Applications of Integration: Areas, Volumes and Mean Values
What Integration Is For
Once you can integrate, a whole range of problems opens up. How much land lies between a river and a straight road? What is the volume of a vase, a bowl or a rocket nose cone? How far does a car travel if its speed keeps changing? What is the average value of an alternating current? All of these are answered by the applications of integration. This tutorial, written for MTH 101/102, Further Mathematics and engineering students, covers areas under curves (including regions below the x-axis), areas between two curves, areas with respect to the y-axis, volumes of revolution, mean values, displacement and distance from velocity, and the work done by a varying force. It assumes you can already evaluate definite integrals, as covered in the integration rules tutorial.
1. Area Under a Curve, Including Regions Below the Axis
The definite integral ∫ f(x) dx from a to b gives the signed area between the curve and the x-axis: parts above the axis count as positive and parts below count as negative.
Worked example 1: a region below the axis
Find the area enclosed by y = x² − 4x and the x-axis.
The curve crosses the axis where x(x − 4) = 0, so at x = 0 and x = 4. Between these, the curve is below the axis.
∫ (x² − 4x) dx from 0 to 4 = [x³/3 − 2x²] from 0 to 4 = 64/3 − 32 = −32/3.
The negative sign just means the region is below the axis. The area is 32/3 ≈ 10.67 square units.
Worked example 2: a curve that crosses the axis
Find the total area between y = x³ − x and the x-axis from x = −1 to x = 1.
Integrating straight across gives ∫ (x³ − x) dx from −1 to 1 = 0, because the region above the axis (from −1 to 0) cancels the region below it (from 0 to 1). That is not the area.
Split the integral at the root x = 0. From 0 to 1: [x⁴/4 − x²/2] = ¼ − ½ = −¼, so that part has area ¼. By symmetry, the part from −1 to 0 also has area ¼.
Total area = ½ square unit.
The rule is to find where the curve crosses the axis, integrate each piece separately, and add the absolute values.
2. Area Between Two Curves
To find the area between curves, suppose y = f(x) lies above y = g(x) from x = a to x = b, the area between them is:
Area = ∫ (f(x) − g(x)) dx from a to b, that is, the integral of (top − bottom).
The limits are usually the x-coordinates where the curves intersect.
Worked example 3
Find the area enclosed between the parabola y = x² and the line y = x + 2.
The curves meet where x² = x + 2, that is x² − x − 2 = 0, so (x − 2)(x + 1) = 0 and x = −1 or x = 2. Between them, the line is above the parabola (try x = 0: the line gives 2 and the parabola 0).
Area = ∫ (x + 2 − x²) dx from −1 to 2 = [x²/2 + 2x − x³/3] from −1 to 2.
At x = 2: 2 + 4 − 8/3 = 10/3. At x = −1: ½ − 2 + ⅓ = −7/6.
Area = 10/3 + 7/6 = 9/2 = 4.5 square units.
The "top minus bottom" formula works even when part of the region is below the x-axis, because the subtraction takes care of the signs.
3. Area with Respect to the y-Axis
Sometimes it is easier to integrate along the y-axis. The area between a curve x = g(y) and the y-axis, from y = c to y = d, is ∫ g(y) dy from c to d.
Worked example 4
Find the area between the curve y = √x, the y-axis and the line y = 3.
Rewrite the curve as x = y². The area = ∫ y² dy from 0 to 3 = [y³/3] from 0 to 3 = 9 square units.
4. Volumes of Revolution
Rotating a region through 360° about an axis sweeps out a solid, and its volume is called a volume of revolution. Slicing that solid into thin discs, each of radius y and thickness dx, gives the volume:
About the x-axis: V = π ∫ y² dx from a to b
About the y-axis: V = π ∫ x² dy from c to d
Worked example 5
The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. Find the volume generated.
V = π ∫ (√x)² dx from 0 to 4 = π ∫ x dx from 0 to 4 = π[x²/2] from 0 to 4 = 8π ≈ 25.1 cubic units. The solid is a bowl-shaped paraboloid.
Worked example 6: proving the cone formula
Rotating the line y = (r/h)x from x = 0 to x = h about the x-axis produces a cone of radius r and height h.
V = π ∫ (r²/h²)x² dx from 0 to h = π(r²/h²) × h³/3 = ⅓πr²h, the familiar formula. In the same way, rotating y = √(r² − x²) from −r to r gives the volume of a sphere, (4/3)πr³.
Worked example 7: the washer method
The region between y = x and y = x² (from x = 0 to x = 1) is rotated about the x-axis. Find the volume.
