CHAPTER 03: CALCULUS – INTEGRATION

MATHEMATICS HIP-HOP ACADEMY • A-LEVEL PURE MATHEMATICS

CHAPTER 03 — INTEGRATION

PART 1: FOUNDATIONS AND ELEMENTARY INTEGRATION

Integration is a central idea of calculus. It reverses differentiation and provides methods for finding accumulated quantities such as areas and total changes.

1. Learning Objectives

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

• Explain integration as the reverse of differentiation.
• Distinguish indefinite and definite integrals.
• Apply the basic rules of integration.
• Integrate algebraic expressions involving positive, fractional and negative powers.
• Find the constant of integration using a given condition.
• Evaluate simple definite integrals.

2. Meaning of Integration

Differentiation finds the rate at which a function changes. Integration reverses this process: it recovers a function from its derivative.

If dy/dx = f(x), then y = ∫ f(x) dx

For example, since the derivative of x³ is 3x², an integral of 3x² is x³.

Remember: Differentiation moves from a function to its derivative; integration works backwards.

3. Indefinite Integrals and the Constant C

An indefinite integral gives a family of functions. We include + C because the derivative of a constant is zero.

∫ f(x) dx = F(x) + C

For example, x² + 1, x² − 7 and x² + 100 all have derivative 2x. Therefore:

∫ 2x dx = x² + C

Always include the constant of integration when evaluating an indefinite integral.

4. Basic Rules of Integration

A. Power Rule

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, where n ≠ −1

Add one to the power, then divide by the new power.

B. Constant Rule

∫ k dx = kx + C

C. Constant Multiple Rule

∫ kf(x) dx = k∫f(x) dx

D. Sum and Difference Rules

∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx

Integrate each term separately.

5. Worked Examples

Example 1: A Simple Power

Evaluate ∫x⁴ dx.

Increase the power: 4 + 1 = 5. Divide by 5.
∫x⁴ dx = x⁵/5 + C.

Answer: x⁵/5 + C

Example 2: A Polynomial

Evaluate ∫(6x² − 4x + 7)dx.

∫6x²dx = 2x³
∫−4x dx = −2x²
∫7dx = 7x
Combine the terms and add C.

Answer: 2x³ − 2x² + 7x + C

Example 3: Fractional Powers

Evaluate ∫(3√x + 2/√x)dx.

Rewrite using indices: 3x^1/2 + 2x^−1/2.
Integrate: 3 × x^3/2/(3/2) + 2 × x^1/2/(1/2) + C.

Answer: 2x^3/2 + 4x^1/2 + C

Example 4: Negative Powers

Evaluate ∫(5x⁻³ − 4x⁻²)dx.

5 × x⁻²/(−2) − 4 × x⁻¹/(−1) + C

Answer: −5/(2x²) + 4/x + C

Example 5: Finding C

Given dy/dx = 6x − 4 and y = 5 when x = 2, find y.

Integrate: y = 3x² − 4x + C.
Substitute x = 2 and y = 5:
5 = 12 − 8 + C, so C = 1.

Answer: y = 3x² − 4x + 1

6. Definite Integration

A definite integral has a lower and an upper limit. It gives a numerical value.

∫ₐᵇ f(x)dx = [F(x)]ₐᵇ = F(b) − F(a)

Find an antiderivative, substitute the upper limit, then subtract the value at the lower limit.

Example 6

Evaluate ∫₁³ 2x dx.

Antiderivative: x².
[x²]₁³ = 3² − 1² = 9 − 1 = 8.

Answer: 8

Example 7

Evaluate ∫₀²(3x² + 2)dx.

Antiderivative: x³ + 2x.
[x³ + 2x]₀² = (8 + 4) − 0 = 12.

Answer: 12

A definite integral gives signed area. Portions below the x-axis contribute negative values. Total geometric area may require splitting the interval.

7. The Special Integral of 1/x

The power rule cannot be used for x⁻¹. Instead:

∫(1/x)dx = ln|x| + C, where x ≠ 0

Example 8

Evaluate ∫(4/x − 3x²)dx.

4∫(1/x)dx − 3∫x²dx = 4ln|x| − x³ + C.

Answer: 4ln|x| − x³ + C

8. Common Mistakes

• Forgetting to increase the power before dividing.
• Dividing by the original power instead of the new power.
• Leaving out + C in an indefinite integral.
• Using the power rule for 1/x.
• Forgetting to use a given condition to calculate C.
• Reversing the order when applying definite limits.
• Assuming a definite integral always equals total geometric area.

9. Practice Exercises

Attempt each question before revealing its answer.

1. Evaluate ∫7x⁶ dx.

Reveal answer

x⁷ + C

2. Evaluate ∫(8x³ − 6x + 5)dx.

Reveal answer

2x⁴ − 3x² + 5x + C

3. Evaluate ∫(4√x + 3/x²)dx.

Reveal answer

(8/3)x^3/2 − 3/x + C

4. Evaluate ∫(6x⁻⁴ + 2x^−1/2)dx.

Reveal answer

−2x⁻³ + 4x^1/2 + C

5. Find y if dy/dx = 10x − 6 and y = 3 when x = 1.

Reveal answer

y = 5x² − 6x + 4

6. Evaluate ∫₀² 3x² dx.

Reveal answer

8

7. Evaluate ∫₁³(2x + 1)dx.

Reveal answer

10

8. Evaluate ∫(5/x + 4x²)dx.

Reveal answer

5ln|x| + (4/3)x³ + C

9. Find ∫(3x² − 4x + 2)dx and verify by differentiation.

Reveal answer

x³ − 2x² + 2x + C. Differentiating gives 3x² − 4x + 2.

10. Evaluate ∫₀¹(4x³ + 2x)dx.

Reveal answer

2

10. Lesson Summary

• Integration reverses differentiation.
• The power rule increases the power by one and divides by the new power, for n ≠ −1.
• Indefinite integrals include + C.
• ∫(1/x)dx = ln|x| + C.
• Definite integrals are evaluated using F(b) − F(a).

NEXT: PART 2 — SUBSTITUTION AND RECOGNISING A FUNCTION WITH ITS DERIVATIVE

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MATHEMATICS HIP-HOP ACADEMY · CHAPTER 03

INTEGRATION

Part 2 · Integration by Substitution and Recognising a Function with Its Derivative

A-LEVEL PURE MATHEMATICS

About this lesson

In Part 1, we integrated expressions directly using elementary rules. Some integrals, however, contain a function together with its derivative, or a complicated expression raised to a power. In such cases, substitution can turn the integral into a simpler one.

Central idea: Look for an inner function and its derivative. If both are present, a suitable substitution may reduce the integral to a standard form.

