Area Under Curves
Use definite integrals to calculate the area between a curve and the x-axis, including handling regions below the axis.
Signed Area vs Actual Area
The definite integral ∫ab f(x) dx gives the signed area between the curve y = f(x) and the x-axis from x = a to x = b. Areas above the x-axis are positive, and areas below the x-axis are negative.
Signed area = ∫ab f(x) dx Actual area = ∫ab |f(x)| dx
Visual: Positive and Negative Regions
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+
−
−
Green regions (above x-axis) contribute positive area. Red regions (below x-axis) contribute negative area to the integral.
Handling Regions Below the x-axis
When a curve dips below the x-axis, the integral over that region gives a negative value. To find the actual (positive) area, you must either:
Method 1: Take the absolute value of the integral for each negative section.
Method 2: Negate the integrand: if f(x) < 0 on [a,b], then Area = −∫ab f(x) dx = ∫ab −f(x) dx.
Example: Area bounded by y = x2 − 4 and the x-axis from x = 0 to x = 3
The curve crosses the x-axis at x = 2 (since x2 − 4 = 0 when x = ±2).
From 0 to 2: f(x) < 0 (below axis). From 2 to 3: f(x) > 0 (above axis).
Area = |∫02 (x2 − 4) dx| + ∫23 (x2 − 4) dx
= |[x3/3 − 4x]02| + [x3/3 − 4x]23
= |(8/3 − 8) − 0| + (9 − 12) − (8/3 − 8)
= |−16/3| + (−3 + 16/3) = 16/3 + 7/3 = 23/3 square units
Composite Areas
For curves that cross the x-axis within the interval, split the integral at each zero (x-intercept) and handle each section separately. The total area is the sum of the absolute values of each section.
Total Area = |∫ac f(x) dx| + |∫cb f(x) dx|
where c is where f(x) = 0 (the x-intercept between a and b)
Common mistake: Do NOT simply compute ∫ab f(x) dx when the curve crosses the x-axis. Positive and negative regions will cancel, giving the wrong answer for the actual area.
Key Vocabulary
Signed Area
The value of a definite integral, where regions below the x-axis are negative.
Actual Area
The total positive area enclosed, calculated by taking absolute values of each section.
x-intercept
A point where the curve crosses the x-axis (f(x) = 0), used to split the integral.
Bounded Region
The enclosed area between a curve, the x-axis, and vertical lines x = a and x = b.
Worked Examples
Find the area under y = x2 from x = 0 to x = 3.
Step 1: Since x2 ≥ 0 for all x, the curve is entirely above the x-axis.
Step 2: Area = ∫03 x2 dx = [x3/3]03 = 27/3 − 0
Answer: Area = 9 square units
Find the area enclosed between y = sin(x), the x-axis, x = 0, and x = 2π.
Step 1: sin(x) is positive on (0, π) and negative on (π, 2π).
Step 2: Area = |∫0π sin(x) dx| + |∫π2π sin(x) dx|
= |[−cos(x)]0π| + |[−cos(x)]π2π| = |−(−1) + 1| + |−1 + (−1)| = 2 + 2
Answer: Area = 4 square units
Find the area between y = x(x − 2) and the x-axis.
Step 1: x-intercepts: x = 0 and x = 2. Between 0 and 2 the parabola is below the axis.
Step 2: Area = −∫02 (x2 − 2x) dx = −[x3/3 − x2]02
= −(8/3 − 4) = −(−4/3) = 4/3
Answer: Area = 4/3 square units
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
Find the area under y = 3x2 from x = 1 to x = 2.
Question 2
If ∫02 f(x) dx = −5, what is the actual area between f(x) and the x-axis on [0, 2]?
Question 3
Why must you split the integral when a curve crosses the x-axis?
Question 4
Find the area enclosed between y = x3 and the x-axis from x = −1 to x = 1.
Question 5
Find the area under y = ex from x = 0 to x = 2.
Key Concepts Summary
- ● Definite integrals give signed area: positive above and negative below the x-axis.
- ● For actual area below the x-axis, negate the integral or take the absolute value.
- ● Always find x-intercepts first to identify where the curve crosses the axis.
- ● Split the integral at each x-intercept and handle each region separately.
- ● Total area is the sum of the absolute values of each section's integral.