Area of the Inscribed Sin()

The integral is evidently:

$A = \int_0^8 \Big| \sqrt{16 - (x-4)^2} \, \sin(2 \pi x) \Big| \, dx \approx 16.04$

Start by trying to construct the equation of a Sin() that would behave exactly as the one shown. You can do this by using facts about graph such as Period being 1 and the amplitude is bound by the semi-circle (sqrt(16-(x-4)^2)) Combining these facts gives us: (sqrt(16-(x-4)^2))sin(2 pi x)

Now Integrate for the Area above y=0: (integrate (0 to .5) (sqrt(16-(x-4)^(2))sin(2π x)))+(integrate (1 to 1.5) (sqrt(16-(x-4)^(2))sin(2π x)))+(integrate (2 to 2.5) (sqrt(16-(x-4)^(2))sin(2π x))) + (integrate (3 to 3.5) (sqrt(16-(x-4)^(2))sin(2π x)))+(integrate (4 to 4.5) (sqrt(16-(x-4)^(2))sin(2π x))) + (integrate (5 to 5.5) (sqrt(16-(x-4)^(2))sin(2π x)))+(integrate (6 to 6.5) (sqrt(16-(x-4)^(2))sin(2π x)))+(integrate (7 to 7.5) (sqrt(16-(x-4)^(2))sin(2π x)))

Which gives us 2.497370126384403091127565555937937506096335388688535814885 + 2.520437102586328362208002484908226307660720089440828740954 + 3.000638812318869440084809167149408983698369872096999590875 = 8.018446041289600893420377207995572797455425350226364146714

Multiply this by 2 in order to get the total positive Area of the shaded regions above and below y=0. 16.036892082579201786840754415991145594910850700452728293428

The nearest Integer is 16