Learning outcomes

• Recognise a function and its derivative within an integrand.
• Choose a suitable substitution and calculate its differential.
• Integrate powers and reciprocal forms after substitution.
• Change the limits when evaluating definite integrals by substitution.
• Check an antiderivative by differentiating it.

1. Recognising a function and its derivative

Consider an integral containing an expression such as (3x² + 2)⁵. The expression inside the brackets is a function of x, and its derivative is 6x. If the integrand contains that derivative (or a constant multiple of it), substitution is likely to work.

If u = g(x), then du/dx = g′(x), so du = g′(x) dx.

This allows us to replace the combination g′(x) dx by du.

Recognition pattern: An integral of the form ∫g′(x)[g(x)]ⁿ dx can be simplified by putting u = g(x), provided the remaining factors match the derivative.

2. The substitution method

1. Identify a suitable inner function and let it equal u.
2. Differentiate: find du/dx.
3. Rearrange to express the required part of dx in terms of du.
4. Rewrite the entire integral using u. Do not leave stray x terms.
5. Integrate with respect to u.
6. For an indefinite integral, substitute back to x and include + C.
7. For a definite integral, either change the limits to u-values or substitute back before applying the original limits.

3. Worked examples

Example 1 — A simple power of a function

Evaluate ∫ 6x(3x² + 4)⁵ dx.

Step 1: Let u = 3x² + 4.

Step 2: du/dx = 6x, hence du = 6x dx.

Step 3: The integral becomes ∫u⁵ du.

Step 4: Integrate: u⁶/6 + C.

Step 5: Substitute back.

Answer: (3x² + 4)⁶/6 + C.

Example 2 — A constant multiple of the derivative

Evaluate ∫ x(2x² + 5)⁴ dx.

Let u = 2x² + 5. Then du = 4x dx, so x dx = du/4.

∫x(2x² + 5)⁴ dx = ¼∫u⁴ du.

= ¼ × u⁵/5 + C.

Answer: (2x² + 5)⁵/20 + C.

Example 3 — Recognising a reciprocal form

Evaluate ∫ [4x/(2x² + 7)] dx.

Let u = 2x² + 7. Then du = 4x dx.

The integral becomes ∫(1/u) du.

∫(1/u) du = ln|u| + C.

Answer: ln(2x² + 7) + C.

Since 2x² + 7 is always positive, the absolute value may be omitted.

Example 4 — Exponential function

Evaluate ∫ 3e^(3x − 1) dx.

Let u = 3x − 1. Then du = 3 dx.

The integral becomes ∫eᵘ du.

Answer: e^(3x − 1) + C.

Example 5 — A trigonometric function

Evaluate ∫ 2 sin(2x + 1) dx.

Let u = 2x + 1. Then du = 2 dx.

The integral becomes ∫sin u du = −cos u + C.

Answer: −cos(2x + 1) + C.

Example 6 — Definite integral with changed limits

Evaluate ∫₀¹ 2x( x² + 1)³ dx.

Let u = x² + 1, so du = 2x dx.

Change the limits: when x = 0, u = 1; when x = 1, u = 2.

The integral becomes ∫₁² u³ du.

= [u⁴/4]₁² = (16/4) − (1/4).

Answer: 15/4.

4. A useful recognition table

• Power pattern: ∫g′(x)[g(x)]ⁿ dx — let u = g(x), then use the power rule.
• Reciprocal pattern: ∫g′(x)/g(x) dx — the result is ln|g(x)| + C.
• Exponential pattern: ∫g′(x)eᵍ⁽ˣ⁾ dx — the result is eᵍ⁽ˣ⁾ + C.
• Sine pattern: ∫g′(x)sin(g(x)) dx — the result is −cos(g(x)) + C.
• Cosine pattern: ∫g′(x)cos(g(x)) dx — the result is sin(g(x)) + C.

These are recognition patterns, not permission to ignore missing factors. If the derivative is present only up to a constant multiple, adjust the integral by that constant.

5. Common mistakes to avoid

• Choosing u but forgetting to calculate du.
• Missing a numerical factor when rearranging du.
• Leaving x in the integral after changing to u.
• Forgetting to substitute back in an indefinite integral.
• Changing the integrand to u but forgetting to change the limits in a definite integral.
• Forgetting + C for an indefinite integral.

6. Practice exercises

Attempt each question before opening its answer.

1. Evaluate ∫ 8x(4x² + 3)³ dx.

Reveal answer

(4x² + 3)⁴/4 + C.

2. Evaluate ∫ 5x(5x² − 2)⁶ dx.

Reveal answer

(5x² − 2)⁷/14 + C.

3. Evaluate ∫ [6x/(3x² + 4)] dx.

Reveal answer

ln(3x² + 4) + C.

4. Evaluate ∫ 4e^(4x + 1) dx.

Reveal answer

e^(4x + 1) + C.

5. Evaluate ∫ 3cos(3x − 2) dx.

Reveal answer

sin(3x − 2) + C.

6. Evaluate ∫ 2x(x² + 5)⁴ dx.

Reveal answer

(x² + 5)⁵/5 + C.

7. Evaluate ∫₀² 2x(x² + 1)² dx.

Reveal answer

[(x² + 1)³/3]₀² = (125 − 1)/3 = 124/3.

8. Evaluate ∫ [2x/(x² + 9)] dx.

Reveal answer

ln(x² + 9) + C.

9. Evaluate ∫ 6(6x + 1)⁵ dx.

Reveal answer

(6x + 1)⁶/6 + C.

10. Evaluate ∫₁² 3x²( x³ + 2)⁴ dx.

Reveal answer

Let u = x³ + 2. Limits: 3 and 10. Answer = [u⁵/5]₃¹⁰ = (100000 − 243)/5 = 19951.

7. Lesson summary

• Substitution simplifies an integral by replacing an inner function with a new variable.
• Always differentiate the substitution to obtain du.
• Match all factors carefully, including constant multipliers.
• For definite integrals, change the limits whenever you integrate entirely in u.
• Differentiate your final answer to check an indefinite integral.

NEXT: PART 3 — TRIGONOMETRIC INTEGRATION

MATHEMATICS HIP-HOP
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MATHEMATICS HIP-HOP ACADEMY • A-LEVEL PURE MATHEMATICS

CHAPTER 03 — INTEGRATION

PART 3 Trigonometric Integration

In this lesson, we learn how to integrate the basic trigonometric functions, handle expressions with angles such as 3x + 1, and use identities to integrate powers and products of sine and cosine.

