Answer:
The sum of the area of the shaded regions = 0.248685
Explanation:
The sum of the area of the shaded region is given as follows;
The point of intersection of the graphs are;
y = x/2
y = sin²x
∴ At the intersection, x/2 = sin²x
sinx = √(x/2)
Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;
x = 0, x ≈ 0.55 or x ≈ 1.85
The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows
1/2 × (0.55)×0.55/2 ≈ 0.075625
The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;
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Therefore;
![A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cint%5Climits%5E%7B0.55%7D_0%20%7Bsin%5E2x%7D%20%5C%2C%20dx%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%20%5Bx%20-sin%28x%29%20%5Ccdot%20cos%28x%29%20%5Cright%5D_0%20%5E%7B0.55%7D)
∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522
The shaded area,
= A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425
Similarly, we have, between points 0.55 and 1.85
A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78
For y = sin²x, we have;
![A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526](https://tex.z-dn.net/?f=A_4%20%3D%20%5Cint%5Climits%5E%7B1.85%7D_%7B0.55%7D%20%7Bsin%5E2x%7D%20%5C%2C%20dx%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%20%5Bx%20-sin%28x%29%20%5Ccdot%20cos%28x%29%20%5Cright%5D_%7B0.55%7D%20%5E%7B1.85%7D%20%5Capprox%201.00526)
The shaded area,
= A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526
The sum of the area of the shaded regions, ∑A =
+
∴ A = 0.023425 + 0.22526 = 0.248685
The sum of the area of the shaded regions, ∑A = 0.248685