Chapter 03 — Integration | Part 5: Applications of Integration (Area, Volume, Motion and Centre of Gravity)

MATHEMATICS HIPHOP ACADEMY

CHAPTER 03 — INTEGRATION

PART 5: APPLICATIONS OF INTEGRATION

Integration is not only a method of reversing differentiation. It is also a powerful mathematical tool for calculating areas, volumes, displacement, velocity, acceleration and centres of gravity.

In this lesson, we apply definite integration to practical mathematical problems and develop the skills needed to solve examination questions involving these applications.

LEARNING OBJECTIVES

By the end of this lesson, you should be able to:

• Calculate the area under a curve using definite integration.
• Find the area enclosed between two curves.
• Calculate volumes of solids formed by rotating curves about coordinate axes.
• Use integration to determine displacement and velocity from acceleration.
• Determine acceleration from a velocity function.
• Find the centre of gravity of uniform rods and simple laminae.
• Interpret definite integrals in practical and examination problems.

SECTION 1: AREA UNDER A CURVE

1.1 Understanding area as an integral

Consider a continuous curve \(y=f(x)\) above the x-axis between \(x=a\) and \(x=b\).

The area between the curve, the x-axis and the vertical lines \(x=a\) and \(x=b\) is given by:

A = ∫_a^b y dx

This formula applies directly when the curve lies above the x-axis throughout the interval.

1.2 Worked Example 1: Area under a curve

Question: Find the area enclosed by the curve \(y=x^2+2\), the x-axis, and the lines \(x=0\) and \(x=2\).

Solution

The required area is:

A = ∫₀²(x² + 2) dx

Integrating:

A = [x³/3 + 2x]₀²

Substituting the limits:

A = (8/3 + 4) − 0

A = 8/3 + 12/3

A = 20/3 square units.

1.3 When a curve lies below the x-axis

A definite integral gives signed area. If a curve lies below the x-axis, its integral over that interval is negative, even though geometrical area is positive.

Therefore, geometrical area is calculated by reversing the sign of the integral where necessary.

Geometrical area = ∫_a^b |f(x)| dx

Worked Example 2: Curve below the x-axis

Question: Find the area between the curve \(y=x-3\), the x-axis and the lines \(x=0\) and \(x=2\).

Solution

For \(0≤x≤2\), the curve is below the x-axis.

Hence:

A = −∫₀²(x−3) dx

A = −[x²/2 − 3x]₀²

A = −[(2−6)−0]

A = 4 square units.

1.4 Area when a curve crosses the x-axis

If a curve crosses the x-axis within the interval, divide the integral at the point where \(y=0\). Calculate each geometrical area separately and add the positive results.

Worked Example 3: Find the total area between \(y=x^2-4\) and the x-axis from \(x=0\) to \(x=3\).

First find where the curve crosses the x-axis:

x² − 4 = 0

x = ±2

In the interval 0 to 3, the crossing occurs at x = 2.

From 0 to 2, the curve is below the x-axis:

A₁ = −∫₀²(x²−4) dx

= −[x³/3 − 4x]₀²

= 16/3

From 2 to 3, the curve is above the x-axis:

A₂ = ∫₂³(x²−4) dx

= [x³/3 − 4x]₂³

= 7/3

Total area = 16/3 + 7/3

Total area = 23/3 square units.

SECTION 2: AREA BETWEEN TWO CURVES

2.1 The general principle

When two curves enclose a region, the area between them is found by integrating the difference between the upper curve and the lower curve.

A = ∫_a^b (upper curve − lower curve) dx

The limits are usually obtained by finding the points where the two curves intersect.

2.2 Finding the points of intersection

At the points of intersection, both curves have the same y-coordinate. Therefore, equate their equations and solve for x.

Worked Example 4: Find the area enclosed by the curves \(y=2x\) and \(y=x^2\).

Solution

At the points of intersection:

2x = x²

x² − 2x = 0

x(x−2) = 0

Therefore, x = 0 or x = 2.

Between these limits, \(2x\) is above \(x²\).

Area = ∫₀²(2x−x²) dx

= [x² − x³/3]₀²

= 4 − 8/3

Area = 4/3 square units.

2.3 Integration with respect to y

Sometimes it is easier to express the curves in terms of y. In such cases, horizontal strips are used and integration is performed with respect to y.

A = ∫_c^d (right boundary − left boundary) dy

Important examination point: With respect to x, subtract lower from upper. With respect to y, subtract left from right.

Worked Example 5: Find the area enclosed by \(x=y^2\) and \(x=2y\).