Learning objectives

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

• Recall standard integrals of trigonometric functions.
• Integrate tangent and explain equivalent logarithmic forms.
• Integrate trigonometric functions with a linear expression inside the angle.
• Use trigonometric identities to integrate expressions involving sin²x, cos²x and sin x cos x.
• Check an integral by differentiating your answer.

1. Standard trigonometric integrals

Integration reverses differentiation. Therefore, we use the derivatives of trigonometric functions to establish their standard integrals. Angles are measured in radians in these formulas.

Integrand	Integral
sin x	−cos x + C
cos x	sin x + C
sec²x	tan x + C
cosec²x	−cot x + C
sec x tan x	sec x + C
cosec x cot x	−cosec x + C

Remember: The constant of integration, C, must be included in every indefinite integral.

Worked Example 1 — Basic integrals

Evaluate: ∫(4 cos x − 3 sin x + 2 sec²x) dx.

= 4∫cos x dx − 3∫sin x dx + 2∫sec²x dx

= 4 sin x + 3 cos x + 2 tan x + C.

Worked Example 2 — A cosecant expression

Evaluate: ∫(5 cosec²x − 2 cosec x cot x) dx.

= 5(−cot x) − 2(−cosec x) + C

Answer: −5 cot x + 2 cosec x + C.

2. Integrating tangent

The integral of tan x is a special result. Rewrite tangent using sine and cosine, then use substitution.

∫ tan x dx = −ln|cos x| + C = ln|sec x| + C

The two forms are equivalent because ln|sec x| = ln|1/cos x| = −ln|cos x|.

Worked Example 3 — Integrating tangent

Evaluate: ∫ tan x dx.

Write tan x = sin x / cos x.

Let u = cos x, so du = −sin x dx.

∫(sin x/cos x) dx = −∫(1/u) du

= −ln|u| + C

Answer: −ln|cos x| + C.

3. Trigonometric functions with a linear angle

When the angle is ax + b, use substitution. The derivative of ax + b is the constant a, so the final answer includes a factor of 1/a.

∫ sin(ax + b) dx = −(1/a) cos(ax + b) + C

∫ cos(ax + b) dx = (1/a) sin(ax + b) + C

∫ sec²(ax + b) dx = (1/a) tan(ax + b) + C

Worked Example 4 — Sine of a linear angle

Evaluate: ∫ sin(3x + 1) dx.

Let u = 3x + 1, hence du = 3 dx and dx = du/3.

= (1/3)∫sin u du

= −(1/3)cos u + C

Answer: −⅓ cos(3x + 1) + C.

Worked Example 5 — Cosine of a linear angle

Evaluate: ∫ 4 cos(2x − 5) dx.

Let u = 2x − 5, so du = 2 dx.

= 4 × (1/2)∫cos u du

= 2 sin u + C

Answer: 2 sin(2x − 5) + C.

4. Using identities to integrate powers

Some trigonometric expressions cannot be integrated directly from the standard table. First use an identity to rewrite the expression into an integrable form.

sin²x = (1 − cos 2x)/2

cos²x = (1 + cos 2x)/2

sin x cos x = ½ sin 2x

Worked Example 6 — Integrating sin²x

Evaluate: ∫ sin²x dx.

Use sin²x = (1 − cos 2x)/2.

= ½∫(1 − cos 2x) dx

= ½[x − ½ sin 2x] + C

Answer: x/2 − sin 2x/4 + C.

Worked Example 7 — Integrating cos²x

Evaluate: ∫ cos²x dx.

Use cos²x = (1 + cos 2x)/2.

= ½∫(1 + cos 2x) dx

= ½[x + ½ sin 2x] + C

Answer: x/2 + sin 2x/4 + C.

Worked Example 8 — Product of sine and cosine

Evaluate: ∫ sin x cos x dx.

Use sin x cos x = ½ sin 2x.

= ½∫sin 2x dx

= ½[−½ cos 2x] + C

Answer: −¼ cos 2x + C.

5. Checking by differentiation

A reliable way to check an indefinite integral is to differentiate the result. You should recover the original integrand.

For example, if F(x) = −⅓ cos(3x + 1), then

F′(x) = −⅓[−sin(3x + 1)] × 3

= sin(3x + 1), which is the original integrand.

6. Common mistakes to avoid

• Forgetting the factor 1/a when integrating sin(ax + b), cos(ax + b) or sec²(ax + b).
• Writing ∫sin x dx = cos x + C. The correct answer has a negative sign.
• Forgetting the absolute value in ln|cos x|.
• Trying to integrate sin²x or cos²x directly without first using a suitable identity.
• Leaving out the constant of integration.
• Using degree-based differentiation formulas without converting to radians.

7. Practice exercises

Attempt each question before opening its answer. Give answers in terms of x and include C for indefinite integrals.

1. Evaluate ∫(3 sin x + 2 cos x) dx.

Reveal answer to Question 1

−3 cos x + 2 sin x + C.

2. Evaluate ∫(4 sec²x − 5 cosec²x) dx.

Reveal answer to Question 2

4 tan x + 5 cot x + C.

3. Evaluate ∫ tan x dx.

Reveal answer to Question 3

−ln|cos x| + C, equivalently ln|sec x| + C.

4. Evaluate ∫ cos(5x) dx.

Reveal answer to Question 4

⅕ sin 5x + C.

5. Evaluate ∫ 6 sin(2x + 3) dx.

Reveal answer to Question 5

−3 cos(2x + 3) + C.

6. Evaluate ∫ sec²(4x − 1) dx.

Reveal answer to Question 6

¼ tan(4x − 1) + C.

7. Evaluate ∫ sin²x dx.

Reveal answer to Question 7

x/2 − ¼ sin 2x + C.

8. Evaluate ∫ cos²x dx.

Reveal answer to Question 8

x/2 + ¼ sin 2x + C.

9. Evaluate ∫ sin x cos x dx.

Reveal answer to Question 9

−¼ cos 2x + C. An equivalent answer is ½ sin²x + C.

10. Evaluate ∫(2 cos 3x − 4 sin 2x) dx.

Reveal answer to Question 10

⅔ sin 3x + 2 cos 2x + C.

Part 3 summary

• Know the standard trigonometric integrals and their signs.
• For a linear angle ax + b, account for the factor 1/a.
• Use identities to rewrite powers and products before integrating.
• Differentiate your answer to check it.

Next: Part 4 — Partial Fractions and Integration by Parts.

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MATHEMATICS HIP-HOP ACADEMY • A-LEVEL PURE MATHEMATICS

CHAPTER 03 — INTEGRATION

PART 4

Partial Fractions and Integration by Parts

Two important integration techniques for handling products of functions and rational algebraic expressions. Learn how to recognise the correct method, show each stage of working and check your final answers.

