Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the centroid of the region bounded by \(y=e^{-x}, x=0, x=2\), and \(y=0\).

Step-by-Step Solution

Answer

The centroid is \(\left( \frac{1 - 2e^{-2}}{1-e^{-2}}, \frac{1 - e^{-4}}{4(1-e^{-2})} \right)\).

Step 1: Identify the Bounded Region

The region is bounded by the curve defined by the equation \(y = e^{-x}\), the vertical lines at \(x = 0\) and \(x = 2\), and the horizontal line \(y = 0\). This region is a section under the exponential decay curve from \(x = 0\) to \(x = 2\).

Step 2: Find the Area of the Region

The area \(A\) of the region can be found by integrating the function \(y = e^{-x}\) from \(x = 0\) to \(x = 2\). \[ A = \int_{0}^{2} e^{-x} \, dx \]Perform the integration:\[ A = \left[-e^{-x}\right]_{0}^{2} = -(e^{-2} - e^{0}) = 1 - e^{-2} \]

Step 3: Calculate the x-coordinate of the Centroid

The x-coordinate of the centroid \( \bar{x} \) is given by the formula:\[ \bar{x} = \frac{1}{A} \int_{0}^{2} x e^{-x} \, dx \]First, use integration by parts:Let \(u = x\) and \(dv = e^{-x} \, dx\). Then, \(du = dx\) and \(v = -e^{-x}\).\[ \bar{x} = \frac{1}{1-e^{-2}} \left([ -xe^{-x} ]_{0}^{2} + \int_{0}^{2} e^{-x} \, dx \right) \]Evaluate the terms:\[ = \frac{1}{1-e^{-2}} \left( ( -2e^{-2} + 0 ) + (1 - e^{-2}) \right) \]\[ = \frac{1}{1-e^{-2}} ( 1 - 2e^{-2} ) \]

Step 4: Calculate the y-coordinate of the Centroid

The y-coordinate of the centroid \( \bar{y} \) is given by the formula:\[ \bar{y} = \frac{1}{A} \int_{0}^{2} \frac{1}{2} (e^{-x})^2 \, dx \]Simplify and compute:\[ \bar{y} = \frac{1}{1 - e^{-2}} \times \frac{1}{2} \int_{0}^{2} e^{-2x} \, dx \]This evaluates to:\[ \bar{y} = \frac{1}{2(1 - e^{-2})} \left[-\frac{1}{2}e^{-2x}\right]_{0}^{2} \]\[ = \frac{1}{4(1 - e^{-2})} (1 - e^{-4}) \]

Step 5: Derive the Centroid Coordinates

Calculate the actual values using the results from the previous steps.\[ \bar{x} = \frac{1 - 2e^{-2}}{1-e^{-2}} \]\[ \bar{y} = \frac{1 - e^{-4}}{4(1-e^{-2})} \]Thus, the coordinates of the centroid are \( (\bar{x}, \bar{y}) \).

Key Concepts

Exponential Functions

Exponential functions involve expressions where a constant base is raised to a variable exponent. They are commonly found in equations that involve growth or decay. For instance, the equation \( y = e^{-x} \) represents an exponential decay function. Here, \( e \) is the base of natural logarithms, and the negative exponent \( -x \) indicates a decreasing function.
Exponential functions have several typical features:

• The graph tends to curve downward if the exponent is negative, as in decay scenarios.
• The function produces a continuous and smooth curve.
• As \( x \) increases, the value of \( y \) approaches zero but never actually reaches it.

These characteristics make exponential functions useful for modeling natural processes like radioactive decay or determining the region under a curve. In solving centroids or other calculus problems involving bounded areas, understanding how exponential functions behave is crucial.

Integration by Parts

Integration by parts is a vital technique for solving integrals, especially when the standard methods don't work easily. It comes from the product rule for differentiation and is very useful when dealing with products of functions.
The formula for integration by parts is:\[\int u \cdot dv = uv - \int v \cdot du\]Here's how you can apply it:

• Identify parts of the integral as \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
• Apply the formula and simplify the expression.

In the given centroid problem, for instance, \( u = x \) and \( dv = e^{-x} dx \). This transforms the difficult integral into a more approachable form. By means of integration by parts, complex exponential integrals often become solvable.

Definite Integrals

Definite integrals are central to calculating areas under curves, which is necessary for finding centroids and other geometric characteristics. Unlike indefinite integrals, which give a function, definite integrals provide a numerical value representing the area between a curve and the x-axis, bounded by two vertical lines.
For definite integrals:

• Identify the limits of integration, which are the bounds where you evaluate the function—these are \( x = 0 \) and \( x = 2 \) in this problem.
• Perform the integration with respect to these limits.
• Subtract the value of the evaluated integral at the lower limit from the value at the upper limit.

Definite integrals like \( \int_{0}^{2} e^{-x} \, dx \) provide essential quantitative metrics, such as area, that feed calculations for more complex geometric properties like centroids.

Centroid Calculation

Calculating the centroid of a bounded region on a coordinate plane involves determining its "center of mass." For a region bound by curves and lines, this requires computations involving both the area and the shape of that region.
Here is the process for centroid calculation:

• Identify the area of the region using a definite integral of the bounding function.
• The x-coordinate of the centroid \( \bar{x} \) is found using \( \frac{1}{A} \int x \cdot f(x) \, dx \).
• Obtain the y-coordinate of the centroid \( \bar{y} \) via \( \frac{1}{A} \int \frac{1}{2} (f(x))^2 \, dx \).
• Simplify these equations to get the numerical values of \( \bar{x} \) and \( \bar{y} \).

So, for the original exercise, using the bounded region calculations, the centroid's coordinates \( (\bar{x}, \bar{y}) \) are calculated by involving integrals of the exponential curve \( e^{-x} \). This enables a precise placement of the centroid in the region, indicating the balance point of the area.