At the intersections:

y² = 2y

y(y−2) = 0

Thus, y = 0 and y = 2.

For this interval, the right boundary is x = 2y and the left boundary is x = y².

A = ∫₀²(2y−y²) dy

= [y² − y³/3]₀²

A = 4/3 square units.

SECTION 3: VOLUMES OF SOLIDS OF REVOLUTION

3.1 Meaning of a solid of revolution

A solid of revolution is formed when a plane region is rotated completely through one full revolution, usually 360°, about a fixed axis.

For example, rotating a rectangle about one of its sides produces a cylinder, while rotating a semicircle about its diameter produces a sphere.

3.2 Rotation about the x-axis

When the region under \(y=f(x)\) is rotated about the x-axis, circular cross-sections are formed.

The radius of each circular cross-section is y. Its area is πy².

V = π∫_a^b y² dx

Worked Example 6: Volume about the x-axis

Question: Find the volume generated when the curve \(y=x\), between \(x=0\) and \(x=3\), is rotated about the x-axis.

Using:

V = π∫₀³ y² dx

Since y = x:

V = π∫₀³ x² dx

= π[x³/3]₀³

= π(27/3)

V = 9π cubic units.

3.3 Rotation about the y-axis

When the region is rotated about the y-axis, the radius of each circular cross-section is the x-coordinate.

If x is expressed as a function of y, the volume is:

V = π∫_c^d x² dy

Alternatively, for a region described using vertical strips, the cylindrical shell method may be used:

V = 2π∫_a^b x y dx

Here, x is the distance from the y-axis and y is the height of each cylindrical shell.

Remember: The disk method uses circular cross-sections. The shell method uses cylindrical shells. Choose the method that gives the simplest integral for the region.

3.4 Volume between two curves

If a region between two curves is rotated about the x-axis, the cross-section may be a washer with an outer radius R and an inner radius r.

V = π∫_a^b (R² − r²) dx

Worked Example 7: The region between \(y=2\) and \(y=x\), for \(0≤x≤2\), is rotated about the x-axis. Find the volume.

The outer radius is 2 and the inner radius is x.

V = π∫₀²(4−x²) dx

= π[4x − x³/3]₀²

= π(8−8/3)

V = 16π/3 cubic units.

SECTION 4: APPLICATIONS TO MOTION

4.1 Displacement, velocity and acceleration

In straight-line motion, displacement, velocity and acceleration are related through differentiation and integration.

Quantity	Meaning	Relationship
Displacement, s	Position relative to an origin	—
Velocity, v	Rate of change of displacement	v = ds/dt
Acceleration, a	Rate of change of velocity	a = dv/dt

By integration:

s = ∫v dt

v = ∫a dt

Integration introduces constants of integration. These constants are determined using initial conditions, such as the initial velocity or initial displacement.

4.2 Finding velocity from acceleration

Worked Example 8: A particle moves in a straight line with acceleration \(a=6t+4\). Its initial velocity is 3 m/s. Find its velocity after 2 seconds.

Since a = dv/dt:

dv/dt = 6t + 4

Integrating:

v = 3t² + 4t + C

At t = 0, v = 3:

3 = 0 + 0 + C

Therefore, C = 3.

Hence:

v = 3t² + 4t + 3

At t = 2:

v = 3(4) + 4(2) + 3

v = 23 m/s.

4.3 Finding displacement from velocity

Worked Example 9: A particle has velocity \(v=4t+2\). If its displacement at t = 0 is 5 metres, find its displacement after 3 seconds.

Since v = ds/dt:

ds/dt = 4t + 2

Integrating:

s = 2t² + 2t + C

At t = 0, s = 5:

5 = C

Therefore:

s = 2t² + 2t + 5

At t = 3:

s = 2(9) + 2(3) + 5

s = 29 metres.

4.4 Finding displacement when acceleration is given

Sometimes acceleration is given as a function of time and displacement is required. In such cases, integrate twice, using the initial conditions to determine both constants.

Worked Example 10: A particle moves with acceleration \(a=12t\). Initially, its velocity is 2 m/s and its displacement is 4 metres. Find its displacement after 2 seconds.

First integrate acceleration:

dv/dt = 12t

v = 6t² + C₁

At t = 0, v = 2, so C₁ = 2.

Thus v = 6t² + 2.

Now integrate velocity:

ds/dt = 6t² + 2

s = 2t³ + 2t + C₂

At t = 0, s = 4, so C₂ = 4.

Thus s = 2t³ + 2t + 4.

