Indefinite & Definite Integrals

CSEC Additional Mathematics Essential Knowledge: Integration is the reverse process of differentiation. It allows us to find functions when given their derivatives, calculate areas under curves, and solve problems involving accumulation. Mastering integrals is crucial for CSEC Add Maths and lays the foundation for higher mathematics.

Key Concept: Integration has two main types: Indefinite Integrals (finding antiderivatives + constant) and Definite Integrals (calculating numerical values representing areas). The fundamental theorem of calculus connects these two concepts.

Part 1: Indefinite Integrals – The Antiderivative

Basic Integration Rules

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Definition of Indefinite Integral

If \(F'(x) = f(x)\), then \(\int f(x) \, dx = F(x) + C\), where \(C\) is the constant of integration.

\[\int f(x) \, dx = F(x) + C\]

The integral sign ∫, the integrand \(f(x)\), and \(dx\) (indicating integration with respect to \(x\)) are essential components.

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Basic Integration Formulas
Function Integral Rule Name
\(\int x^n \, dx\) \(\frac{x^{n+1}}{n+1} + C, \quad n \neq -1\) Power Rule
\(\int k \, dx\) \(kx + C\) Constant Rule
\(\int e^x \, dx\) \(e^x + C\) Exponential Rule
\(\int \frac{1}{x} \, dx\) \(\ln|x| + C\) Logarithmic Rule
\(\int \sin x \, dx\) \(-\cos x + C\) Trigonometric Rules
\(\int \cos x \, dx\) \(\sin x + C\) Trigonometric Rules

Remember: Always add \(+C\) for indefinite integrals! This constant represents all possible vertical shifts of the antiderivative.

📝 Example 1: Basic Integration

Find \(\int (3x^2 + 4x – 5) \, dx\)

1
Apply the power rule to each term separately:
2
\(\int 3x^2 \, dx = 3 \cdot \frac{x^{3}}{3} = x^3\)
3
\(\int 4x \, dx = 4 \cdot \frac{x^{2}}{2} = 2x^2\)
4
\(\int (-5) \, dx = -5x\)
5
Combine and add constant: \(x^3 + 2x^2 – 5x + C\)
📝 Example 2: Integrating with Fractional/Negative Powers

Find \(\int \left( \sqrt{x} + \frac{1}{x^2} \right) dx\)

1
Rewrite in power form: \(\sqrt{x} = x^{1/2}\), \(\frac{1}{x^2} = x^{-2}\)
2
\(\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}\)
3
\(\int x^{-2} \, dx = \frac{x^{-1}}{-1} = -x^{-1} = -\frac{1}{x}\)
4
Combine: \(\frac{2}{3}x^{3/2} – \frac{1}{x} + C\)

Part 2: Definite Integrals – Area Under Curves

A

The Definite Integral Concept

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Definition of Definite Integral

The definite integral \(\int_a^b f(x) \, dx\) represents the net signed area between the curve \(y = f(x)\), the x-axis, and the vertical lines \(x = a\) and \(x = b\).

\[\int_a^b f(x) \, dx = F(b) – F(a)\]

where \(F(x)\) is any antiderivative of \(f(x)\). This is the Fundamental Theorem of Calculus.

Area = ∫f(x)dx from a to b
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Properties of Definite Integrals
Property Formula Meaning
Reversal of Limits \(\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\) Swapping limits changes sign
Zero Width \(\int_a^a f(x) \, dx = 0\) No area under a point
Additivity \(\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx\) Can split integration interval
Constant Multiple \(\int_a^b k f(x) \, dx = k \int_a^b f(x) \, dx\) Constants can be factored out
Sum/Difference \(\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx\) Integrate term by term
📝 Example 3: Evaluating a Definite Integral

Evaluate \(\int_1^3 (2x + 1) \, dx\)

1
Find the antiderivative: \(F(x) = x^2 + x\)
2
Apply the Fundamental Theorem: \(F(3) – F(1)\)
3
\(F(3) = 3^2 + 3 = 9 + 3 = 12\)
4
\(F(1) = 1^2 + 1 = 1 + 1 = 2\)
5
Subtract: \(12 – 2 = 10\)
6
Answer: \(\int_1^3 (2x + 1) \, dx = 10\)
📝 Example 4: Definite Integral with Substitution (CSEC Level)

Evaluate \(\int_0^2 3x(1 + x^2)^2 \, dx\)

