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. \[ y=9 \cos x, y=9-\frac{18 x}{\pi} \] Area \( =9-\frac{9 \pi}{4} \)

To find the area of the region enclosed by the curves \( y = 9 \cos x \) and \( y = 9 - \frac{18x}{\pi} \), follow these steps: 1. **Sketch the Curves** 2. **Determine the Points of Intersection** 3. **Choose the Appropriate Variable for Integration** 4. **Set Up and Evaluate the Integral** 5. **Compute the Area** Let's go through each step in detail. --- ### 1. Sketch the Curves **Curve 1:** \( y = 9 \cos x \) - This is a cosine function with amplitude 9. - Period: \( 2\pi \) - Oscillates between \( y = 9 \) and \( y = -9 \). **Curve 2:** \( y = 9 - \frac{18x}{\pi} \) - This is a linear function with slope \( -\frac{18}{\pi} \) and y-intercept at \( y = 9 \). - Decreases as \( x \) increases. **Graph Overview:** - The cosine curve starts at \( y = 9 \) when \( x = 0 \) and oscillates down and up. - The linear function starts at \( y = 9 \) when \( x = 0 \) and decreases linearly. - The curves intersect where \( 9 \cos x = 9 - \frac{18x}{\pi} \). --- ### 2. Determine the Points of Intersection Set the two equations equal to find the points of intersection: \[ 9 \cos x = 9 - \frac{18x}{\pi} \] Simplify: \[ \cos x = 1 - \frac{2x}{\pi} \] This equation is transcendental and doesn't have an algebraic solution, so we'll solve it numerically or graphically. **Approach:** - Use graphical methods or numerical methods (like the Newton-Raphson method) to approximate the solutions. **Approximated Solution:** Upon solving numerically, the curves intersect at two points: 1. At \( x = 0 \): \[ y = 9 \cos 0 = 9 \quad \text{and} \quad y = 9 - \frac{18 \times 0}{\pi} = 9 \] 2. At \( x = \frac{\pi}{2} \): \[ y = 9 \cos \left( \frac{\pi}{2} \right) = 0 \quad \text{and} \quad y = 9 - \frac{18 \times \frac{\pi}{2}}{\pi} = 9 - 9 = 0 \] Therefore, the points of intersection are at \( x = 0 \) and \( x = \frac{\pi}{2} \). --- ### 3. Choose the Appropriate Variable for Integration To find the area between the two curves from \( x = 0 \) to \( x = \frac{\pi}{2} \), we can integrate with respect to \( x \). **Reasoning:** - The functions are expressed as \( y \) in terms of \( x \). - The limits of integration are easily defined in terms of \( x \). --- ### 4. Set Up and Evaluate the Integral The area \( A \) between two curves \( y = f(x) \) and \( y = g(x) \) from \( x = a \) to \( x = b \) is: \[ A = \int_{a}^{b} \left| f(x) - g(x) \right| dx \] In our case: \[ f(x) = 9 \cos x \quad \text{and} \quad g(x) = 9 - \frac{18x}{\pi} \] Since \( 9 \cos x \geq 9 - \frac{18x}{\pi} \) in the interval \( [0, \frac{\pi}{2}] \), the area is: \[ A = \int_{0}^{\frac{\pi}{2}} \left( 9 \cos x - \left( 9 - \frac{18x}{\pi} \right) \right) dx \] Simplify the integrand: \[ 9 \cos x - 9 + \frac{18x}{\pi} \] So the integral becomes: \[ A = \int_{0}^{\frac{\pi}{2}} \left( 9 \cos x - 9 + \frac{18x}{\pi} \right) dx \] **Compute the Integral:** \[ A = \int_{0}^{\frac{\pi}{2}} 9 \cos x \, dx - \int_{0}^{\frac{\pi}{2}} 9 \, dx + \int_{0}^{\frac{\pi}{2}} \frac{18x}{\pi} \, dx \] Evaluate each integral separately: 1. **First Integral:** \[ \int_{0}^{\frac{\pi}{2}} 9 \cos x \, dx = 9 \sin x \Big|_{0}^{\frac{\pi}{2}} = 9(1 - 0) = 9 \] 2. **Second Integral:** \[ \int_{0}^{\frac{\pi}{2}} 9 \, dx = 9x \Big|_{0}^{\frac{\pi}{2}} = 9 \left( \frac{\pi}{2} - 0 \right) = \frac{9\pi}{2} \] 3. **Third Integral:** \[ \int_{0}^{\frac{\pi}{2}} \frac{18x}{\pi} \, dx = \frac{18}{\pi} \cdot \frac{x^2}{2} \Big|_{0}^{\frac{\pi}{2}} = \frac{9}{\pi} \left( \left( \frac{\pi}{2} \right)^2 - 0 \right) = \frac{9}{\pi} \cdot \frac{\pi^2}{4} = \frac{9\pi}{4} \] **Combine the Results:** \[ A = 9 - \frac{9\pi}{2} + \frac{9\pi}{4} = 9 - \frac{9\pi}{2} + \frac{9\pi}{4} \] Simplify the terms involving \( \pi \): \[ -\frac{9\pi}{2} + \frac{9\pi}{4} = -\frac{9\pi}{4} \] Therefore: \[ A = 9 - \frac{9\pi}{4} \] --- ### 5. Compute the Area The calculated area is: \[ A = 9 - \frac{9\pi}{4} \] --- ### **Conclusion** The area of the region enclosed by the curves \( y = 9 \cos x \) and \( y = 9 - \frac{18x}{\pi} \) is: \[ \boxed{9 - \frac{9\pi}{4}} \] This matches the area provided in your question. --- ### **Visualization** Here's a sketch of the region:  *(Note: Please create a graph using graphing software or tools like Desmos to visualize the curves and the enclosed area.)* --- If you need further assistance with graphing or understanding the integration steps, feel free to ask!