At t = 2:

s = 2(8) + 4 + 4

s = 24 metres.

4.5 Distance and displacement

Displacement is the change in position and may be positive, negative or zero. Distance is the total length travelled and is never negative.

If velocity changes sign during an interval, the particle changes direction. To find total distance, identify the times when v = 0 and calculate the magnitude of displacement on each interval.

Worked Example 11: A particle has velocity \(v=2t-4\) m/s for \(0≤t≤3\). Find its displacement and total distance travelled.

Displacement:

∫₀³(2t−4) dt

= [t²−4t]₀³

= 9−12

Displacement = −3 m.

To find the total distance, first solve v = 0:

2t−4 = 0, giving t = 2.

From 0 to 2, velocity is negative:

Distance₁ = −∫₀²(2t−4)dt = 4 m.

From 2 to 3, velocity is positive:

Distance₂ = ∫₂³(2t−4)dt = 1 m.

Total distance = 4 + 1.

Total distance = 5 m.

SECTION 5: CENTRE OF GRAVITY

5.1 Meaning of centre of gravity

The centre of gravity of a body is the point through which its resultant weight may be considered to act.

For a uniform body in a uniform gravitational field, the centre of gravity coincides with its centre of mass.

5.2 Centre of gravity of a system of particles

For particles of masses \(m_1,m_2,\ldots,m_n\) positioned at distances \(x_1,x_2,\ldots,x_n\) along a straight line, the centre of mass is:

x̄ = (Σmx)/(Σm)

This is the total moment about the origin divided by the total mass.

Worked Example 12: Three particles of masses 2 kg, 3 kg and 5 kg are positioned at x = 1 m, 4 m and 6 m respectively. Find the centre of mass.

x̄ = Σmx/Σm

= [(2×1)+(3×4)+(5×6)]/(2+3+5)

= (2+12+30)/10

x̄ = 4.4 m.

5.3 Centre of gravity of a uniform rod

A uniform rod has the same mass per unit length throughout its length. Its centre of gravity is therefore at its midpoint.

For a rod extending from x = 0 to x = L:

x̄ = L/2

5.4 Centre of gravity of a non-uniform rod

If the linear mass density of a thin rod varies with position, let its density be λ(x). A small element of length dx has mass:

dm = λ(x) dx

The total mass is:

M = ∫_a^b λ(x) dx

The centre of mass is obtained by dividing the total moment by the total mass:

x̄ = [∫_a^b xλ(x) dx] / [∫_a^b λ(x) dx]

Worked Example 13: A thin rod extends from x = 0 to x = 2 metres. Its linear density is λ(x) = 2x kg/m. Find its centre of mass.

First find the total mass:

M = ∫₀² 2x dx

= [x²]₀² = 4 kg.

Now find the moment about the origin:

Moment = ∫₀² x(2x) dx

= 2∫₀² x² dx

= 2[x³/3]₀²

= 16/3 kg m.

Therefore:

x̄ = (16/3)/4

x̄ = 4/3 metres.

5.5 Centre of gravity of a uniform lamina

A lamina is a thin, flat sheet of material. For a uniform lamina, mass is proportional to area. Its centre of gravity is therefore the centroid of its area.

For a region bounded by y = f(x), the x-axis and the lines x = a and x = b, with f(x) non-negative:

A = ∫_a^b y dx

The x-coordinate of its centroid is:

x̄ = [∫_a^b xy dx] / A

The y-coordinate of its centroid is:

ȳ = [½∫_a^b y² dx] / A

These expressions apply to the region between the curve and the x-axis. For other regions, the appropriate area and moments must be used.

Worked Example 14: Find the centroid of the uniform lamina bounded by \(y=x\), the x-axis, x = 0 and x = 2.

First calculate its area:

A = ∫₀²x dx

= [x²/2]₀² = 2.

Calculate x̄:

x̄ = (1/A)∫₀²x(x) dx

= ½∫₀²x² dx

= ½[8/3]

x̄ = 4/3.

Calculate ȳ:

ȳ = (1/2A)∫₀²x² dx

= ¼[8/3]

ȳ = 2/3.

Therefore, the centroid is:

(4/3, 2/3).

SECTION 6: EXAMINATION STRATEGIES

Common mistakes to avoid

• Forgetting to split an area integral when a curve crosses the x-axis.
• Subtracting the lower curve from the upper curve incorrectly.
• Using the wrong limits of integration.
• Forgetting to square the radius when calculating volume by the disk method.
• Confusing displacement with total distance travelled.
• Forgetting the constant of integration when finding velocity or displacement.
• Failing to use initial conditions to determine constants.
• Dividing the moment by the wrong total mass or area when finding a centre of gravity.
• Giving square units for volume instead of cubic units.