1
Let \(u = 1 + x^2\), then \(du = 2x \, dx\)
2
Adjust for the 3x dx: We have \(3x \, dx = \frac{3}{2} \cdot 2x \, dx = \frac{3}{2} du\)
3
Change limits: When \(x = 0\), \(u = 1\). When \(x = 2\), \(u = 1 + 4 = 5\)
4
Rewrite integral: \(\int_1^5 u^2 \cdot \frac{3}{2} du = \frac{3}{2} \int_1^5 u^2 \, du\)
5
Integrate: \(\frac{3}{2} \left[ \frac{u^3}{3} \right]_1^5 = \frac{1}{2} \left[ u^3 \right]_1^5\)
6
Evaluate: \(\frac{1}{2} (125 – 1) = \frac{124}{2} = 62\)

Part 3: Applications – Area Under/Between Curves

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Calculating Areas with Integration

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Area Under a Curve

To find the area between \(y = f(x)\) and the x-axis from \(x = a\) to \(x = b\):

\[\text{Area} = \int_a^b |f(x)| \, dx\]

Important: If the curve goes below the x-axis, the definite integral gives net signed area (positive above, negative below). For total area, take the absolute value or integrate sections separately.

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Area Between Two Curves

If \(f(x) \geq g(x)\) on \([a, b]\), the area between them is:

\[\text{Area} = \int_a^b [f(x) – g(x)] \, dx\]
1
Find intersection points (limits of integration) by solving \(f(x) = g(x)\)
2
Determine which function is on top (\(f(x)\)) and which is on bottom (\(g(x)\))
3
Set up the integral: \(\int_{\text{left}}^{\text{right}} (\text{top} – \text{bottom}) \, dx\)
4
Evaluate the definite integral
📝 Example 5: Area Under a Curve (CSEC Past Paper Style)

Find the area bounded by the curve \(y = x^2 – 4x + 3\), the x-axis, and the lines \(x = 0\) and \(x = 3\).

1
Sketch or analyze: The parabola intersects x-axis when \(x^2 – 4x + 3 = 0\) ⇒ \((x-1)(x-3) = 0\) ⇒ \(x = 1, 3\)
2
The curve goes below x-axis between 1 and 3. We need total area, so integrate in two parts.
3
Area from 0 to 1: \(\int_0^1 (x^2 – 4x + 3) dx = \left[ \frac{x^3}{3} – 2x^2 + 3x \right]_0^1\)
4
\(\left( \frac{1}{3} – 2 + 3 \right) – 0 = \frac{1}{3} + 1 = \frac{4}{3}\)
5
Area from 1 to 3: The curve is below, so area = \(-\int_1^3 (x^2 – 4x + 3) dx\)
6
\(-\left[ \frac{x^3}{3} – 2x^2 + 3x \right]_1^3 = -\left[ (9 – 18 + 9) – (\frac{1}{3} – 2 + 3) \right]\)
7
\(-\left[ 0 – (\frac{1}{3} + 1) \right] = -\left[ -\frac{4}{3} \right] = \frac{4}{3}\)
8
Total area: \(\frac{4}{3} + \frac{4}{3} = \frac{8}{3}\) square units
📝 Example 6: Area Between Two Curves

Find the area enclosed by the curves \(y = x^2\) and \(y = 2x – x^2\).

1
Find intersection points: Set \(x^2 = 2x – x^2\) ⇒ \(2x^2 – 2x = 0\) ⇒ \(2x(x-1) = 0\) ⇒ \(x = 0, 1\)
2
Determine which is on top: Test point \(x = 0.5\): \(2x – x^2 = 1 – 0.25 = 0.75\), \(x^2 = 0.25\). So \(2x – x^2\) is above \(x^2\) on [0,1]
3
Set up integral: Area = \(\int_0^1 [(2x – x^2) – x^2] dx = \int_0^1 (2x – 2x^2) dx\)
4
Integrate: \(\left[ x^2 – \frac{2x^3}{3} \right]_0^1\)
5
Evaluate: \(\left(1 – \frac{2}{3}\right) – 0 = \frac{1}{3}\) square units

Part 4: Integration Techniques for CSEC

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Essential Integration Methods

1. Term-by-Term Integration

For polynomials and sums: Integrate each term separately using the power rule.

Example: \(\int (3x^2 + 2x + 1) dx = x^3 + x^2 + x + C\)

CSEC Tip: Rewrite roots as fractional powers and reciprocals as negative powers first.

