01 · A horizontal line
Find the area between y = 10 − x² and y = 1.
Hint
The intersections have x = ±3; the gap is 9 − x².
Worked solution
Area = ∫−33(9 − x²) dx = 36.
Understand · explore · practise
Area between curves using integration
Find area between a curve and a line or two curves. Solve intersections, use upper minus lower, split changing boundaries and combine integrals with triangles and trapeziums.
Before you startDefinite integrals, simultaneous equations, sign checks and coordinate geometry
01 / Integrate the vertical gap
Upper minus lower is the strip’s height.
Area = ∫ab[upper curve − lower curve] dx
Use this when the order stays the same between a and b. A thin strip has height equal to the difference in y-values, regardless of whether either curve lies above the x-axis.
The model compares y = 7x − x² + c with y = 3x + c between 0 and 4. Move the strip and shift both curves together. Their heights change, but their gap and enclosed area stay the same.
Inspect the gap between the curvesMove at your pace
Common shift c = 0. At x = 2, upper y = 10, lower y = 6, and their vertical gap is 4. The enclosed area from 0 to 4 is 32/3 square units for every shown shift. Gold shading is the full enclosed region; the gold vertical segment is one strip height.
Watch vertical strips fill the gap between a line and a curve
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Find the boundaries
Set the two y-values equal.
For y = 7x − x² and y = 3x:
7x − x² = 3x
x(4 − x) = 0 ⇒ x = 0 or 4
Between the intersections, the parabola lies above the line: their difference is x(4 − x) > 0.
Area = ∫04[(7x − x²) − 3x] dx
= [2x² − x³/3]04
= 32/3
The line is part of the boundary. Integrating only the parabola would include the triangle below the line as well.
03 / Both curves can be below the axis
The upper curve has the greater y-value.
Subtract 15 from both curves in the first example. The upper curve 7x − x² − 15 and the lower line 3x − 15 are both below the axis throughout the region.
(7x − x² − 15) − (3x − 15) = 4x − x²
The area is still 32/3. Do not take absolute values of both y-values before subtracting: that can reverse their order. Compare the actual coordinates.
04 / Two curves, or a change of order
Check every relevant intersection.
Compare y = x² − 4x + 3 with y = x²/2 − x + 3. The second curve is higher between x = 0 and 6:
Upper − lower = −x²/2 + 3x
Area = ∫06(−x²/2 + 3x) dx = 18
When the upper curve changes
For y = x³/4 and y = x on −2 ≤ x ≤ 2, intersections occur at −2, 0 and 2. The cubic is higher on (−2,0); the line is higher on (0,2).
Area = ∫−20(x³/4 − x) dx
+ ∫02(x − x³/4) dx
= 1 + 1 = 2
A single signed difference over the whole symmetric interval is zero and would miss both regions.
05 / A boundary that switches
Below both curves means choosing the lower one.
In the first quadrant, find the region below both y = x² + 2 and y = 8 − x, above the x-axis and to the right of the y-axis.
The curves meet at x = 2 in this quadrant. From 0 to 2 the quadratic is lower; from 2 to 8 the line is lower. The line reaches the axis at 8.
Area = ∫02(x² + 2) dx
+ ∫28(8 − x) dx
= 20/3 + 18 = 74/3
The second part is also a triangle of base 6 and height 6. The enclosed lens between the two curves is a different region, with area 125/6 between their intersections −3 and 2.
06 / Use the specified boundary
A region bounded partly by the axis is not always the whole lens.
Take y = x(x − 4) and y = 2x. They meet at O(0,0) and C(6,12). The quadratic also meets the axis at B(4,0). We want the region enclosed by the line segment OC, the curve from C to B and the axis from B to O.
Area = ∫042x dx
+ ∫46[2x − x(x − 4)] dx
= 16 + 28/3 = 76/3
Alternatively take the large triangle under OC, with area 36, and subtract ∫ from 4 to 6 of x(x − 4), which is 32/3. The part of the quadratic below the axis is outside the requested region.
07 / A chord above a curve
Find the line equation, or use its trapezium area.
On y = x + 4/x², take A(1,5) and B(2,3). Their chord has slope −2, so AB is y = 7 − 2x. It lies above the curve between the endpoints.
Area under chord = ½(5 + 3)(2 − 1) = 4
Area under curve = [x²/2 − 4/x]12 = 7/2
Area between = 4 − 7/2 = 1/2
You can obtain the same result by integrating 7 − 2x − (x + 4/x²). The interval 1 to 2 does not cross the reciprocal’s singularity.
08 / Curves containing roots
Solve intersections without introducing invalid inputs.
Compare y = 5√x − x3/2 + 2 and y = 2 + x/2, for x ≥ 0. Write t = √x, so t ≥ 0. Their difference factors as:
5t − t³ − t²/2 = t(2 − t)(t + 5/2)
The valid roots are t = 0 and 2, giving x = 0 and 4. The root t = −5/2 is not a square root. The curve is above the line on (0,4).
Area = ∫04(5x1/2 − x3/2 − x/2) dx
= [(10/3)x3/2 − (2/5)x5/2 − x²/4]04
= 148/15
A cube-root curve, a line and the axis
Let f(x) = x2/3 − 3x−1/3 + 2 for x > 0. It crosses the axis at (1,0) and passes through A(8,9/2). Join A to B(5,0). The line AB is y = (3/2)(x − 5).
