Practice Maths

Area Under a Curve

Key Terms

When f(x) ≥ 0: Area = ∫ab f(x) dx (integral gives the geometric area directly)
When f(x) ≤ 0: Area = −∫ab f(x) dx (take the negative to get a positive area)
When f(x) changes sign: split the integral at x-intercepts and add absolute values
Always sketch the curve first to identify which parts are above/below the x-axis
The signed area ∫ab f(x) dx may not equal the geometric area if f changes sign
Area formulas:
If f(x) ≥ 0 on [a,b]:   A = ∫ab f(x) dx

If f(x) ≤ 0 on [a,b]:   A = −∫ab f(x) dx = ∫ab |f(x)| dx

If f changes sign at x = c (a < c < b):
A = |∫ac f(x) dx| + |∫cb f(x) dx|

Or equivalently: A = ∫ab |f(x)| dx
Worked Example 1: Find the area enclosed by y = x(x − 3) and the x-axis.

Step 1: Find x-intercepts: x(x−3) = 0 ⇒ x = 0 or x = 3
Step 2: Determine sign: parabola opens up; below x-axis on [0,3]
Step 3: Area = −∫03 x(x−3) dx = −∫03 (x²−3x) dx
= −[x³/3 − 3x²/2]03
= −[(9 − 27/2) − 0]
= −[−9/2] = 9/2 = 4.5 units²
Worked Example 2: Find the total area enclosed by y = sin x and the x-axis for x ∈ [0, 2π].

Step 1: sin x = 0 at x = 0, π, 2π
On [0,π]: sin x ≥ 0. On [π, 2π]: sin x ≤ 0
Step 2: A = ∫0π sin x dx + |∫π2π sin x dx|
= [−cos x]0π + |[−cos x]π2π|
= (1+1) + |(−1−1)| = 2 + 2 = 4 units²
Hot Tip: ALWAYS draw a sketch before setting up an area integral. A single definite integral from a to b can give zero even if there is a non-zero area (positive and negative parts cancel). Split at x-intercepts and add absolute values.

Signed Area vs Geometric Area

The definite integral ∫ab f(x) dx computes a signed area: regions above the x-axis contribute positively, regions below contribute negatively. This is mathematically precise but not always what we mean when we say "area".

The geometric area (always positive) is what you'd measure physically — it's the total amount of region enclosed between the curve and the x-axis, regardless of which side of the axis the curve lies on.

Example: ∫02π sin x dx = 0. This doesn't mean there's no area! The positive region [0,π] cancels the negative region [π,2π]. The geometric area is 4 sq units.

x y +A −A 0 π 2π y = sin x
The blue region (+A) and red region (−A) are equal in size. Their signed areas cancel in the integral, but the total geometric area is 2A.

Finding Area When f(x) ≥ 0

This is the simplest case. If you can confirm f(x) ≥ 0 throughout [a,b], the definite integral directly gives the area:

A = ∫ab f(x) dx

Strategy:

  1. Sketch the function to confirm it stays above (or on) the x-axis.
  2. Identify the limits a and b (often given as x-intercepts or specific x-values).
  3. Evaluate the definite integral.
  4. State the area with units (units²).

Example: Area under y = 4 − x² from x = −2 to x = 2:
f(−2) = 0, f(0) = 4, f(2) = 0. The parabola opens downward, so f(x) ≥ 0 on [−2, 2].
A = ∫−22 (4−x²) dx = [4x − x³/3]−22 = (8−8/3)−(−8+8/3) = 32/3 units²

Finding Area When f(x) ≤ 0

When the curve is entirely below the x-axis, the definite integral gives a negative number. To find the geometric area, take the negative of the integral:

A = −∫ab f(x) dx

Example: Find the area between y = x² − 4 and the x-axis from x = 0 to x = 2.
At x=0: f(0) = −4 < 0. At x=2: f(2) = 0. Since x²−4 ≤ 0 on [0,2]:
A = −∫02 (x²−4) dx = −[x³/3 − 4x]02 = −(8/3−8) = −(8/3−24/3) = 16/3 units²

Finding Area When f Changes Sign

This is the most common exam scenario — and where most errors occur. When the curve crosses the x-axis within [a,b]:

  1. Find all x-intercepts within [a,b] (set f(x) = 0).
  2. Split the integral at each x-intercept.
  3. Evaluate each piece separately.
  4. Take the absolute value of each piece and add them up.

Example: Find the area between y = x(x−2)(x−4) and the x-axis from x = 0 to x = 4.
Roots at x = 0, 2, 4. Positive on [0,2], negative on [2,4] (check: at x=1, f=1(−1)(−3)=3>0; at x=3, f=3(1)(−1)=−3<0).
A = ∫02 f(x)dx + |∫24 f(x)dx| = 4 + 4 = 8 units²

Exam trap: If asked to find ∫04 x(x−2)(x−4)dx, the answer is 0 (signed areas cancel). If asked for the total area enclosed, the answer is 8. Read the question carefully!

Strategy Guide and Common Errors

Step-by-step approach for area problems:

  1. Sketch the curve. Mark x-intercepts.
  2. Identify all regions (above/below x-axis).
  3. Set up separate integrals for each region.
  4. Evaluate and take absolute values.
  5. Sum for total area. Label with units².

Common errors:

  • Forgetting to split at x-intercepts when f changes sign.
  • Getting a negative answer and forgetting to negate it.
  • Using the wrong limits (forgetting that limits come from intersection with x-axis, not y-axis).
  • Confusing "find the integral" with "find the area" — they're different when f goes negative.

Mastery Practice

See Answers ➔
  1. Find the area enclosed by y = x² − 1 and the x-axis.
  2. Find the area enclosed by y = 3x − x² and the x-axis.
  3. Calculate ∫−12 (x³ − x) dx. Then find the actual area enclosed by y = x³ − x and the x-axis between x = −1 and x = 2.
  4. Find the area between y = cos x and the x-axis from x = 0 to x = π/2.
  5. Find the total area enclosed by y = x(x − 2)(x + 1) and the x-axis.
  6. The curve y = ex − 2 intersects the x-axis at x = ln 2. Find the area enclosed by the curve, the x-axis, and the lines x = 0 and x = ln 2.
  7. Find the area of the region bounded by y = 4 − x² and the x-axis. Include a sketch showing the region.
  8. A student calculates ∫02π sin x dx = 0 and concludes "the area enclosed by y = sin x and the x-axis on [0, 2π] is zero". Identify and correct the error, and state the correct total area.
  9. Find the total area enclosed between y = x³ − 3x and the x-axis. You must find all x-intercepts, determine the sign of f in each sub-interval, and show full working.
  10. A particle's velocity (in m/s) at time t seconds is given by v(t) = t² − 4t + 3 for 0 ≤ t ≤ 4.
    • (a) Find the x-intercepts of v(t) in the given domain.
    • (b) Calculate the total distance travelled by the particle (total area under |v(t)|).
    • (c) Calculate the displacement (signed integral) and explain the difference.