Learning objectives

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

• derive and apply the integration by parts formula from the product rule;
• choose suitable expressions for u and dv;
• integrate products involving algebraic, exponential, trigonometric and logarithmic functions;
• apply integration by parts more than once when necessary;
• evaluate definite integrals using integration by parts;
• recognise rational functions suitable for partial fraction decomposition;
• resolve fractions with distinct and repeated linear factors;
• integrate the resulting partial fractions;
• use polynomial division when a rational fraction is improper.

SECTION A — INTEGRATION BY PARTS

Integration by parts is a technique used when an integrand is the product of two functions and direct integration or substitution is not convenient. It is especially useful when differentiating one factor makes it simpler.

1. Deriving the formula from the product rule

Recall the product rule for differentiation. If u and v are functions of x, then:

d(uv)/dx = u(dv/dx) + v(du/dx)

Rearranging gives:

u(dv/dx) = d(uv)/dx − v(du/dx)

Integrating both sides with respect to x:

∫u(dv/dx) dx = uv − ∫v(du/dx) dx

In differential notation, the formula is written:

∫u dv = uv − ∫v du

Important: Integration by parts is the reverse of the product rule. It changes one product integral into a product term minus another integral.

2. Choosing u and dv

The success of integration by parts depends on choosing u and dv carefully. A useful approach is:

• Choose u as the factor that becomes simpler when differentiated.
• Choose dv as the remaining factor that can be integrated easily.
• Differentiate u to obtain du.
• Integrate dv to obtain v.
• Substitute into ∫u dv = uv − ∫v du.

A useful guide for selecting u is the LIATE order:

Letter	Type of function	Example
L	Logarithmic	ln x
I	Inverse trigonometric	tan⁻¹x
A	Algebraic	x, x², 3x+1
T	Trigonometric	sin x, cos x
E	Exponential	eˣ

LIATE is a guide rather than an inflexible rule. The best choice is the one that makes the remaining integral simpler.

3. Worked examples: one application

Worked Example 1

Evaluate:

∫xeˣ dx

Solution

Choose:

u = x,    dv = eˣ dx

Then:

du = dx,    v = eˣ

Using the formula:

∫u dv = uv − ∫v du

∫xeˣ dx = xeˣ − ∫eˣ dx

= xeˣ − eˣ + C

Answer:

∫xeˣ dx = eˣ(x − 1) + C

Worked Example 2

Evaluate:

∫x cos x dx

Solution

Choose:

u = x,    dv = cos x dx

Therefore:

du = dx,    v = sin x

Applying the formula:

∫x cos x dx = x sin x − ∫sin x dx

= x sin x + cos x + C

Answer: x sin x + cos x + C

Worked Example 3

Evaluate:

∫ln x dx,   x > 0

Solution

Write the integral as:

∫ln x × 1 dx

Choose:

u = ln x,    dv = dx

Then:

du = 1/x dx,    v = x

Using integration by parts:

∫ln x dx = x ln x − ∫x(1/x) dx

= x ln x − ∫1 dx

= x ln x − x + C

Answer: x ln x − x + C

4. Repeated integration by parts

Some integrals require the integration by parts formula to be applied more than once. This commonly happens with a polynomial multiplied by an exponential or trigonometric function.

The polynomial becomes simpler each time it is differentiated. Eventually, its derivative becomes zero, allowing the integral to be completed.

Worked Example 4

Evaluate:

∫x²eˣ dx

First application

Choose:

u = x²,    dv = eˣ dx

Then:

du = 2x dx,    v = eˣ

Therefore:

∫x²eˣ dx = x²eˣ − 2∫xeˣ dx

Second application

From Worked Example 1:

∫xeˣ dx = eˣ(x − 1)

Substitute:

∫x²eˣ dx = x²eˣ − 2eˣ(x − 1) + C

Factor out eˣ:

= eˣ[x² − 2x + 2] + C

Answer: eˣ(x² − 2x + 2) + C

Worked Example 5

Evaluate:

∫x² sin x dx

Solution

Choose:

u = x²,    dv = sin x dx

Then:

du = 2x dx,    v = −cos x

Applying integration by parts:

I = −x² cos x + 2∫x cos x dx

From Worked Example 2:

∫x cos x dx = x sin x + cos x

Therefore:

I = −x² cos x + 2(x sin x + cos x) + C

Answer: −x² cos x + 2x sin x + 2cos x + C

5. Definite integration by parts

For definite integrals, the boundary values are applied to the product uv. The formula becomes:

∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du

There is no constant of integration in the final value of a definite integral.

Worked Example 6

Evaluate:

∫₀¹ xeˣ dx

Solution

From Worked Example 1:

∫xeˣ dx = eˣ(x − 1)

Apply the limits:

= [eˣ(x − 1)]₀¹

= e¹(1 − 1) − e⁰(0 − 1)

= 0 − (−1)

Answer: 1

Common examination errors

• Forgetting the negative sign when integrating sin x.
• Choosing u and dv without checking whether the remaining integral becomes simpler.
• Forgetting to integrate dv when finding v.
• Applying the formula but omitting the second integral.
• For definite integrals, failing to substitute both limits correctly.

SECTION B — PARTIAL FRACTIONS AND INTEGRATION

Partial fractions provide a way of expressing a rational function as a sum of simpler fractions. These simpler fractions are often much easier to integrate.

6. Understanding partial fractions

A rational function is a quotient of two polynomials. For example:

(5x + 1) / [(x + 1)(x + 2)]

If the denominator can be factorised, the fraction may be expressed as a sum of fractions with simpler denominators.

For example:

(5x + 1) / [(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

Once A and B have been found, the original fraction can be integrated term by term.

First check the degrees. Before decomposing a rational function, check whether the degree of the numerator is less than the degree of the denominator. If the numerator degree is equal to or greater than the denominator degree, perform polynomial division first.

7. Partial fractions with distinct linear factors

When the denominator contains different linear factors, each factor is represented by a separate fraction.

For example:

1/[(x + a)(x + b)] = A/(x + a) + B/(x + b)

The constants are found by multiplying through by the original denominator and comparing coefficients or substituting suitable values of x.

Worked Example 7

Evaluate:

∫ (5x + 1)/[(x + 1)(x + 2)] dx

Solution

Resolve into partial fractions:

(5x + 1)/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

Multiply through by (x + 1)(x + 2):

5x + 1 = A(x + 2) + B(x + 1)

Expand:

5x + 1 = Ax + 2A + Bx + B

Collect coefficients:

A + B = 5

2A + B = 1

Subtract the first equation from the second:

A = −4

Therefore B = 9.

