01 · Find new intersections
Where do y = x² and y = 3x meet?
Hint
Solve x² = 3x.
Worked solution
x = 0 and x = 3, giving points (0,0) and (3,9).
Understand · explore · practise
Area between curves
Find areas between curves using intersections and upper minus lower. Handle changing curve order, touching intersections, exact exponential, logarithmic and trigonometric areas, and area ratios.
Before you startDefinite integrals, solving intersections, signed area and graph sketches.
01 / Integrate the vertical gap
Use the upper curve minus the lower curve.
Area = ∫ₐᵇ [upper function − lower function] dx
If the ordering changes, split the interval and change the subtraction order.
The height of a narrow vertical strip is the gap between the curves. Its position above or below the x-axis does not matter: a gap is still upper minus lower. Find the intersections and establish the ordering before integrating.
Check which curve is aboveExplore
Integral in the chosen order = 10.666667.
Geometric area between the curves = 10.666667.
On [−1, 3], g is above f.
The chosen order is nonnegative throughout this interval.
Green bands mean your chosen subtraction is positive; amber bands mean it is negative. Use the upper curve minus the lower on each piece. Numeric areas use exact polynomial primitives, displayed as decimals.
02 / Find the bounded region first
Intersections supply the limits when no vertical boundaries are given.
Find the area enclosed by y = x² and y = 2x + 3.Worked example
x² = 2x + 3 → (x − 3)(x + 1) = 0
The intersections have x = −1 and x = 3.
At x = 0, the line is 3 and the parabola is 0
The line lies above between the two intersections.
∫₋₁³ (2x + 3 − x²) dx
Use line minus parabola.
[x² + 3x − x³/3]₋₁³ = 32/3
Report square units.
Watch: vertical gaps build the area
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Establish the ordering on each interval
One test point works between consecutive intersections for continuous curves.
Find the area between y = x² and y = 3x.Worked example
The limits are 0 and 3
These are the intersection x-values.
At x = 1, 3x > x²
The line is the upper curve throughout (0,3).
∫₀³ (3x − x²) dx
Integrate the gap.
[3x²/2 − x³/3]₀³ = 9/2
The result is positive.
02 · Correct the order
On [0,1], which is above: y = x or y = x²?
Hint
Compare them at x = 1/2.
Worked solution
x is above x², so the gap is x − x².
03 · A simple bounded area
Find the area enclosed by y = x and y = x².
Hint
Use limits 0 and 1.
Worked solution
∫₀¹ (x − x²) dx = 1/6 square unit.
04 / Split when the upper curve changes
A single subtraction can cancel separate geometric regions.
Find total area between y = x² and y = 2x + 3 from x = −2 to x = 4.Worked example
Intersections inside the interval occur at −1 and 3
These divide the interval into three pieces.
Parabola above on [−2,−1] and [3,4]; line above on [−1,3]
Test the signs of 2x + 3 − x².
Outer areas are 7/3 each; middle area is 32/3
Use the appropriate order on each piece.
Total area = 46/3
Integrating line minus parabola without splitting would give 6 instead.
04 · A changed ordering
For y = x and y = x³ on [−1,1], where does the ordering change inside the interval?
Hint
Solve x = x³ and examine the interior root.
Worked solution
At x = 0. On (−1,0), x³ is above x; on (0,1), x is above x³.
05 · Use odd symmetry carefully
Find the total area between y = x and y = x³ on [−1,1].
Hint
The two regions have equal area.
Worked solution
2∫₀¹ (x − x³) dx = 1/2 square unit.
05 / The x-axis is not the boundary unless stated
Two negative curves can enclose a positive gap.
Find the area between y = −x² − 1 and y = −5 on [−2,2].Worked example
−x² − 1 ≥ −5 on this interval
The parabola is the upper curve.
Gap = (−x² − 1) − (−5) = 4 − x²
Subtracting the lower negative function adds 5.
∫₋₂² (4 − x²) dx = 32/3
No extra sign reversal is needed because the vertical gap is nonnegative.
06 · Both curves below zero
If the upper curve has y = −2 and the lower has y = −7 at a point, what is the strip height?
Hint
Subtract the lower value.
Worked solution
−2 − (−7) = 5.
07 · Fixed vertical boundaries
Find the area between y = x + 2 and y = x − 1 from x = 0 to x = 4.
Hint
The vertical gap is constant.
Worked solution
∫₀⁴ 3 dx = 12 square units.
06 / Use exact exponential primitives
A tangent contact need not reverse the ordering.
Find the area between y = eˣ and y = 1 + x on [0,1].Worked example
eˣ ≥ 1 + x, with equality at x = 0
For x > 0, the difference has derivative eˣ − 1 > 0 and starts at zero.
