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To find the area of the region enclosed by the curves given by \( y = 7\cos(x) \) and \( y = 7 - 7\cos(x) \) over the interval \( 0 \leq x \leq \pi \), we will follow these steps:

### 1. Find the points of intersection:
The points of intersection occur where the two curves are equal:
\[
7\cos(x) = 7 - 7\cos(x)
\]
\[
2\cos(x) = 1
\]
\[
\cos(x) = \frac{1}{2}
\]
This equation holds true at \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \). However, since our interval is \( 0 \leq x \leq \pi \), the relevant point of intersection is \( x = \frac{\pi}{3} \).

### 2. Set up the integral:
The area between two curves \( y = f(x) \) and \( y = g(x) \) from \( x = a \) to \( x = b \) is given by:
\[
A = \int_{a}^{b} |f(x) - g(x)| \, dx
\]
For this problem, we need to consider the integral from \( 0 \) to \( \frac{\pi}{3} \) and from \( \frac{\pi}{3} \) to \( \pi \) separately since the roles of the curves swap at the point \( x = \frac{\pi}{3} \).

- For \( 0 \leq x \leq \frac{\pi}{3} \):
\[
A_1 = \int_{0}^{\frac{\pi}{3}} \left[7 - 7\cos(x) - 7\cos(x)\right] dx = \int_{0}^{\frac{\pi}{3}} \left[7 - 14\cos(x)\right] dx
\]

- For \( \frac{\pi}{3} \leq x \leq \pi \):
\[
A_2 = \int_{\frac{\pi}{3}}^{\pi} \left[7\cos(x) - (7 - 7\cos(x))\right] dx = \int_{\frac{\pi}{3}}^{\pi} \left[14\cos(x) - 7\right] dx
\]

### 3. Evaluate the integrals:

#### \( A_1 \):
\[
A_1 = \int_{0}^{\frac{\pi}{3}} \left(7 - 14\cos(x)\right) dx = 7x - 14\sin(x) \Big|_{0}^{\frac{\pi}{3}} = 7\left(\frac{\pi}{3}\right) - 14\left(\frac{\sqrt{3}}{2}\right) - \left(7(0) - 14(0)\right)
\]
\[
A_1 = \frac{7\pi}{3} - 7\sqrt{3}
\]

#### \( A_2 \):
\[
A_2 = \int_{\frac{\pi}{3}}^{\pi} \left(14\cos(x) - 7\right) dx = 14\sin(x) - 7x \Big|_{\frac{\pi}{3}}^{\pi} = 14\sin(\pi) - 7\pi - \left(14\sin\left(\frac{\pi}{3}\right) - 7\left(\frac{\pi}{3}\right)\right)
\]
\[
A_2 = -7\pi + 7\sqrt{3} + \frac{7\pi}{3}
\]

### 4. Total Area:
\[
A = A_1 + A_2 = \left(\frac{7\pi}{3} - 7\sqrt{3}\right) + \left(-7\pi + 7\sqrt{3} + \frac{7\pi}{3}\right) = \frac{7\pi}{3} + \frac{7\pi}{3} - 7\pi = \frac{7\pi}{3}
\]

Thus, the area of the region is:
\[
\boxed{\frac{7\pi}{3}}
\]

Would you like to ask for details or have any other questions?

Here are some related questions to deepen your understanding:
1. How do the cosine function's properties affect the shape of the curves?
2. What is the significance of the interval \(0 \leq x \leq \pi\) in this problem?
3. Can this approach be applied to find areas between other trigonometric functions?
4. What would the area be if the interval changed to \(0 \leq x \leq 2\pi\)?
5. How would the solution change if the curves were \( y = a\cos(x) \) and \( y = b - b\cos(x) \) for different values of \(a\) and \(b\)?

**Tip:** Always check the points of intersection carefully, as they dictate how the integral should be set up, especially if the curves switch roles within the given interval.

Learn how to calculate the area of the region enclosed by two cosine curves using integration. This problem involves finding the area between curves defined by trigonometric functions over a specified interval. Suitable for undergraduate students studying calculus and trigonometry.

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