Area Bounded by y = sin x and the x-Axis Between x = 0 and x = 3π/2
Understanding the Given Area Under the Curve Problem
This question asks us to calculate the total geometrical area enclosed between the curve y = sin x and the x-axis over the interval from x = 0 to x = 3π/2. The problem is based on definite integration, but an important feature of the sine curve must be considered before writing the integral.
The curve y = sin x does not remain above the x-axis throughout the given interval. From x = 0 to x = π, the value of sin x is positive, so the curve lies above the x-axis. However, from x = π to x = 3π/2, the value of sin x is negative, so the curve lies below the x-axis.
Since the question asks for the area bounded by the curve and the x-axis, both regions must contribute positively to the final answer. Therefore, the interval must be divided at x = π, where the sine curve crosses the x-axis.
Behavior of the Curve y = sin x on the Given Interval
The standard sine curve begins at the origin because:
sin 0 = 0
It rises above the x-axis and reaches its maximum value at x = π/2:
sin(π/2) = 1
The curve then returns to the x-axis at x = π:
sin π = 0
After x = π, the sine function becomes negative. At x = 3π/2, it reaches:
sin(3π/2) = −1
Therefore, the interval from 0 to 3π/2 contains one complete positive region and one negative region. For calculating geometrical area, the magnitude of both regions must be added.
Formula for the Area Between a Curve and the x-Axis
If a function f(x) remains above the x-axis on an interval [a, b], the area is:
Area = ∫ab f(x) dx
If the function lies below the x-axis, the definite integral becomes negative. However, geometrical area cannot be negative. Therefore, the magnitude of the integral must be taken.
In general:
Total area = ∫ |f(x)| dx
For the present problem, the sine function changes sign at x = π. Therefore, the total area is calculated by dividing the interval into two parts:
Total area = Area from 0 to π + Area from π to 3π/2
Step-by-Step Solution
Step 1: Identify Where the Curve Crosses the x-Axis
The curve crosses the x-axis wherever:
sin x = 0
Within the given interval 0 ≤ x ≤ 3π/2, the relevant zero points are:
x = 0 and x = π
The point x = π is especially important because the sign of sin x changes there. Therefore, the total area must be calculated separately over the intervals:
0 ≤ x ≤ π
and:
π ≤ x ≤ 3π/2
Step 2: Calculate the Area from x = 0 to x = π
Between 0 and π, the sine curve lies above the x-axis. Therefore, the first area is:
A1 = ∫0π sin x dx
The antiderivative of sin x is:
∫ sin x dx = −cos x
Therefore:
A1 = [−cos x]0π
Applying the limits:
A1 = −cos π − (−cos 0)
Using:
cos π = −1
and:
cos 0 = 1
we obtain:
A1 = −(−1) + 1
Therefore:
A1 = 2 square units
Step 3: Calculate the Area from x = π to x = 3π/2
Between π and 3π/2, the sine curve lies below the x-axis. Therefore, the definite integral of sin x over this interval will be negative.
The signed integral is:
∫π3π/2 sin x dx
Using the antiderivative −cos x:
= [−cos x]π3π/2
Applying the limits:
= −cos(3π/2) − (−cos π)
Since:
cos(3π/2) = 0
and:
cos π = −1
we get:
= 0 − 1
Therefore, the signed integral is:
−1
However, the question asks for geometrical area, which must be positive. Hence:
A2 = |−1|
Therefore:
A2 = 1 square unit
Step 4: Add the Areas of Both Regions
The total area bounded by the curve and the x-axis is the sum of the two positive geometrical areas:
Total area = A1 + A2
Substituting the calculated values:
Total area = 2 + 1
Therefore:
Total area = 3 square units
Why the Direct Integral from 0 to 3π/2 Does Not Give the Required Area
If we directly evaluate the definite integral over the complete interval, we obtain:
∫03π/2 sin x dx
Using the antiderivative −cos x:
= [−cos x]03π/2
Therefore:
= −cos(3π/2) + cos 0
= 0 + 1
= 1
This value represents the net signed area, not the total geometrical area. The positive region contributes +2, while the region below the x-axis contributes −1. Therefore, the direct integral gives:
2 − 1 = 1
However, the question asks for the area bounded by the curve and the x-axis. Geometrical area is always positive, so the correct calculation is:
2 + 1 = 3
Alternative Solution Using the Absolute Value of sin x
The total geometrical area can also be written as:
Area = ∫03π/2 |sin x| dx
Since sin x is positive between 0 and π and negative between π and 3π/2:
|sin x| = sin x, for 0 ≤ x ≤ π
and:
|sin x| = −sin x, for π ≤ x ≤ 3π/2
Therefore:
Area = ∫0π sin x dx − ∫π3π/2 sin x dx
Substituting the values of the two integrals:
Area = 2 − (−1)
Hence:
Area = 3 square units
Geometrical Interpretation of the Result
The sine curve forms a complete positive arch between x = 0 and x = π. The area of this region is 2 square units. From x = π to x = 3π/2, the curve forms half of a negative arch below the x-axis, whose geometrical area is 1 square unit.
Since both regions are enclosed between the curve and the x-axis, both must be counted positively. Therefore, the complete bounded area is the sum of 2 square units and 1 square unit.
This distinction between signed integral and total geometrical area is essential whenever a curve crosses the x-axis within the interval of integration.
Final Answer
The area bounded by y = sin x and the x-axis between x = 0 and x = 3π/2 is 3 square units.
Answer: 3
