Find the area bounded by the curve and the lines y=sinx, y= 1/2, x=5pi/6 and x=pi/6
Note that sin(5pi/6)=sin(pi/6)=1/2.
Area=Integral(from pi/6 to 5pi/6)(sinx-1/2)dx=(-cosx-x/2)I(pi/6 to 5pi/6)=
(-cos(5pi/6)-5pi/12)-(-cos(pi/6)-pi/12)=
=sqrt(3)-pi/3=0.685
To find the area bounded by the curve y = sinx, the line y = 1/2, and the lines x = 5π/6 and x = π/6, we need to integrate the function between the appropriate x-values.
Step 1: Determine the intersection points of the curve y = sinx and the line y = 1/2.
Set y = sinx equal to y = 1/2:
sinx = 1/2
To find the x-values, we can use the inverse sine function:
x = arcsin(1/2)
x = π/6 or x = 5π/6
Step 2: Determine the limits of integration for calculating the area.
The x-values where the intersection occurs (π/6 and 5π/6) will serve as the limits of integration.
Step 3: Set up the integral to calculate the area.
The area can be calculated using the formula:
A = ∫[a, b] (f(x) - g(x)) dx
Where a and b are the limits of integration, f(x) is the curve, and g(x) is the line.
In this case, f(x) = sinx, g(x) = 1/2, and the limits of integration are π/6 and 5π/6. Therefore, the integral to calculate the area is:
A = ∫[π/6, 5π/6] (sinx - 1/2) dx
Step 4: Evaluate the integral.
A = ∫[π/6, 5π/6] (sinx - 1/2) dx
Integrating sinx gives [-cosx]:
A = [-cosx] ∣[π/6, 5π/6] - [(1/2)x]
Now we can substitute the limits of integration:
A = -cos(5π/6) - (-cos(π/6)) - [(1/2)(5π/6 - π/6)]
A = -(-√3/2) - (-√3/2) - [(1/2)(4π/6)]
A = √3/2 + √3/2 - (2π/6)
A = √3 - (2π/6)
Simplifying further:
A = √3 - π/3
Therefore, the area bounded by the curve y = sinx, the line y = 1/2, and the lines x = 5π/6 and x = π/6 is √3 - π/3.
To find the area bounded by the curve y = sinx, y = 1/2, x = 5π/6, and x = π/6, we can use the concept of definite integrals.
Step 1: Determine the points of intersection
Let's start by finding the x-values where the curve y = sinx intersects the line y = 1/2:
sinx = 1/2
Solving this equation for the values of x, we get:
x = π/6 and x = 5π/6
These are the x-values where the curve and the line intersect.
Step 2: Set up the definite integral
To find the area bounded by the curve and the lines, we need to integrate the difference between the two functions (sinx - 1/2) from the x-values of intersection (π/6 to 5π/6):
Area = ∫[π/6, 5π/6] (sinx - 1/2) dx
Step 3: Evaluate the definite integral
Now we need to evaluate the integral. Integrate the function (sinx - 1/2) with respect to x from π/6 to 5π/6:
Area = [ -cosx - (1/2)x ] evaluated from π/6 to 5π/6
Substituting the upper and lower limits into the expression, we get:
Area = [ -cos(5π/6) - (1/2)(5π/6) ] - [ -cos(π/6) - (1/2)(π/6) ]
Step 4: Simplify and calculate
Evaluating the expression, we get:
Area = [ -sqrt(3)/2 - (5π/12) ] - [ -1/2 - π/12 ]
Simplifying further:
Area = [ -sqrt(3)/2 - 5π/12 ] - [ -1/2 - π/12 ]
Area = [-sqrt(3)/2 + π/12 - (-1/2) - 5π/12]
Area = [-sqrt(3)/2 + π/12 + 1/2 - 5π/12]
Step 5: Calculate the final result
Combining the terms, we get:
Area = (-sqrt(3) + π - 5π) / 12 + 1/2
Area = (-sqrt(3) - 4π) / 12 + 1/2
Area = -sqrt(3)/12 - π/3 + 1/2
Hence, the area bounded by the curve y = sinx, y = 1/2, x = 5π/6, and x = π/6 is (-sqrt(3)/12 - π/3 + 1/2) square units.