https://r2.erweima.ai/mathbot/image/17253103729341ju0ncip.jpg

The image you provided shows the following equations:

\[
y = e^x
\]
\[
y = 2
\]

These represent two curves: the first is an exponential function \( y = e^x \), and the second is a horizontal line \( y = 2 \). To calculate the centroid of the region bounded by these curves, we can follow these steps:

### 1. **Find the Intersection Points:**

First, let's find the points where the curves intersect. This occurs when \( e^x = 2 \).

\[
e^x = 2 \implies x = \ln(2)
\]

So, the region of interest is bounded between \( x = 0 \) and \( x = \ln(2) \).

### 2. **Area of the Region:**

The area \( A \) of the region between the curves can be found by integrating the difference between the functions over the interval from \( x = 0 \) to \( x = \ln(2) \):

\[
A = \int_0^{\ln(2)} (2 - e^x) \, dx
\]

### 3. **Calculate the Centroid ( \( \bar{x}, \bar{y} \) ):**

The coordinates of the centroid \((\bar{x}, \bar{y})\) can be found using the following formulas:

\[
\bar{x} = \frac{1}{A} \int_{0}^{\ln(2)} x(2 - e^x) \, dx
\]

\[
\bar{y} = \frac{1}{A} \int_{0}^{\ln(2)} \frac{2 + e^x}{2} \cdot (2 - e^x) \, dx
\]

### 4. **Integrate and Simplify:**

Let's calculate these integrals and then the centroid coordinates. I'll do this step by step.

#### 4.1. **Area \( A \) Calculation:**

\[
A = \int_0^{\ln(2)} (2 - e^x) \, dx = \left[ 2x - e^x \right]_0^{\ln(2)}
\]

\[
A = \left( 2\ln(2) - e^{\ln(2)} \right) - \left( 2(0) - e^0 \right)
\]

\[
A = 2\ln(2) - 2 + 1 = 2\ln(2) - 1
\]

#### 4.2. **\( \bar{x} \) Calculation:**

\[
\bar{x} = \frac{1}{2\ln(2) - 1} \int_{0}^{\ln(2)} x(2 - e^x) \, dx
\]

The integral will be solved as follows:

\[
\bar{x} = \frac{1}{2\ln(2) - 1} \left[ \ln(2) \cdot \left( 2\ln(2) - 1 \right) - \left( \frac{x e^x}{2} - \frac{e^x}{4} \right) \right]_0^{\ln(2)}
\]

This will give the \( \bar{x} \) coordinate.

#### 4.3. **\( \bar{y} \) Calculation:**

\[
\bar{y} = \frac{1}{2\ln(2) - 1} \int_{0}^{\ln(2)} \frac{2 + e^x}{2} \cdot (2 - e^x) \, dx
\]

This involves the integration:

\[
\bar{y} = \frac{1}{2\ln(2) - 1} \left[ \frac{2x + e^x}{2} \cdot \left( 2x - e^x \right) \right]_0^{\ln(2)}
\]

This will give the \( \bar{y} \) coordinate.

### Final Calculations

I will now calculate these to determine the exact coordinates of the centroid.

Let's proceed.The calculations yield the following results:

- **Area \( A \):** \( A = 2\ln(2) - 1 \)
- **Centroid \( \bar{x} \):** \( \bar{x} = \frac{-2\ln(2) + (\ln(2))^2 + 1}{2\ln(2) - 1} \)
- **Centroid \( \bar{y} \):** \( \bar{y} = \frac{-\frac{3}{4} + 2\ln(2)}{2\ln(2) - 1} \)

These expressions provide the exact coordinates of the centroid for the region bounded by the curves \( y = e^x \) and \( y = 2 \) between \( x = 0 \) and \( x = \ln(2) \).

Would you like a detailed step-by-step explanation, or do you have any specific questions?

Here are some related questions you might find interesting:

1. How would the centroid change if the upper curve was \( y = 3 \) instead of \( y = 2 \)?
2. What are the physical interpretations of the centroid in various contexts?
3. How would you find the centroid if the region was bounded by different curves?
4. Can you generalize the procedure to find the centroid of a region bounded by any two functions?
5. How would you graphically represent the centroid on the original plot?

**Tip:** The area calculation is essential in determining the centroid, as it normalizes the moment integrals to give the average position of the area distribution.

Learn how to calculate the centroid for the region bounded by the curves y = e^x and y = 2 using integral calculus. This detailed guide includes step-by-step calculations of the area, centroid coordinates, and integrals involved. Suitable for university-level mathematics students.

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