Each slice is a washer (a disc with a hole), with outer radius x and inner radius x². So V = π ∫ (outer² − inner²) dx:
V = π ∫ (x² − x⁴) dx from 0 to 1 = π(⅓ − ⅕) = 2π/15 cubic units.
Worked example 8: rotation about the y-axis
The region between y = x², the y-axis and the line y = 4 is rotated about the y-axis. Find the volume.
x² = y, so V = π ∫ y dy from 0 to 4 = π[y²/2] from 0 to 4 = 8π cubic units.
5. The Mean Value of a Function
The average value of f(x) over the interval from a to b is the height of the rectangle with the same area:
Mean value = (1 ÷ (b − a)) ∫ f(x) dx from a to b
Worked example 9
Find the mean value of f(x) = x² over the interval from 0 to 3.
Mean = ⅓ × [x³/3] from 0 to 3 = ⅓ × 9 = 3.
Worked example 10: alternating current
Find the mean value of sin x over half a cycle, from 0 to π.
Mean = (1/π) × [−cos x] from 0 to π = (1/π) × 2 = 2/π ≈ 0.637. This is why the average of a rectified a.c. signal is about 0.637 of its peak value.
6. Displacement and Distance from Velocity
Velocity is the rate of change of displacement, so integrating velocity over time gives the displacement. To find the total distance travelled, split the integral wherever the velocity changes sign, as with areas below the axis.
Worked example 11
A particle moves with velocity v = 3t² − 12t + 9 m/s. Find its displacement and the total distance it travels between t = 0 and t = 4 s.
Displacement: s(t) = t³ − 6t² + 9t, so the displacement = s(4) − s(0) = 64 − 96 + 36 = 4 m.
Distance: v = 3(t − 1)(t − 3), which is zero at t = 1 and t = 3, so the particle changes direction twice. s(0) = 0, s(1) = 4, s(3) = 0 and s(4) = 4.
Distance = 4 + 4 + 4 = 12 m. The particle goes forward 4 m, back 4 m, then forward 4 m again.
7. Work Done by a Varying Force
When a force is constant, work = force × distance. When the force varies with position, the work done is the area under the force–distance graph: W = ∫ F dx.
Worked example 12: stretching a spring
A spring obeys Hooke's law, F = kx, with k = 200 N/m. Find the work done in stretching it from an extension of 0.1 m to 0.3 m.
W = ∫ 200x dx from 0.1 to 0.3 = 100(0.3² − 0.1²) = 100 × 0.08 = 8 J.
Common Mistakes in Applications of Integration
- Integrating straight across a root when the area is asked for, so that the regions above and below the axis cancel.
- Subtracting the curves the wrong way round. Use top minus bottom. A negative result means the curves were swapped.
- Forgetting π, or forgetting to square y, in volumes of revolution.
- Using (outer − inner)² instead of outer² − inner² in the washer method.
- Confusing displacement with distance. Distance needs the integral split wherever the velocity is zero.
Applications of Integration: Practice Questions
- Find the area enclosed by y = 6x − x² and the x-axis.
- Find the area between y = x² and y = 2x.
- The line y = x from x = 0 to x = 3 is rotated about the x-axis. Find the volume generated, and check it with the cone formula.
- Find the volume generated when the region under y = x² from x = 0 to x = 2 is rotated about the x-axis.
- Find the mean value of f(x) = 2x + 1 over the interval from 1 to 3.
- A body moves with velocity v = 4 − 2t m/s. Find its displacement and the distance travelled in the first 3 seconds.
Answers: 1. The roots are 0 and 6, and ∫ (6x − x²) dx from 0 to 6 = 108 − 72 = 36 square units. 2. They meet at x = 0 and 2, so the area is ∫ (2x − x²) dx from 0 to 2 = 4 − 8/3 = 4/3. 3. π ∫ x² dx from 0 to 3 = 9π. A cone of radius 3 and height 3 has volume ⅓π × 9 × 3 = 9π. 4. π ∫ x⁴ dx from 0 to 2 = 32π/5. 5. ½ × [x² + x] from 1 to 3 = ½ × 10 = 5. 6. Displacement = [4t − t²] from 0 to 3 = 3 m. The velocity is zero at t = 2, so the body goes 4 m forward and then 1 m back: distance = 5 m.
Conclusion
Integration measures accumulated quantities. Areas come from ∫ y dx, split at roots and taking absolute values for regions below the axis, and from ∫ (top − bottom) dx between curves. Volumes of revolution come from π ∫ y² dx (or π ∫ x² dy about the y-axis), with washers for hollow solids. The mean value of a function is (1/(b − a)) ∫ f dx. Integrating velocity gives displacement (and, split at turning points, distance), and integrating force gives work. These are the core applications in MTH 101/102 and engineering mathematics.