The required region follows the curve from x = 1 to 8, the segment AB and the axis back to 1. Its area is the integral under the curve minus the triangle under AB:
∫18f(x) dx = 191/10
Triangle area = ½ × 3 × 9/2 = 27/4
Required area = 247/20
The curve is above the chord segment: it is concave down on x > 0, and its value at x = 5 is positive. A sketch helps identify the part bounded by the axis.
09 / Recover the equations first
Intersection coordinates impose algebraic conditions.
The curve y = p + 8x − x² meets y = qx + 12 at x = 2 and x = 5. Their difference has leading coefficient −1 and these two roots:
−x² + (8 − q)x + (p − 12)
= −(x − 2)(x − 5) = −x² + 7x − 10
q = 1, p = 2
The first intersection is A(2,14). A horizontal line through A meets the curve again at C(6,14). The small region between AC and the curve therefore has area:
∫26[(2 + 8x − x²) − 14] dx = 32/3
The original sloping line helped determine the curve; it is not the final region’s lower boundary.
10 / Equal areas can determine a line
Combine a curved part with a triangle.
For f(x) = x²(x + 3), the finite region between the negative x-axis and the curve has area 27/4. Another region follows the curve from O to A(1,4), then a straight line to B(b,0), then the positive axis back to O, with b > 1.
Area under curve from 0 to 1 = 5/4
Triangle from x = 1 to b has area ½(b − 1)4
Second area = 5/4 + 2(b − 1)
Equating the two areas gives 5/4 + 2(b − 1) = 27/4, hence b = 15/4. Check b > 1 before accepting the geometry.
11 / Your turn
Name the upper and lower boundary on every interval.
Use exact values and sketch the requested region before calculating.
02 · A sloping line
Find the enclosed area between y = 8x − x² and y = 3x.
Hint
The gap is 5x − x² and its zeros are 0 and 5.
Worked solution
Area = ∫05(5x − x²) dx = 125/6.
03 · Below zero
Find the area enclosed by y = −x² − 2 and y = −6.
Hint
The upper minus lower difference is 4 − x².
Worked solution
Limits −2 and 2; area = 32/3.
04 · Two quadratics
Find the enclosed area between y = x² and y = 3x² − 6x.
Hint
The first curve is higher on (0,3).
Worked solution
Area = ∫03(6x − 2x²) dx = 9.
05 · Swapping order
Find the total area between y = x³ and y = 4x on −2 ≤ x ≤ 2.
Hint
Split at zero. Each half has area 4.
Worked solution
Total area = 8.
The signed difference across the whole interval is zero, which is not the geometric area.
06 · A reciprocal chord
Find the area between y = 2/x² and its chord joining the points at x = 1 and x = 2.
Hint
The endpoint heights are 2 and 1/2.
Worked solution
Trapezium area = ½(2 + 1/2)(1) = 5/4
Curve integral = [−2/x]12 = 1
Gap area = 1/4.
07 · Below both boundaries
Find the first-quadrant region below both y = x² + 3 and y = 9 − x, bounded by the axes.
Hint
The positive intersection is x = 2. Switch from curve to line there.
Worked solution
Area = ∫02(x² + 3) dx
+ ∫29(9 − x) dx
= 26/3 + 49/2 = 199/6.
08 · A region excluding the lower lobe
For y = x(x − 2) and y = x, find the region enclosed by the line from O to (3,3), the curve back to (2,0), and the x-axis back to O.
Hint
Subtract the curve integral from 2 to 3 from the large triangle.
Worked solution
Area = 9/2 − 4/3 = 19/6.
09 · A root and a line
Find the area between y = √x and y = x/2 for x ≥ 0.
Hint
The intersections are 0 and 4; the root curve is higher between them.
Worked solution
Area = ∫04(√x − x/2) dx
= 16/3 − 4 = 4/3.
10 · Unknown equations
The curve y = p + 10x − x² meets y = qx + 18 at x = 1 and 6. Find p and q, then the area between the curve and the horizontal chord through the first intersection.
Hint
The difference is −(x − 1)(x − 6). The first height is 21.
Worked solution
p = 12, q = 3.
The horizontal chord y = 21 meets the curve at x = 1 and 9.
Area = ∫19(−x² + 10x − 9) dx = 256/3.
11 · An equal-area line
For f(x) = x²(x + 2), the negative-side enclosed area equals the region formed by the curve from O to A(1,3), the line to B(b,0), and the positive axis, with b > 1. Find b.
Hint
The negative-side area is 4/3; the integral from 0 to 1 is 11/12.
Worked solution
11/12 + (3/2)(b − 1) = 4/3
b = 23/18.
This satisfies b > 1.
12 · A common vertical shift
Two curves enclose a region of area 7. Both are shifted down by 20. What is the corresponding enclosed area?
Hint
Subtract the two shifted y-values.
Worked solution
The area remains 7. Their difference and their intersection x-values are unchanged, even if the region moves below the x-axis.
12 / Recap
The boundary determines the integral.
- Solve intersections before choosing the limits.
- Integrate upper minus lower, even below the axis.
- Split when the curves exchange order or the boundary changes.
- Use triangle or trapezium areas when they simplify a straight boundary.
- Recover unknown equations before integrating the final region.
- Reject algebraic roots that do not describe the requested geometry.
Section 1 of 12 · Integrate the vertical gap