find the area between the curves of the following functions: y = 7 cos(x),  y = 7 − 7 cos(x),  0 ≤ x ≤ 𝜋

To find the area between the curves \( y = 7\cos(x) \) and \( y = 7 - 7\cos(x) \) over the interval \( 0 \leq x \leq \pi \), we proceed step by step.

### 1. Find the Intersection of the Curves
The area between the curves is computed by integrating the difference of the functions where one curve is above the other. First, we check where they intersect by setting \( 7\cos(x) = 7 - 7\cos(x) \):

\[
7\cos(x) = 7 - 7\cos(x)
\]
\[
2\cos(x) = 1
\]
\[
\cos(x) = \frac{1}{2}
\]

This occurs at \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \). However, since we're only concerned with \( 0 \leq x \leq \pi \), we take the intersection point at \( x = \frac{\pi}{3} \).

### 2. Set Up the Area Integrals
We now split the problem into two integrals: 
1. From \( 0 \) to \( \frac{\pi}{3} \), where \( y = 7\cos(x) \) is above \( y = 7 - 7\cos(x) \).
2. From \( \frac{\pi}{3} \) to \( \pi \), where \( y = 7 - 7\cos(x) \) is above \( y = 7\cos(x) \).

#### a. First Integral: \( 0 \leq x \leq \frac{\pi}{3} \)
Here, the area is given by:

\[
A_1 = \int_0^{\frac{\pi}{3}} \left( [7 - 7\cos(x)] - [7\cos(x)] \right) \, dx
= \int_0^{\frac{\pi}{3}} \left( 7 - 14\cos(x) \right) \, dx
\]

#### b. Second Integral: \( \frac{\pi}{3} \leq x \leq \pi \)
Here, the area is:

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

### 3. Evaluate the Integrals
Since both integrals are of the same form, we calculate them together. The integrals of \( 7 \) and \( \cos(x) \) are straightforward:

\[
\int 7 \, dx = 7x
\]
\[
\int \cos(x) \, dx = \sin(x)
\]

Thus:

\[
A_1 = \left[ 7x - 14\sin(x) \right]_0^{\frac{\pi}{3}}
= \left( 7 \cdot \frac{\pi}{3} - 14 \sin\left(\frac{\pi}{3}\right) \right) - \left( 0 - 14 \sin(0) \right)
\]
\[
= \frac{7\pi}{3} - 14 \cdot \frac{\sqrt{3}}{2} = \frac{7\pi}{3} - 7\sqrt{3}
\]

Similarly,

\[
A_2 = \left[ 7x - 14\sin(x) \right]_{\frac{\pi}{3}}^{\pi}
= \left( 7\pi - 14 \sin(\pi) \right) - \left( 7 \cdot \frac{\pi}{3} - 14 \sin\left(\frac{\pi}{3}\right) \right)
\]
\[
= 7\pi - \frac{7\pi}{3} - 7\sqrt{3}
= \frac{14\pi}{3} - 7\sqrt{3}
\]

### 4. Add the Two Areas
The total area is the sum of \( A_1 \) and \( A_2 \):

\[
A = A_1 + A_2 = \left( \frac{7\pi}{3} - 7\sqrt{3} \right) + \left( \frac{14\pi}{3} - 7\sqrt{3} \right)
= \frac{21\pi}{3} - 14\sqrt{3}
= 7\pi - 14\sqrt{3}
\]

Thus, the area between the curves is:

\[
A = 7\pi - 14\sqrt{3}
\]

Would you like to explore the details of any steps or need further clarification?

### 5 Related Questions:
1. How do we determine which curve is above the other without solving for intersection points?
2. What are the general conditions for finding areas between curves?
3. Can we apply a similar approach to find volumes of solids between curves?
4. How does the symmetry of functions impact the calculation of areas between curves?
5. How would the result change if the interval were different?

### Tip:
Always check the limits of integration and the order of the curves before setting up integrals for areas.

find the area between the curves of the following functions: y = 7 cos(x), y = 7 − 7 cos(x), 0 ≤ x ≤ 𝜋

This problem involves calculating the area between two trigonometric curves, y = 7 cos(x) and y = 7 - 7 cos(x), using integral calculus. The solution includes finding the intersection points and evaluating the integrals for the area. Suitable for college-level calculus students.

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