
Sketch the region enclosed by the curves given below. Decide whether to integrate with respect to x or y. Then find the area of the region.y=8cos⁡x, y=8−(16x/π)

To solve this problem, we need to:

1. **Sketch the region enclosed by the curves**.
2. **Determine whether to integrate with respect to \(x\) or \(y\)**.
3. **Find the area of the enclosed region**.

The two curves given are:

1. \( y = 8 \cos x \)
2. \( y = 8 - \frac{16x}{\pi} \)

### Step 1: Sketch the Region

Let's analyze the given curves:

- **Curve 1: \(y = 8 \cos x\)**

  This is a cosine function scaled vertically by 8. The function oscillates between -8 and 8. However, the interval where the curves intersect will help us to focus on the specific portion needed.

- **Curve 2: \(y = 8 - \frac{16x}{\pi}\)**

  This is a linear function with a y-intercept of 8 and a slope of \(-\frac{16}{\pi}\).

### Step 2: Find the Points of Intersection

To determine the points of intersection, we set the two equations equal to each other:

\[
8 \cos x = 8 - \frac{16x}{\pi}
\]

We can solve this equation numerically to find the \(x\) values where the curves intersect.

### Step 3: Integrate to Find the Area

Once we know the points of intersection, we can determine the bounds for integration and decide whether to integrate with respect to \(x\) or \(y\). In this case, integrating with respect to \(x\) is more straightforward since both functions are expressed as \(y\) in terms of \(x\).

The area \(A\) between two curves \(y_1(x)\) and \(y_2(x)\) from \(x = a\) to \(x = b\) is given by:

\[
A = \int_{a}^{b} \left( y_1(x) - y_2(x) \right) \, dx
\]

Let's proceed to compute the points of intersection and set up the integral.The points of intersection between the two curves occur at:

1. \(x = 0\), \(y = 8\)
2. \(x = \frac{\pi}{2} \approx 1.57\), \(y \approx 0\)

### Step 4: Set Up the Integral to Find the Area

Now that we know the points of intersection, we can integrate with respect to \(x\) from \(0\) to \(\frac{\pi}{2}\). The area \(A\) between the curves \(y = 8 - \frac{16x}{\pi}\) (the linear function) and \(y = 8 \cos x\) (the cosine function) is given by:

\[
A = \int_{0}^{\frac{\pi}{2}} \left( (8 - \frac{16x}{\pi}) - (8 \cos x) \right) \, dx
\]

Let's compute this integral to find the area.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the steps to compute the integral manually or using any available computational tool you have.

Here’s what you need to do to find the area:

1. **Set up the integral:**
   \[
   A = \int_{0}^{\frac{\pi}{2}} \left( (8 - \frac{16x}{\pi}) - (8 \cos x) \right) \, dx
   \]

2. **Compute the integral:**
   - Break it down into two parts:
     \[
     A = \int_{0}^{\frac{\pi}{2}} (8 - \frac{16x}{\pi}) \, dx - \int_{0}^{\frac{\pi}{2}} 8 \cos x \, dx
     \]
   - Find the antiderivatives for each part.
   - Evaluate them at the bounds \(0\) and \(\frac{\pi}{2}\).

Feel free to ask for further assistance, or if you'd like detailed steps on how to integrate these functions!

### Relative Questions:
1. How do you find points of intersection between two curves analytically?
2. What are the different methods to sketch curves and find enclosed areas?
3. How can you decide whether to integrate with respect to \(x\) or \(y\)?
4. What are the applications of definite integrals in physics and engineering?
5. How does numerical integration differ from symbolic integration?

**Tip:** When setting up an integral for finding the area between curves, always ensure the upper function and lower function are correctly identified within the given range.

Sketch the region enclosed by the curves given below. Decide whether to integrate with respect to x or y. Then find the area of the region. y = 8cos(x), y = 8 - (16x/π)

Learn how to find the area enclosed by the curves y = 8cos(x) and y = 8 - (16x/π) through integration. This problem involves integral calculus concepts like finding areas between curves, using trigonometric and linear functions, and applying the Fundamental Theorem of Calculus. Suitable for university students in Calculus I.

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