∫₀¹ (eˣ − 1 − x) dx
Use exponential minus line.
[eˣ − x − x²/2]₀¹ = e − 5/2
Keep the exact positive result.
The curves touch at zero and eˣ remains above 1 + x on both sides. An intersection does not automatically mean the subtraction order changes.
08 · Area across a touching point
Find the area between eˣ and 1 + x on [−1,1].
Hint
The exponential is above throughout; integrate the same gap across zero.
Worked solution
e − e⁻¹ − 2 square units.
09 · Explain the negative side
Why is eˣ − 1 − x positive for x < 0?
Hint
Its derivative is negative there and its value at zero is zero.
Worked solution
As x increases towards zero, the difference decreases to zero; therefore it is positive to the left.
07 / Split logarithmic curves at exact intersections
Check the logarithm’s domain as well as the ordering.
Find total area between y = ln x and y = 1 on [1,e²].Worked example
ln x = 1 gives x = e
The entire interval lies in x > 0.
On [1,e], integrate 1 − ln x
A primitive is 2x − x ln x; this area is e − 2.
On [e,e²], integrate ln x − 1
A primitive is x ln x − 2x; this area is e.
Total area = 2e − 2
Both positive areas add.
10 · Logarithm and the axis
Find the area between y = ln x and y = 0 from x = 1 to x = e.
Hint
ln x is nonnegative on this interval.
Worked solution
[x ln x − x]₁ᵉ = 1 square unit.
08 / Find exact trig intersections
Use radians and check both subintervals.
Find the area between y = sin x and y = cos x on [0,π/2].Worked example
sin x = cos x gives x = π/4
This is the only interior intersection.
Cosine is above before π/4; sine is above after it
The ordering reverses.
∫₀^(π/4) (cos x − sin x) dx = √2 − 1
Use sin x + cos x as the primitive.
The second region has the same area
By symmetry about x = π/4.
Total = 2√2 − 2
Do not use the signed gap integral over the whole interval.
11 · Check the midpoint ordering
On (π/4,π/2), which function is larger, sin x or cos x?
Hint
Consider the endpoint π/2 and continuity between intersections.
Worked solution
sin x.
12 · A horizontal trig boundary
Find the area between y = sin x and y = 1 on [0,π].
Hint
The line is above sine, touching at π/2.
Worked solution
∫₀^π (1 − sin x) dx = π − 2.
09 / Name each region before forming a ratio
Areas in a ratio must describe the intended geometric pieces.
On [0,1], compare the region between y = x and y = x² with the region under y = x² above the axis.Worked example
Between the curves: ∫₀¹ (x − x²) dx = 1/6
This is the upper region.
Under the parabola: ∫₀¹ x² dx = 1/3
This is the lower region.
Ratio = (1/6):(1/3) = 1:2
Put the regions in the requested order.
13 · A region split by a vertical line
Under y = x(4 − x) on [0,4], find the ratio of the area on [0,1] to the area on [1,4].
Hint
The first area is 5/3 and the total is 32/3.
Worked solution
The remaining area is 9, so the ratio is 5:27.
10 / Turn a stated area into an equation
Check any parameter restriction after solving.
A horizontal line y = h lies above y = x² on [0,3], where h ≥ 9. Their area is 18. Find h.Worked example
∫₀³ (h − x²) dx = 18
The condition h ≥ 9 establishes the ordering.
3h − 9 = 18
Integrate the constant h and the parabola.
h = 9
This satisfies the restriction and gives contact at x = 3.
14 · A parameterised constant gap
Two parallel curves have constant vertical gap k > 0 on an interval of width 5. Their area is 12. Find k.
Hint
Area is gap times interval width.
Worked solution
k = 12/5.
15 · Check before using an equation
Why was h ≥ 9 needed in the worked example?
Hint
The maximum of x² on [0,3] is 9.
Worked solution
It ensures h − x² is nonnegative throughout, so one subtraction order gives the geometric area.
11 / Find intersections, test ordering, integrate the gap
Touching and crossing intersections behave differently.
- Identify the actual horizontal bounds of the region.
- Solve the curve-intersection equation.
- Check which function is upper on each interval.
- Split only when ordering or the formula requires it.
- Integrate upper minus lower and add positive areas.
- Define each region clearly in ratio or parameter questions.
16 · A final enclosed region
Find the area enclosed by y = 4 − x² and y = 0.
Hint
The intersections are −2 and 2; the parabola is above between them.
Worked solution
∫₋₂² (4 − x²) dx = 32/3 square units.
Section 1 of 11 · Integrate the vertical gap