So:

(5x + 1)/[(x + 1)(x + 2)] = −4/(x + 1) + 9/(x + 2)

Integrate:

= −4ln|x + 1| + 9ln|x + 2| + C

Answer: −4ln|x + 1| + 9ln|x + 2| + C

Worked Example 8

Evaluate:

∫ (3x + 7)/(x² + 3x + 2) dx

Solution

First factorise the denominator:

x² + 3x + 2 = (x + 1)(x + 2)

Write:

(3x + 7)/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

Multiply through:

3x + 7 = A(x + 2) + B(x + 1)

Comparing coefficients:

A + B = 3

2A + B = 7

Subtracting gives:

A = 4,    B = −1

Hence:

3x + 7 over (x + 1)(x + 2) = 4/(x + 1) − 1/(x + 2)

Integrating:

= 4ln|x + 1| − ln|x + 2| + C

Answer: 4ln|x + 1| − ln|x + 2| + C

8. Partial fractions with repeated linear factors

When a linear factor is repeated, every power of that factor must be included in the decomposition.

For example, if the denominator contains (x − a)², write:

A/(x − a) + B/(x − a)²

For a factor repeated three times, the decomposition includes all three powers:

A/(x − a) + B/(x − a)² + D/(x − a)³

Do not omit any of the powers of a repeated factor. Leaving out a term generally makes it impossible to obtain the correct decomposition.

Worked Example 9

Evaluate:

∫ (x + 3)/(x + 1)² dx

Solution

Write the partial fraction form:

(x + 3)/(x + 1)² = A/(x + 1) + B/(x + 1)²

Multiply through by (x + 1)²:

x + 3 = A(x + 1) + B

Expand:

x + 3 = Ax + A + B

Compare coefficients of x:

A = 1

Compare constant terms:

A + B = 3

1 + B = 3, so B = 2.

Therefore:

(x + 3)/(x + 1)² = 1/(x + 1) + 2/(x + 1)²

Integrate:

∫1/(x + 1) dx + 2∫(x + 1)⁻² dx

= ln|x + 1| − 2/(x + 1) + C

Answer: ln|x + 1| − 2/(x + 1) + C

9. Irreducible quadratic factors

An irreducible quadratic is a quadratic expression that cannot be factorised into real linear factors.

When the denominator contains an irreducible quadratic, the numerator over that quadratic is generally linear. Sometimes the numerator can be separated into a multiple of the derivative of the denominator plus a constant.

Worked Example 10

Evaluate:

∫ (2x + 3)/(x² + 4) dx

Solution

Separate the numerator:

∫(2x + 3)/(x² + 4) dx

= ∫2x/(x² + 4) dx + 3∫1/(x² + 4) dx

First integral

Let u = x² + 4.

Then du = 2x dx.

Therefore:

∫2x/(x² + 4) dx = ln(x² + 4)

Second integral

Use the standard result:

∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) + C

Here a = 2.

Therefore:

3∫dx/(x² + 4) = (3/2)tan⁻¹(x/2)

Combining both parts:

Answer: ln(x² + 4) + (3/2)tan⁻¹(x/2) + C

10. Improper rational fractions: division first

A rational fraction is improper when the degree of its numerator is equal to or greater than the degree of its denominator.

In this situation, perform polynomial division before attempting partial fraction decomposition.

Worked Example 11

Evaluate:

∫ (x² + 1)/(x − 1) dx

Solution

The numerator has degree 2 while the denominator has degree 1.

Divide x² + 1 by x − 1:

x² + 1 = (x − 1)(x + 1) + 2

Therefore:

(x² + 1)/(x − 1) = x + 1 + 2/(x − 1)

Now integrate each term:

∫[x + 1 + 2/(x − 1)] dx

= x²/2 + x + 2ln|x − 1| + C

Answer: x²/2 + x + 2ln|x − 1| + C

11. Choosing the correct integration method

Form of integrand	Suitable method
Product of two functions, such as xeˣ	Integration by parts
Polynomial multiplied by sine or cosine	Repeated integration by parts when needed
Logarithmic function such as ln x	Integration by parts
Proper rational fraction with factorable denominator	Partial fractions
Repeated linear factors	Partial fractions with every repeated power
Numerator degree at least denominator degree	Polynomial division first

Checking your answer Differentiate the final expression. For an indefinite integral, the derivative of your answer should reproduce the original integrand. This is one of the most reliable ways to check an integration result.

SECTION C — PRACTICE EXERCISES

Attempt the questions independently before opening the answer key. Show your choice of method, intermediate steps and final result.

Exercise A — Integration by parts

1. Evaluate ∫xe^(2x) dx.

2. Evaluate ∫x sin x dx.

3. Evaluate ∫x ln x dx, for x > 0.

4. Evaluate ∫x² cos x dx.

5. Evaluate ∫₀ᵖⁱ x cos x dx.

Exercise B — Partial fractions and integration

6. Evaluate ∫(2x + 5)/[(x + 1)(x + 3)] dx.

7. Evaluate ∫(4x + 1)/(x² − x − 2) dx.

8. Evaluate ∫(x + 2)/(x − 1)² dx.

9. Evaluate ∫(3x + 2)/(x² + 1) dx.

10. Evaluate ∫(x² + 4)/(x + 1) dx.

SECTION D — ANSWERS TO PRACTICE EXERCISES

Use these answers to check your work after attempting all the questions. For examination preparation, reproduce the full working independently.

Reveal answers 1–5: Integration by parts

1. ∫xe^(2x) dx

Answer: e^(2x)(x/2 − 1/4) + C

2. ∫x sin x dx

Answer: −x cos x + sin x + C

3. ∫x ln x dx

Answer: (x²/2)ln x − x²/4 + C

4. ∫x² cos x dx

Answer: x² sin x + 2x cos x − 2sin x + C

5. ∫₀ᵖⁱ x cos x dx

Answer: −2

Reveal answers 6–10: Partial fractions

6. ∫(2x + 5)/[(x + 1)(x + 3)] dx

Answer: (3/2)ln|x + 1| + (1/2)ln|x + 3| + C

7. ∫(4x + 1)/(x² − x − 2) dx

Answer: 3ln|x − 2| + ln|x + 1| + C

8. ∫(x + 2)/(x − 1)² dx

Answer: ln|x − 1| − 3/(x − 1) + C

9. ∫(3x + 2)/(x² + 1) dx

Answer: (3/2)ln(x² + 1) + 2tan⁻¹x + C

10. ∫(x² + 4)/(x + 1) dx

Answer: x²/2 − x + 5ln|x + 1| + C

FINAL EXAMINATION CHECKLIST

• State u, dv, du and v clearly when using integration by parts.
• Choose u so that differentiation makes it simpler where possible.
• Remember the negative sign when integrating sine.
• Apply integration by parts again when the remaining integral requires it.
• For definite integrals, substitute the upper limit and subtract the lower-limit value.
• Factorise the denominator before resolving a suitable rational fraction.
• Include every required term when a linear factor is repeated.
• Perform polynomial division before decomposing an improper fraction.
• Use absolute values in logarithmic expressions where the domain requires them.
• Differentiate your final indefinite integral to check the result.