Useful formula summary

Application	Formula
Area under a curve	∫ y dx
Area between curves	∫(upper − lower) dx
Volume about x-axis	π∫ y² dx
Volume about y-axis using disks	π∫ x² dy
Volume using cylindrical shells	2π∫ xy dx
Velocity	v = ∫a dt
Displacement	s = ∫v dt
Centre of mass of particles	x̄ = Σmx/Σm
Centre of mass of a rod	x̄ = ∫xλ dx / ∫λ dx

SECTION 7: PRACTICE EXERCISES

Attempt all the questions before opening the answers. Show all working clearly and include the correct units where necessary.

Exercise 1: Areas

1. Find the area under y = 3x² + 1 between x = 0 and x = 2.
2. Find the total area between y = x² − 9 and the x-axis from x = 0 to x = 4.
3. Find the area enclosed by y = 3x and y = x².
4. Find the area enclosed by x = y² and x = 4y.

Open answers and solutions

1. ∫₀²(3x²+1)dx = [x³+x]₀² = 10 square units.

2. The curve crosses at x = 3. Total area = 18 + 10/3 = 64/3 square units.

3. Intersections: x = 0 and x = 3. Area = ∫₀³(3x−x²)dx = 9/2 square units.

4. Intersections: y = 0 and y = 4. Area = ∫₀⁴(4y−y²)dy = 32/3 square units.

Exercise 2: Volumes

1. Find the volume formed when y = 2x, from x = 0 to x = 2, is rotated about the x-axis.
2. Find the volume formed when y = √x, from x = 0 to x = 4, is rotated about the x-axis.
3. The region between y = 3 and y = x, for 0 ≤ x ≤ 3, is rotated about the x-axis. Find the volume.

Open answers and solutions

1. V = π∫₀²4x²dx = 32π/3 cubic units.

2. V = π∫₀⁴x dx = 8π cubic units.

3. V = π∫₀³(9−x²)dx = 18π cubic units.

Exercise 3: Motion

1. A particle has acceleration a = 8t + 2. Its initial velocity is 4 m/s. Find its velocity at t = 3 seconds.
2. A particle has velocity v = 6t − 4. Its initial displacement is 2 metres. Find its displacement after 4 seconds.
3. A particle has velocity v = t² − 4t + 3 for 0 ≤ t ≤ 4. Find its displacement and total distance travelled over this interval.

Open answers and solutions

1. v = 4t² + 2t + 4. At t = 3, v = 46 m/s.

2. s = 3t² − 4t + 2. At t = 4, s = 34 m.

3. Displacement = ∫₀⁴(t²−4t+3)dt = 4/3 m. The velocity changes sign at t = 1 and t = 3. Total distance = 4 m.

Exercise 4: Centre of Gravity

1. Two particles of masses 4 kg and 6 kg are positioned at x = 2 m and x = 7 m. Find their centre of mass.
2. A rod extends from x = 0 to x = 3 metres and has density λ(x) = x kg/m. Find its centre of mass.
3. Find the centroid of the uniform lamina bounded by y = 2x, the x-axis, x = 0 and x = 3.

Open answers and solutions

1. x̄ = [(4×2)+(6×7)]/10 = 5 m.

2. M = ∫₀³x dx = 9/2. Moment = ∫₀³x² dx = 9. Therefore x̄ = 2 m.

3. A = 9. x̄ = (1/9)∫₀³2x²dx = 2 m. ȳ = (1/18)∫₀³4x²dx = 2 m. Centroid = (2, 2).

FINAL REVISION CHECKLIST

Before moving to Part 6, make sure you can:

• Calculate areas using definite integration.
• Handle curves above and below the x-axis.
• Find areas enclosed by two curves.
• Calculate volumes of revolution using disks and washers.
• Use cylindrical shells where appropriate.
• Obtain velocity and displacement by integration.
• Distinguish between distance and displacement.
• Find the centre of mass of particles and rods.
• Calculate the centroid of a simple uniform lamina.

CONCLUSION

Integration provides a link between algebraic mathematics and practical applications. Through definite integration, we can calculate geometrical areas and volumes. Through integration in motion, we can determine velocity and displacement from acceleration. Through moments and mass distributions, we can locate centres of gravity.

Mastering these applications prepares you for more advanced mathematical problems and examination questions.

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