1}$ ✓

Geometric interpretation

This result shows that:

• For $n = 1$: The triangle under the line has area $\frac{1}{2}$ of the rectangle (well-known)
• For $n = 2$: The area under the parabola is $\frac{1}{3}$ of the rectangle
• For $n = 3$: The area under the cubic is $\frac{1}{4}$ of the rectangle
• In general: As $n$ increases, the fraction decreases, meaning the curve "bends away" more from the rectangle

Summary

For the power function $y = kx^n$ where $k > 0$ and $n \geq 0$, the area under the curve from $x = 0$ to $x = a$ equals $\frac{1}{n+1}$ times the area of the rectangle with vertices at $(0, 0)$, $(a, 0)$, $(a, ka^n)$, and $(0, ka^n)$.

Proof:

\[A_{\text{curve}} = \int_0^a kx^n \, dx = \frac{ka^{n+1}}{n+1}\]

\[A_{\text{rect}} = a \cdot ka^n = ka^{n+1}\]

\[\frac{A_{\text{curve}}}{A_{\text{rect}}} = \frac{ka^{n+1}/(n+1)}{ka^{n+1}} = \frac{1}{n+1}\]

\[\boxed{A_{\text{curve}} = \frac{1}{n+1} \cdot A_{\text{rect}}}\]

Key insights:

• This is a fundamental relationship for power functions
• The ratio $\frac{1}{n+1}$ depends only on the exponent $n$, not on $k$ or $a$
• As $n \to \infty$, the ratio approaches 0 (curve becomes more concentrated near $x = a$)
• This generalizes Archimedes' result for the parabola ($n = 2$, ratio = $\frac{1}{3}$)