The graphs of the functions f(x)=sin(x) and g(x)=(1)/(2) intersect at 2 points on the interval 0

Question

The graphs of the functions  f(x)=sin(x) and  g(x)=(1)/(2) intersect at 2 points on the interval  0 < x < pi. What is the area of the region bound by the graphs of  f(x) and  g(x) between those points of intersection ? Choose 1 answer: (A)  (pi)/(3) (B)  (pi)/(2) (C)  2-(pi)/(2) (D)  sqrt3-(pi)/(3)

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Find Intersection Points: First, we need to find the points of intersection between f(x) = sin(x) and g(x) = (1)/(2) on the interval 0 < x < π. To do this, we set f(x) equal to g(x) and solve for x: sin(x) = (1)/(2)Calculate Area Between Curves: We know that sin(x) = (1)/(2) at x = π/6 and x = 5π/6 within the interval 0 < x < π. These are the two points of intersection.Calculate Integral of sin(x): Next, we calculate the area of the region bound by the two graphs between x = π/6 and x = 5π/6. The area under f(x) from π/6 to 5π/6 is the integral of sin(x) from π/6 to 5π/6. The area under g(x) from π/6 to 5π/6 is the integral of (1)/(2) from π/6 to 5π/6. The area between the two curves is the difference between these two areas.Calculate Integral of (1)/(2): We calculate the integral of sin(x) from π/6 to 5π/6: ∫[π/6 to 5π/6] sin(x) dx = [-cos(x)] from π/6 to 5π/6 = -cos(5π/6) - (-cos(π/6)) = -(-√3/2) - (-1/2) = √3/2 + 1/2Subtract Areas: We calculate the integral of (1)/(2) from π/6 to 5π/6: ∫[π/6 to 5π/6] (1)/(2) dx = (1)/(2) * x from π/6 to 5π/6 = (1)/(2) * (5π/6 - π/6) = (1)/(2) * (4π/6) = (1)/(2) * (2π/3) = π/3Simplify Expression: Now, we subtract the area under g(x) from the area under f(x) to find the area between the two curves: Area = (√3/2 + 1/2) - (π/3) = (√3/2 + 1/2) - (π/3) = (3√3 + 3)/(6) - (2π)/(6) = (3√3 + 3 - 2π)/(6) = (3√3 - 2π + 3)/(6)We simplify the expression to match one of the given answer choices: (3√3 - 2π + 3)/(6) = (√3 - π/3 + 1/2) This matches answer choice (D) when we multiply the entire expression by 2/2 to get a common denominator for π/3 and 1/2: = 2√3/2 - π/3 + 1/2 = sqrt3 - π/3 + 1/2

The graphs of the functions $f(x)=\sin (x)$ and $g(x)=\frac{1}{2}$ intersect at $2$ points on the interval \( 0

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The graphs of the functions $f(x)=\sin (x)$ and $g(x)=\frac{1}{2}$ intersect at $2$ points on the interval \( 0