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;
[tex]\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx[/tex]
Therefore;
[tex]A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}[/tex]
∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522
The shaded area, [tex]A_{1 shaded}[/tex] = 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;
[tex]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[/tex]
The shaded area, [tex]A_{2 shaded}[/tex] = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526
The sum of the area of the shaded regions, ∑A = [tex]A_{1 shaded}[/tex] + [tex]A_{2 shaded}[/tex]
∴ A = 0.023425 + 0.22526 = 0.248685
The sum of the area of the shaded regions, ∑A = 0.248685