Part 4 summary

Integration by parts is useful for products of functions, while partial fractions simplify suitable rational functions into forms that can be integrated term by term.

The central skill is recognising the structure of an integrand and selecting the method that makes the calculation manageable.

Practise both methods carefully, show every important algebraic step and verify your answers by differentiation.

MATHEMATICS HIP-HOP ACADEMY
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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.

MATHEMATICS HIPHOP ACADEMY
Making Mathematics Easier, One Concept at a Time.

MATHEMATICS HIPHOP ACADEMY

CHAPTER 03 — INTEGRATION

PART 6: DIFFERENTIAL EQUATIONS

A differential equation is an equation that involves an unknown function and one or more of its derivatives. Differential equations are useful because many real-life situations involve quantities that change with respect to time or another variable.

For example, population may increase with time, the temperature of a cooling object may decrease, and the velocity of a moving particle may change continuously.

In this lesson, we learn how to form, solve and interpret simple first-order differential equations using integration.

LEARNING OBJECTIVES

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

• Explain the meaning of a differential equation.
• Identify the order of a differential equation.
• Form simple differential equations from statements involving rates of change.
• Recognise differential equations with separable variables.
• Solve first-order differential equations by separating variables.
• Use initial conditions to obtain particular solutions.
• Apply integration techniques when solving differential equations.
• Interpret solutions in practical situations, including growth and decay.
• Recognise the limitations of a mathematical model.

SECTION 1: INTRODUCTION TO DIFFERENTIAL EQUATIONS

1.1 What is a differential equation?

A differential equation is an equation that contains an unknown function and at least one derivative of that function.

Consider the equation:

dy/dx = 3x²

This is a differential equation because it contains the derivative of y with respect to x.

To solve it, integrate both sides with respect to x:

y = ∫3x² dx = x³ + C

Here, C is an arbitrary constant of integration.

1.2 Meaning of order

The order of a differential equation is the order of the highest derivative appearing in the equation.

Differential equation	Highest derivative	Order
dy/dx + 2y = 0	First derivative	First order
d²y/dx² + 3dy/dx + y = 0	Second derivative	Second order
d³y/dx³ − 4y = 0	Third derivative	Third order

Remember: The order depends on the highest derivative present, not on the power of a derivative.

1.3 First-order differential equations

A first-order differential equation contains only the first derivative of the unknown function as its highest derivative.

A common form is:

dy/dx = f(x,y)

Some first-order equations can be rearranged so that all the terms involving y are on one side and all the terms involving x are on the other. Such equations are called differential equations with separable variables.

1.4 General solution and particular solution

A general solution contains an arbitrary constant and represents a family of possible solution curves.

A particular solution is obtained when additional information, such as an initial condition, is used to determine the constant.

Worked Example 1: General and particular solutions

Question: Solve dy/dx = 4x. Hence find the particular solution when y = 7 at x = 1.

Solution

Integrate both sides:

y = ∫4x dx

y = 2x² + C

This is the general solution.

Use y = 7 when x = 1:

7 = 2(1)² + C

7 = 2 + C

C = 5

Therefore, the particular solution is:

y = 2x² + 5.

SECTION 2: FORMING DIFFERENTIAL EQUATIONS

2.1 Understanding rates of change

The derivative of a quantity represents its rate of change with respect to another variable.

For example, if P represents population and t represents time, then:

dP/dt = rate of change of population with time

If s represents displacement and t represents time:

ds/dt = velocity

2.2 Translating statements into equations

Mathematical statements involving rates of change can be translated into differential equations.

Statement	Differential equation
Rate of change of y is constant k	dy/dx = k
Rate of change of y is proportional to y	dy/dx = ky
Rate of decrease of y is proportional to y	dy/dx = −ky, where k is positive
Rate of change of P is proportional to P²	dP/dt = kP²
Rate of change of y is proportional to x	dy/dx = kx

The constant k is called the constant of proportionality.

2.3 Worked Example 2: Forming a differential equation

Question: The rate of change of a quantity y with respect to x is proportional to the square of y. Form the differential equation.

Solution

The rate of change of y with respect to x is dy/dx.

It is proportional to y².

Introducing the constant of proportionality k:

dy/dx = ky².

2.4 Eliminating an arbitrary constant

A differential equation may also be formed from a family of curves by differentiating the equation and eliminating the arbitrary constant.

Worked Example 3: Forming a differential equation from a family of curves

Question: Form the differential equation of the family of curves y = Cx², where C is an arbitrary constant.

Given:

y = Cx²

Differentiate with respect to x:

dy/dx = 2Cx

From the original equation:

C = y/x²

Substitute into the differentiated equation:

dy/dx = 2x(y/x²)

Therefore:

x(dy/dx) = 2y.

The arbitrary constant has been eliminated.

SECTION 3: SEPARATION OF VARIABLES

3.1 Meaning of separable variables

A differential equation is separable if it can be rearranged so that all expressions involving y, together with dy, appear on one side, while all expressions involving x, together with dx, appear on the other side.

A common form is:

dy/dx = f(x)g(y)

Separating the variables gives:

dy/g(y) = f(x) dx

We can then integrate both sides:

∫[1/g(y)] dy = ∫f(x) dx

3.2 The general procedure

1. Write the differential equation clearly.
2. Separate the variables.
3. Integrate both sides, using one arbitrary constant.
4. Simplify the resulting equation.
5. Use any given initial condition to find the constant.
6. Check the solution in the original equation.

3.3 Worked Example 4: A simple separable equation

Question: Solve dy/dx = 3xy².

Solution

Separate the variables:

dy/y² = 3x dx

Integrate both sides:

∫y⁻² dy = ∫3x dx

−1/y = 3x²/2 + C

This is an acceptable implicit general solution.

Multiplying by −1 and absorbing the sign into the arbitrary constant:

1/y = A − 3x²/2.

Equivalently:

y = 1/(A − 3x²/2).

Here A is an arbitrary constant. The constant solution y = 0 must also be checked in the original equation because division by y² excluded it during separation.

3.4 Worked Example 5: A solution involving logarithms

Question: Solve dy/dx = 2xy, given that y = 3 when x = 0.

