1. **State the problem:** We need to find the sum of the areas enclosed by the graphs of $$f(x)=2\cos\left(\frac{\pi x}{4}\right)$$ and $$g(x)=x-6$$ between $$x=4$$ and $$x=8$$. 2. **Understand the problem:** The enclosed area between two curves $$f(x)$$ and $$g(x)$$ from $$a$$ to $$b$$ is given by $$\text{Area} = \int_a^b |f(x)-g(x)| \, dx$$ We must find where the graphs intersect between 4 and 8 to split the integral if needed. 3. **Find intersection points:** Solve $$2\cos\left(\frac{\pi x}{4}\right) = x - 6$$ for $$x$$ in $$[4,8]$$. Using a graphing calculator or numerical methods, approximate intersections: - At $$x=4$$, $$f(4)=2\cos(\pi) = 2(-1) = -2$$, $$g(4)=4-6=-2$$, so they intersect at $$x=4$$. - Check near $$x=6$$: $$f(6)=2\cos\left(\frac{3\pi}{2}\right)=2\times 0=0$$ $$g(6)=0$$ So they intersect at $$x=6$$. - Check at $$x=8$$: $$f(8)=2\cos(2\pi)=2(1)=2$$ $$g(8)=8-6=2$$ So they intersect at $$x=8$$. 4. **Determine which function is on top between intersections:** - For $$x \in (4,6)$$, test $$x=5$$: $$f(5)=2\cos\left(\frac{5\pi}{4}\right)=2\times (-\frac{\sqrt{2}}{2})=-\sqrt{2} \approx -1.414$$ $$g(5)=5-6=-1$$ Since $$g(5) > f(5)$$, $$g(x)$$ is above $$f(x)$$ on $$[4,6]$$. - For $$x \in (6,8)$$, test $$x=7$$: $$f(7)=2\cos\left(\frac{7\pi}{4}\right)=2\times \frac{\sqrt{2}}{2}=\sqrt{2} \approx 1.414$$ $$g(7)=7-6=1$$ Since $$f(7) > g(7)$$, $$f(x)$$ is above $$g(x)$$ on $$[6,8]$$. 5. **Set up the integral for the total enclosed area:** $$\text{Area} = \int_4^6 (g(x)-f(x)) \, dx + \int_6^8 (f(x)-g(x)) \, dx$$ 6. **Calculate each integral:** - First integral: $$\int_4^6 (x-6 - 2\cos\left(\frac{\pi x}{4}\right)) \, dx = \int_4^6 (x-6) \, dx - 2 \int_4^6 \cos\left(\frac{\pi x}{4}\right) \, dx$$ Calculate each part: $$\int_4^6 (x-6) \, dx = \left[ \frac{x^2}{2} - 6x \right]_4^6 = \left( \frac{36}{2} - 36 \right) - \left( \frac{16}{2} - 24 \right) = (18 - 36) - (8 - 24) = -18 - (-16) = -18 + 16 = -2$$ For the cosine integral, use substitution: Let $$u = \frac{\pi x}{4}$$, then $$du = \frac{\pi}{4} dx$$, so $$dx = \frac{4}{\pi} du$$. Change limits: When $$x=4$$, $$u=\pi$$. When $$x=6$$, $$u=\frac{3\pi}{2}$$. So: $$\int_4^6 \cos\left(\frac{\pi x}{4}\right) dx = \int_{\pi}^{\frac{3\pi}{2}} \cos(u) \frac{4}{\pi} du = \frac{4}{\pi} \int_{\pi}^{\frac{3\pi}{2}} \cos(u) du = \frac{4}{\pi} [\sin(u)]_{\pi}^{\frac{3\pi}{2}} = \frac{4}{\pi} (\sin(\frac{3\pi}{2}) - \sin(\pi)) = \frac{4}{\pi} (-1 - 0) = -\frac{4}{\pi}$$ Multiply by -2: $$-2 \times -\frac{4}{\pi} = \frac{8}{\pi}$$ So the first integral is: $$-2 + \frac{8}{\pi}$$ - Second integral: $$\int_6^8 (2\cos\left(\frac{\pi x}{4}\right) - (x-6)) \, dx = 2 \int_6^8 \cos\left(\frac{\pi x}{4}\right) dx - \int_6^8 (x-6) dx$$ Calculate each part: For the cosine integral, with substitution as before: When $$x=6$$, $$u=\frac{3\pi}{2}$$. When $$x=8$$, $$u=2\pi$$. $$2 \int_6^8 \cos\left(\frac{\pi x}{4}\right) dx = 2 \times \frac{4}{\pi} \int_{\frac{3\pi}{2}}^{2\pi} \cos(u) du = \frac{8}{\pi} [\sin(u)]_{\frac{3\pi}{2}}^{2\pi} = \frac{8}{\pi} (\sin(2\pi) - \sin(\frac{3\pi}{2})) = \frac{8}{\pi} (0 - (-1)) = \frac{8}{\pi}$$ For the linear integral: $$\int_6^8 (x-6) dx = \left[ \frac{x^2}{2} - 6x \right]_6^8 = \left( \frac{64}{2} - 48 \right) - \left( \frac{36}{2} - 36 \right) = (32 - 48) - (18 - 36) = -16 - (-18) = -16 + 18 = 2$$ So the second integral is: $$\frac{8}{\pi} - 2$$ 7. **Sum the two areas:** $$\text{Total Area} = \left(-2 + \frac{8}{\pi}\right) + \left(\frac{8}{\pi} - 2\right) = -4 + \frac{16}{\pi}$$ 8. **Calculate numerical value:** $$\pi \approx 3.1416$$ $$\text{Total Area} \approx -4 + \frac{16}{3.1416} = -4 + 5.092 = 1.092$$ Rounded to three decimal places: $$\boxed{1.092}$$