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Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates of the centroid. \(y=e^{x}, \quad y=0, \quad x=0, \quad x=1\)

Short Answer

Expert verified
Centroid coordinates: \(\left( \frac{1}{e-1} , \frac{e^2 - 1}{4(e-1)} \right)\).

Step by step solution

01

Understand the Problem

The task is to find the region bounded by the curves and find the centroid of that region. Specifically, we're looking at the bounded region under the curve \(y = e^x\) on the interval \([0, 1]\), bounded below by \(y = 0\) (the x-axis), and vertically by \(x = 0\) and \(x = 1\).
02

Sketch the Region

To sketch the region, draw the line \(x=0\), the line \(x=1\), the horizontal line \(y=0\), and the curve \(y=e^x\) between \(x=0\) and \(x=1\). The region of interest is the area beneath the curve \(y=e^x\) from \(x=0\) to \(x=1\).
03

Estimate the Location of Centroid

The centroid of a region is its balancing point. Visually, the centroid will be located below the midpoint \((x = 0.5)\) on the curve \(y=e^x\) because more area is concentrated nearer to \(x = 0\).
04

Calculate the Area of the Region

The area \(A\) of the region is found using the integral: \[ A = \int_{0}^{1} e^x \, dx \] Evaluating this integral gives: \[ A = \left[ e^x \right]_{0}^{1} = e^1 - e^0 = e - 1 \]
05

Find the x-coordinate of the Centroid

The x-coordinate \( \bar{x} \) of the centroid for a vertically simple region is given by: \[ \bar{x} = \frac{1}{A} \int_{0}^{1} x \cdot e^x \, dx \]Using integration by parts, the integral becomes: \[ \bar{x} = \frac{1}{e-1} \left[ xe^x \Big|_{0}^{1} - \int_{0}^{1} e^x \, dx \right] = \frac{1}{e-1} \left[ (1 \cdot e^1) - (0 \cdot e^0) - (e^1 - e^0) \right] = \frac{1}{e-1} \](Evaluating the integrals)\[ \frac{1}{e-1}(e - (e-1)) = \frac{1}{e-1} \]
06

Find the y-coordinate of the Centroid

The y-coordinate \( \bar{y} \) of the centroid is given by: \[ \bar{y} = \frac{1}{A} \cdot \frac{1}{2} \int_{0}^{1} (e^x)^2 \, dx \]This simplifies to: \[ \bar{y} = \frac{1}{2(e-1)} \int_{0}^{1} e^{2x} \, dx \]Evaluating the integral, we have: \[ \bar{y} = \frac{1}{2(e-1)} \cdot \left[ \frac{1}{2} e^{2x} \right]_{0}^{1} = \frac{1}{4(e-1)} (e^2 - 1) \] \[ \bar{y} = \frac{e^2 - 1}{4(e-1)} \]
07

Compile the Coordinates of the Centroid

The centroid's coordinates are \(\left( \frac{1}{e-1} , \frac{e^2 - 1}{4(e-1)} \right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centroid Calculation
Calculating the centroid of a region refers to finding its geometric center, which is essentially the mean position of all the points within the region of interest. In calculus, the centroid plays a vital role because it represents the balancing point of a shape or object. For bounded regions defined by curves, the process involves working with both x and y coordinates separately.
The x-coordinate of the centroid, denoted as \( \bar{x} \), is calculated using the formula:
  • \( \bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) \, dx \)
Here, \( A \) is the area under the curve, and \( f(x) \) represents the function defining the curve over the interval \([a, b]\).
Similarly, the y-coordinate, \( \bar{y} \), is determined by:
  • \( \bar{y} = \frac{1}{A} \cdot \frac{1}{2} \int_{a}^{b} [f(x)]^2 \, dx \)
This process captures the symmetry and weight distribution along the vertical axis.
Ultimately, finding the centroid is about understanding how the region beneath a curve is distributed. When done correctly, you end up with a point \((\bar{x}, \bar{y})\) that represents the centroid's exact location.
Integral Calculus
Integral calculus is a branch of calculus concerned with finding the integral, or the antiderivative, of functions. This process is the reverse of differentiation and captures the concept of accumulation, such as areas under curves.
In the context of centroid calculation, integral calculus helps compute the exact areas and coordinates involved. Formally, the process uses definite integrals:
  • The integral \( \int_{a}^{b} f(x) \, dx \) calculates the area under the curve \( y = f(x) \) from \( x = a \) to \( x = b \).
  • Using integration by parts simplifies more complex integrals, particularly when products of functions appear in our expressions.
Integration by parts is based on the formula:
  • \( \int u \, dv = uv - \int v \, du \)

Integral calculus is a powerful tool in analyzing and dealing with functions where direct calculation isn't practical. This includes physical applications like finding the centroids of irregularly bounded regions.
Bounded Regions
Bounded regions refer to areas confined within specific boundaries, often defined by a combination of curves and lines in the plane. These regions can take various forms depending on the functions and the given intervals.
Understanding the geometric boundaries is crucial in exercises involving sketching and calculating centroids:
  • For a given region, identify curves and lines defining it. Common examples include horizontal or vertical lines and curves represented by functions like \( y = e^x \).
  • Sketching the region involves drawing these curves and the lines to understand the shape visually.
  • The bounded region always lies beneath, between, or above these lines and curves based on the given constraints.
This exercise requires a good grasp of the intersection points and intercepts of the curves to ensure the region is correctly identified.
By visualizing the bounded region, students get a better idea of how the areas translate into more tangible shapes, paving the way to deeper insights, especially when estimating the centroid's location.

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