Solution

Separate the variables:

dy/y = 2x dx

Integrate:

∫(1/y)dy = ∫2x dx

ln|y| = x² + C

Exponentiating:

|y| = e^x²+C

Combining the constants gives:

y = Ae^x²

Use the condition y = 3 when x = 0:

3 = Ae⁰

A = 3

Therefore:

y = 3e^x².

3.5 Worked Example 6: Using algebra before integration

Question: Solve dy/dx = (2x + 1)/(3y²), given that y = 1 when x = 0.

Solution

Separate the variables:

3y² dy = (2x + 1) dx

Integrate:

∫3y² dy = ∫(2x + 1) dx

y³ = x² + x + C

Use y = 1 when x = 0:

1 = 0 + 0 + C

C = 1

Thus:

y³ = x² + x + 1

y = ∛(x² + x + 1).

SECTION 4: DIFFERENTIAL EQUATIONS INVOLVING PARTIAL FRACTIONS

4.1 Why partial fractions may be required

Some separable differential equations produce a rational expression when the variables are separated. In such cases, partial fractions may be needed before integration.

Worked Example 7: Using partial fractions

Question: Solve:

dy/dx = y(1 − y), given that y = 1/2 when x = 0.

Solution

Separate the variables:

dy/[y(1−y)] = dx

Resolve into partial fractions:

1/[y(1−y)] = 1/y + 1/(1−y)

Integrate:

∫[1/y + 1/(1−y)]dy = ∫1 dx

ln|y| − ln|1−y| = x + C

Combine the logarithms:

ln|y/(1−y)| = x + C

Exponentiate:

y/(1−y) = Ae^x

Use y = 1/2 when x = 0:

(1/2)/(1/2) = A

A = 1

Therefore:

y/(1−y) = e^x

Rearrange:

y = e^x(1−y)

y + ye^x = e^x

y(1+e^x) = e^x

y = e^x/(1+e^x).

Important: The integration required in a differential equation may involve any technique already studied in this Integration chapter, including substitution, trigonometric integration and partial fractions.

SECTION 5: INITIAL CONDITIONS AND PARTICULAR SOLUTIONS

5.1 What is an initial condition?

An initial condition gives the value of the unknown function at a particular value of the independent variable.

For example:

y = 4 when x = 2

This information is used to determine the arbitrary constant in the general solution.

5.2 General solution versus particular solution

General solution	Particular solution
Contains an arbitrary constant.	Constant has been determined.
Represents a family of curves.	Represents a specific solution.
Found by solving the differential equation.	Found by using additional information.

5.3 Worked Example 8: Finding a particular solution

Question: Solve dy/dx = 4y/x, given that y = 6 when x = 1. Assume x is positive.

Separate the variables:

dy/y = 4 dx/x

Integrate:

ln|y| = 4ln|x| + C

Since x is positive:

ln|y| = 4ln x + C

Exponentiating and combining constants:

y = Ax⁴

Use y = 6 when x = 1:

6 = A(1)⁴

A = 6

Therefore:

y = 6x⁴.

5.4 Interpreting a family of solution curves

The general solution of a differential equation may represent a family of curves. Each value of the arbitrary constant gives a different member of that family.

An initial condition selects the particular curve passing through the specified point.

Graphical interpretation: If a general solution contains the constant A, changing A changes the position or shape of the solution curve. A given initial condition selects the appropriate value of A.

SECTION 6: APPLICATIONS OF DIFFERENTIAL EQUATIONS

6.1 Exponential growth

Suppose the rate of increase of a quantity is proportional to the quantity itself.

Let P be the quantity and t be time. The differential equation is:

dP/dt = kP, where k is positive.

Separating and integrating:

dP/P = k dt

ln|P| = kt + C

Therefore, the general form is:

P = Ae^kt

If the initial quantity is P₀ at t = 0, then A = P₀.

P = P₀e^kt

6.2 Worked Example 9: Population growth

Question: A population P increases at a rate proportional to its size. Initially, the population is 500. If the constant of proportionality is 0.04 per year, find the population after 5 years.

The differential equation is:

dP/dt = 0.04P

Separate and integrate:

dP/P = 0.04dt

ln P = 0.04t + C

Therefore:

P = Ae^0.04t

Initially P = 500 when t = 0:

500 = A

Thus:

P = 500e^0.04t

At t = 5:

P = 500e^0.2

P ≈ 610.7.

The model predicts a population of approximately 611 individuals after 5 years.

6.3 Exponential decay

If the rate of decrease of a quantity is proportional to the quantity itself, the differential equation is:

dN/dt = −kN, where k is positive.

Separating and integrating:

dN/N = −kdt

ln|N| = −kt + C

Therefore:

N = N₀e^−kt

This model is used in situations such as the decrease of a substance over time, under suitable assumptions.

6.4 Worked Example 10: Exponential decay

Question: A quantity N decreases at a rate proportional to its size. Initially N = 800, and the constant of proportionality is 0.1 per hour. Find N after 4 hours.

The differential equation is:

dN/dt = −0.1N

Separate and integrate:

dN/N = −0.1dt

ln N = −0.1t + C

Therefore:

N = Ae^−0.1t

At t = 0, N = 800:

A = 800

Thus:

N = 800e^−0.1t

At t = 4:

N = 800e^−0.4

N ≈ 536.3.

6.5 Worked Example 11: Finding the constant of proportionality

Question: A population grows according to dP/dt = kP. Initially P = 200. After 3 years, P = 300. Find k and hence find the population after 6 years.

Separate and integrate:

dP/P = kdt

ln P = kt + C

Therefore:

P = Ae^kt

At t = 0, P = 200, so A = 200.

Hence:

P = 200e^kt

Use P = 300 when t = 3:

300 = 200e^3k

3/2 = e^3k

Taking logarithms:

ln(3/2) = 3k

k = ln(3/2)/3

k ≈ 0.1352 per year.

At t = 6:

P = 200e^6k

Since e^3k = 3/2:

P = 200(3/2)²

P = 450.

6.6 Interpreting solutions in context

A mathematical solution must be interpreted in relation to the situation being modelled.

• Check whether the quantity should increase or decrease.
• Use the correct units for time and the quantity.
• Check whether the answer is physically or practically reasonable.
• Consider whether the model remains realistic over a long period.

Limitations of exponential models: A population cannot necessarily grow exponentially forever because food, space and other resources may be limited. A simple decay model may also become inaccurate if the conditions affecting the rate change.

SECTION 7: CHECKING A SOLUTION

7.1 Substitution into the original equation

After finding a solution, differentiate it and substitute the derivative and the function into the original differential equation. The equation should be satisfied.

