A factory is making a machine part whose shape is bounded by the curves f(x) = cosh(x), g(x) = x,
x = 0, and x = 3. Sketch these curves and shade the area of the part. Use integration to determine
the area of this part. If the part’s mass density is 7 grams/(unit area), what is the mass of the
part?

To solve this problem, let's break it down step-by-step:

### 1. **Understanding the Curves:**
   - We are given two functions:
     \[
     f(x) = \cosh(x) \quad \text{and} \quad g(x) = x
     \]
     as well as the vertical boundaries \( x = 0 \) and \( x = 3 \).
   
   The region of interest is the area bounded by these curves between \( x = 0 \) and \( x = 3 \).

### 2. **Sketching the Curves:**
   - The function \( \cosh(x) \) (the hyperbolic cosine) is symmetric and grows exponentially for large values of \( x \). It looks similar to an upward-opening parabola but grows faster.
   - The function \( g(x) = x \) is a straight line passing through the origin with a slope of 1.
   - You are asked to find the area between these two curves from \( x = 0 \) to \( x = 3 \).

### 3. **Setting Up the Integral:**
   The area between two curves is given by the formula:
   \[
   \text{Area} = \int_{a}^{b} \left( f(x) - g(x) \right) \, dx
   \]
   Here, \( f(x) = \cosh(x) \) and \( g(x) = x \), and the bounds are \( x = 0 \) to \( x = 3 \). Therefore, the area is:
   \[
   \text{Area} = \int_{0}^{3} \left( \cosh(x) - x \right) \, dx
   \]

### 4. **Computing the Integral:**
   We can now compute the integral:
   \[
   \text{Area} = \int_{0}^{3} \cosh(x) \, dx - \int_{0}^{3} x \, dx
   \]

   The first integral is straightforward because the derivative of \( \sinh(x) \) is \( \cosh(x) \):
   \[
   \int \cosh(x) \, dx = \sinh(x)
   \]
   So:
   \[
   \int_{0}^{3} \cosh(x) \, dx = \sinh(3) - \sinh(0) = \sinh(3)
   \]

   The second integral is a basic polynomial integral:
   \[
   \int x \, dx = \frac{x^2}{2}
   \]
   So:
   \[
   \int_{0}^{3} x \, dx = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2}
   \]

   Putting these together:
   \[
   \text{Area} = \sinh(3) - \frac{9}{2}
   \]

   Using the value \( \sinh(3) \approx 10.0179 \), we get:
   \[
   \text{Area} \approx 10.0179 - 4.5 = 5.5179 \, \text{square units}
   \]

### 5. **Finding the Mass:**
   The mass density is given as 7 grams per unit area. Therefore, the mass of the part is:
   \[
   \text{Mass} = \text{density} \times \text{area} = 7 \times 5.5179 \approx 38.6253 \, \text{grams}
   \]

### Conclusion:
- The area of the machine part is approximately **5.518 square units**.
- The mass of the machine part is approximately **38.63 grams**.

Would you like more details on the integration steps or visual representation of the curves?

Here are 5 related questions:
1. How does the graph of \( \cosh(x) \) differ from \( e^x \)?
2. How do you calculate the mass of an object with variable density?
3. What are the properties of hyperbolic functions like \( \cosh(x) \) and \( \sinh(x) \)?
4. How do you find the volume of a 3D solid using integration?
5. What is the relationship between area under a curve and definite integrals?

**Tip:** When integrating, it's helpful to break complex regions into smaller, more manageable integrals.

A factory is making a machine part whose shape is bounded by the curves f(x) = cosh(x), g(x) = x, x = 0, and x = 3. Sketch these curves and shade the area of the part. Use integration to determine the area of this part. If the part’s mass density is 7 grams/(unit area), what is the mass of the part?

Find the area of a machine part bounded by cosh(x) and x between 0 and 3, using integration techniques. The part's mass is calculated based on a given mass density of 7 grams/unit area. Learn how to apply definite integrals to real-world engineering problems.

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