Answer:
7√3 + 7π/6 ≈ 15.790
Step-by-step explanation:
First, find the intersection:
7 cos(2x) = 7 − 7 cos(2x)
14 cos(2x) = 7
cos(2x) = 1/2
2x = π/3
x = π/6
From the graph:
On 0 ≤ x ≤ π/6, 7 cos(2x) > 7 − 7 cos(2x).
On π/6 ≤ x ≤ π/2, 7 − 7 cos(2x) > 7 cos(2x).
So the integral is:
[tex]\int\limits^\frac{\pi}{6} _0 {[7cos(2x)-(7-7cos(2x))]} \, dx + \int\limits^\frac{\pi}{2} _\frac{\pi}{6} {[(7-7cos(2x))-7cos(2x)]} \, dx[/tex]
[tex]\int\limits^\frac{\pi}{6} _0 {[7cos(2x)-7+7cos(2x))]} \, dx + \int\limits^\frac{\pi}{2} _\frac{\pi}{6} {[7-7cos(2x)-7cos(2x)]} \, dx[/tex]
[tex]\int\limits^\frac{\pi}{6} _0 {[14cos(2x)-7]} \, dx + \int\limits^\frac{\pi}{2} _\frac{\pi}{6} {[7-14cos(2x)]} \, dx[/tex]
[tex][7sin(2x)-7x]|^\frac{\pi}{6} _0 + [7x-7sin(2x)]|^\frac{\pi}{2} _\frac{\pi}{6}[/tex]
[tex][7sin(\frac{\pi }{3} )-\frac{7\pi }{6} ] - [7sin(0)-0 ] + [\frac{7\pi }{2} -7sin(\pi )] - [\frac{7\pi }{6} -7sin(\frac{\pi }{3} )][/tex]
[tex]7sin(\frac{\pi }{3} )-\frac{7\pi }{6} + \frac{7\pi }{2} -7sin(\pi ) - \frac{7\pi }{6} +7sin(\frac{\pi }{3} )[/tex]
[tex]14sin(\frac{\pi }{3} )-\frac{7\pi }{3} + \frac{7\pi }{2}[/tex]
[tex]7\sqrt{3} + \frac{7\pi }{6}[/tex]
