Question

Let \(n \geq 1\) be constant, and consider the region bounded by \(f(x)=x^{n},\) the \(x\) -axis, and \(x=1\). Find the centroid of this region. As \(n \rightarrow \infty\), what does the region look like, and where is its centroid?

Short answer

The centroid of the region bounded by the curve \(f(x) = x^n\), the x-axis, and \(x = 1\) is \((\frac{n+1}{n+2},\frac{n+1}{2n+2})\). As \(n \rightarrow \infty\), the centroid is \((1,0.5)\), resembling a vertical line at \(x = 1\).

Step-by-step solution

Calculating the Area under f(x)

The area 'A' under the curve \( f(x) = x^n \) from 0 to 1 is given by: \( A = \int_{0}^{1} x^n dx \), computing this we get \( A = \frac{1}{n+1} \).

Calculating the centroid coordinates

Next, calculate the x-coordinate of the centroid \(\bar{x}\) using the formula: \(\bar{x}= \frac{1}{A} \int_{0}^{1} x. f(x) dx \) which calculates to \(\bar{x}= \frac{1}{\frac{1}{n+1}} \int_{0}^{1} x^{n+1} dx = \frac{n+1}{n+2} \). Similarly, calculate the y-coordinate of the centroid \(\bar{y}\) using the formula: \( \bar{y}= \frac{1}{2A} \int_{0}^{1} (f(x))^2 dx \) which calculates to \(\bar{y} = \frac{1}{2\frac{1}{n+1}} \int_{0}^{1} x^{2n} dx = \frac{n+1}{2n+2} \).

Describing the region and its centroid as \( n \rightarrow \infty \)

As \( n \rightarrow \infty \), the graph of the function \( x^n \) approaches the x-axis for \(0<x<1\), and equals 1 at \(x=1\), resembling a vertical line. The centroid \(\bar{x}\) approaches 1, and \(\bar{y}\) approaches 0.5.