Worked Example 12: Verifying a solution

Question: Verify that y = 5e^2x satisfies dy/dx = 2y.

Given:

y = 5e^2x

Differentiate:

dy/dx = 10e^2x

Now calculate 2y:

2y = 2(5e^2x)

2y = 10e^2x

Therefore:

dy/dx = 2y

The given function satisfies the differential equation.

SECTION 8: EXAMINATION STRATEGIES

Common mistakes to avoid

• Failing to separate the variables correctly.
• Integrating with respect to the wrong variable.
• Forgetting the constant of integration.
• Writing an incorrect sign when integrating 1/(1−y).
• Forgetting that integrating 1/y gives ln|y|.
• Confusing the general solution with the particular solution.
• Failing to substitute the initial condition correctly.
• Forgetting to determine the constant of proportionality when required.
• Dividing by a factor that may be zero without checking for a constant solution.
• Giving an answer without interpreting it in the context of the question.

Useful formula summary

Situation	Mathematical form
General first-order equation	dy/dx = f(x,y)
Separable equation	dy/dx = f(x)g(y)
Separated form	dy/g(y) = f(x)dx
Proportional growth	dP/dt = kP
Growth solution	P = P₀ekt
Proportional decay	dN/dt = −kN
Decay solution	N = N₀e−kt
Particular solution	Use the given condition to determine C or A

SECTION 9: PRACTICE EXERCISES

Attempt the following questions before opening the solutions. Show all working clearly.

Exercise 1: Basic Concepts and Formation

1. State the order of the differential equation d²y/dx² + 5dy/dx + 6y = 0.
2. Form a differential equation if the rate of change of y with respect to x is proportional to y.
3. Form the differential equation of the family y = Cx³, eliminating C.
4. Explain the difference between a general solution and a particular solution.

Open answers and solutions

1. Second order, because the highest derivative is d²y/dx².

2. dy/dx = ky, where k is the constant of proportionality.

3. Differentiate: dy/dx = 3Cx². Since C = y/x³, substitute to obtain x(dy/dx) = 3y.

4. A general solution contains an arbitrary constant and represents a family of solutions. A particular solution is obtained by using additional information to determine that constant.

Exercise 2: Separation of Variables

1. Solve dy/dx = 5x².
2. Solve dy/dx = 4xy.
3. Solve dy/dx = 6x/y, given y = 2 when x = 1.
4. Solve dy/dx = (3x²)/(2y), given y = 2 when x = 0.

Open answers and solutions

1. Integrating gives y = 5x³/3 + C.

2. dy/y = 4x dx. Therefore ln|y| = 2x² + C, or y = Ae^2x².

3. y dy = 6x dx. Hence y²/2 = 3x² + C. Using (1,2), C = −1. Therefore y² = 6x² − 2. The positive branch applies near the given initial value.

4. 2y dy = 3x² dx. Hence y² = x³ + C. Using y = 2 at x = 0 gives C = 4. Therefore y² = x³ + 4.

Exercise 3: Particular Solutions

1. Solve dy/dx = 3y, given y = 4 when x = 0.
2. Solve dy/dx = 2x(y + 1), given y = 1 when x = 0.
3. Solve dy/dx = y/x, given y = 6 when x = 2, assuming x is positive.

Open answers and solutions

1. y = Ae^3x. Using y = 4 at x = 0 gives y = 4e^3x.

2. dy/(y+1) = 2x dx. Thus ln|y+1| = x² + C. The condition gives A = 2. Therefore y = 2e^x² − 1.

3. dy/y = dx/x. Hence ln y = ln x + C and y = Ax. Using y = 6 when x = 2 gives A = 3. Therefore y = 3x.

Exercise 4: Applications to Growth and Decay

1. A quantity P grows at a rate proportional to P. Initially P = 100 and k = 0.2 per hour. Find P after 3 hours.
2. A substance decreases according to dN/dt = −0.05N. Initially N = 600. Find N after 10 hours.
3. A population grows according to dP/dt = kP. It increases from 400 to 500 in 2 years. Find k.
4. Explain one limitation of using an exponential growth model for a population over a very long period.

Open answers and solutions

1. P = 100e^0.2t. At t = 3, P = 100e^0.6 ≈ 182.2.

2. N = 600e^−0.05t. At t = 10, N = 600e^−0.5 ≈ 363.9.

3. P = 400e^kt. At t = 2, 500 = 400e^2k. Hence k = ½ln(5/4) ≈ 0.1116 per year.

4. Resources such as food, water and space may become limited, so the population may not continue growing at a constant proportional rate.

Exercise 5: Examination-Style Mixed Questions

1. Solve dy/dx = 2y/(x + 1), given y = 4 when x = 1. Assume x + 1 is positive.
2. Solve dy/dx = y(2 − y), given y = 1 when x = 0.
3. A quantity Q satisfies dQ/dt = kQ. Given that Q = 50 when t = 0 and Q = 80 when t = 4, find an expression for Q in terms of t.
4. Verify that y = 3e^−2x satisfies dy/dx = −2y.

Open answers and solutions

1. dy/y = 2dx/(x+1). Integrating gives ln y = 2ln(x+1) + C. The condition gives A = 1. Therefore y = (x+1)².

2. dy/[y(2−y)] = dx. Partial fractions give ½[1/y + 1/(2−y)]. Integrating gives ½ln|y/(2−y)| = x + C. Using y = 1 at x = 0 gives the ratio constant as 1. Therefore y = 2e^2x/(1+e^2x).

3. Q = 50e^kt. Using Q = 80 at t = 4 gives k = ¼ln(8/5). Therefore Q = 50e^(t/4)ln(8/5), equivalently Q = 50(8/5)^t/4.

4. Differentiating gives dy/dx = −6e^−2x. Also, −2y = −6e^−2x. Thus dy/dx = −2y, as required.

FINAL REVISION CHECKLIST

Before completing this topic, make sure you can:

• Identify and explain differential equations.
• Determine the order of a differential equation.
• Form equations from statements about rates of change.
• Eliminate arbitrary constants by differentiation.
• Separate variables correctly.
• Integrate both sides using appropriate techniques.
• Use initial conditions to find particular solutions.
• Apply partial fractions where necessary.
• Model simple growth and decay situations.
• Interpret the answer and recognise limitations of a model.

CONCLUSION

Differential equations provide a mathematical way of describing how quantities change. By separating variables and integrating, we can obtain general solutions and use initial conditions to determine particular solutions.

The methods in this lesson connect the integration techniques studied throughout this chapter to practical applications such as population growth, decay and other changing quantities.

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