Question

Find the centroid of the region bounded by \(y=e^{-x}, x=0, x=2\), and \(y=0 .\) Hint: Use the fact that \(\int x e^{-x} d x=-x e^{-x}-e^{-x}+C\)

Short answer

The centroid is approximately \( \left(0.42265, 0.1573\right) \).

Step-by-step solution

Identify the Bounding Curves and Region

The region is bounded by the curve \( y = e^{-x} \), the lines \( x = 0 \) and \( x = 2 \), and the line \( y = 0 \). This means we have a region bounded by an exponential decay function over the interval \( x \in [0,2] \).

Find the Area of the Region

The area under the curve from \( x = 0 \) to \( x = 2 \) is given by the integral \( A = \int_{0}^{2} e^{-x} \, dx \). Evaluate this using the antiderivative \( -e^{-x} \), yielding:\[A = \left[-e^{-x}\right]_{0}^{2} = -(e^{-2}) - (-(e^{0})) = 1 - \frac{1}{e^2}\].

Calculate the x-component of the Centroid

The x-coordinate of the centroid, \( \bar{x} \), is given by \( \bar{x} = \frac{1}{A} \int_{0}^{2} x e^{-x} \, dx \). Using the provided integral result:\[ \int_{0}^{2} x e^{-x} \, dx = \left[-xe^{-x} - e^{-x} \right]_{0}^{2} = \left[(-2e^{-2} - e^{-2}) - (0 - 1)\right] = -3e^{-2} + 1\]Thus, \( \bar{x} = \frac{-3e^{-2} + 1}{1 - \frac{1}{e^2}} \).

Calculate the y-component of the Centroid

The y-coordinate, \( \bar{y} \), is given by \( \bar{y} = \frac{1}{A} \int_{0}^{2} \frac{1}{2} (e^{-x})^2 \, dx \). First, find the integral:\[ \int_{0}^{2} \frac{1}{2} e^{-2x} \, dx = \frac{1}{2} \left[ -\frac{1}{2}e^{-2x} \right]_{0}^{2} = \frac{1}{2} \left( -\frac{1}{2}e^{-4} + \frac{1}{2} \right) = \frac{1}{4}(1 - e^{-4})\]Then, \( \bar{y} = \frac{\frac{1}{4}(1 - e^{-4})}{1 - \frac{1}{e^2}} \).

Compute the Exact Centroid Coordinates

Substitute the numerators and denominators from Steps 3 and 4. The exact centroid coordinates \( (\bar{x}, \bar{y}) \) are:\[\bar{x} = \frac{-3e^{-2} + 1}{1 - \frac{1}{e^2}}, \quad \bar{y} = \frac{\frac{1}{4}(1 - e^{-4})}{1 - \frac{1}{e^2}}\]

Key concepts

Integrals

Integrals are fundamental tools in calculus that allow us to find the area under curves. In this exercise, we use integrals to determine the area of the region under the exponential curve from \(x = 0\) to \(x = 2\). The integral of a function over an interval gives us the accumulated value, which, when dealing with functions like \(y = e^{-x}\), represents the area between the curve and the x-axis. Calculating this involves finding the antiderivative of the function and evaluating it at the specified bounds. This process helps us understand the total space taken up by this function along the x-axis within defined limits.

Exponential Function

Exponential functions, such as \(y = e^{-x}\), describe rapid growth or decay processes. In this problem, \(y = e^{-x}\) represents an exponential decay, creating a curve that starts at \(y = 1\) when \(x = 0\) and approaches \(y = 0\) as \(x\) increases. These functions have unique properties, where their derivatives are proportional to the value of the function itself. This makes them quite essential for modeling situations involving exponential changes, like decay dynamics. The region bounded by this function helps us examine how the area changes as it stretches along the x-axis up to a point like \(x=2\).

Area Under a Curve

Finding the area under a curve is a significant application of integrals. It involves calculating the space contained between a curve and the x-axis over a specific interval. In this scenario, to determine the area under the curve \(y = e^{-x}\) between \(x = 0\) and \(x = 2\), you evaluate the integral \(\int_{0}^{2} e^{-x} \, dx\). An antiderivative, such as \(-e^{-x}\), helps in performing this evaluation over the bounds. The calculated area provides not just the measure of space but serves as a crucial component when determining the coordinates of the centroid of a region.

Coordinates of a Centroid

The coordinates of a centroid represent the 'center of mass' of a region. For a 2D surface bound by curves, these coordinates \((\bar{x}, \bar{y})\) can be found by using specific integral formulas. The x-coordinate \(\bar{x}\) is calculated by weighting each segment of the area by its respective x-value, divided by the total area. The formula is \(\bar{x} = \frac{1}{A} \int_{0}^{2} x e^{-x} \, dx\). Similarly, \(\bar{y}\) evaluates the average vertical position using \(\bar{y} = \frac{1}{A} \int_{0}^{2} \frac{1}{2} (e^{-x})^2 \, dx\). These formulas integrate the idea of balancing the entire region over coordinate axes to pinpoint the centroid's exact location. Understanding these concepts is pivotal in many fields involving balance and center of gravity.