2. Substitution Method

For composite functions: Let \(u = g(x)\), then \(du = g'(x) dx\).

Example: \(\int 2x e^{x^2} dx\): Let \(u = x^2\), \(du = 2x dx\) → \(\int e^u du = e^u + C = e^{x^2} + C\)

CSEC Level: Usually simple substitutions like \(u = ax + b\) or \(u = x^2\).

3. Integration of Trigonometric Functions

Memorize basic integrals: \(\int \sin x dx = -\cos x + C\), \(\int \cos x dx = \sin x + C\)

CSEC Focus: Usually combined with substitution, e.g., \(\int \sin(2x) dx = -\frac{1}{2}\cos(2x) + C\)

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Common Integration Formulas for CSEC
Function Integral Notes
\(\int (ax + b)^n dx\) \(\frac{(ax + b)^{n+1}}{a(n+1)} + C, \quad n \neq -1\) Use substitution or formula
\(\int e^{kx} dx\) \(\frac{1}{k} e^{kx} + C\)
\(\int \frac{1}{ax + b} dx\) \(\frac{1}{a} \ln|ax + b| + C\)
\(\int \sin(ax + b) dx\) \(-\frac{1}{a} \cos(ax + b) + C\)
\(\int \cos(ax + b) dx\) \(\frac{1}{a} \sin(ax + b) + C\)
📝 Example 7: Using Integration Formulas

Find \(\int (2x + 1)^3 dx\)

1
Method 1 (Expand): \((2x+1)^3 = 8x^3 + 12x^2 + 6x + 1\)
2
Integrate: \(2x^4 + 4x^3 + 3x^2 + x + C\)
3
Method 2 (Formula): \(\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\)
4
Here \(a=2, b=1, n=3\): \(\frac{(2x+1)^4}{2 \cdot 4} + C = \frac{(2x+1)^4}{8} + C\)
5
Note: Both answers are equivalent (differ by constant).

Part 5: Solving Differential Equations (CSEC Level)

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Basic Differential Equations

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Solving \(\frac{dy}{dx} = f(x)\)

The simplest differential equation: Integrate both sides with respect to \(x\).

\[\frac{dy}{dx} = f(x) \quad \Rightarrow \quad y = \int f(x) \, dx\]

If given an initial condition (e.g., \(y = y_0\) when \(x = x_0\)), use it to find the constant \(C\).

📝 Example 8: Solving a Differential Equation

Solve \(\frac{dy}{dx} = 3x^2 – 2x\), given that \(y = 4\) when \(x = 1\).

1
Integrate: \(y = \int (3x^2 – 2x) dx = x^3 – x^2 + C\)
2
Use initial condition: When \(x = 1\), \(y = 4\): \(4 = 1^3 – 1^2 + C = 0 + C\)
3
So \(C = 4\)
4
Particular solution: \(y = x^3 – x^2 + 4\)
📝 Example 9: Real-World Application (CSEC Past Paper)

The rate of change of the area \(A\) of a culture of bacteria is given by \(\frac{dA}{dt} = 3t^2 + 2t\), where \(t\) is time in hours. If \(A = 5\) when \(t = 1\), find \(A\) in terms of \(t\).

1
Integrate: \(A = \int (3t^2 + 2t) dt = t^3 + t^2 + C\)
2
Use condition: When \(t = 1\), \(A = 5\): \(5 = 1^3 + 1^2 + C = 2 + C\)
3
So \(C = 3\)
4
Answer: \(A = t^3 + t^2 + 3\)

Quiz: Test Your Integration Skills

Indefinite & Definite Integrals Quiz
Question 1: Find \(\int (4x^3 – 3x^2 + 2x – 1) dx\)
Answer:
Term-by-term integration:
\(\int 4x^3 dx = x^4\)
\(\int -3x^2 dx = -x^3\)
\(\int 2x dx = x^2\)
\(\int -1 dx = -x\)
So: \(x^4 – x^3 + x^2 – x + C\)
Question 2: Evaluate \(\int_0^{\pi/2} \cos x \, dx\)
Answer:
\(\int \cos x dx = \sin x + C\)
So \(\int_0^{\pi/2} \cos x dx = \left[ \sin x \right]_0^{\pi/2} = \sin(\pi/2) – \sin(0) = 1 – 0 = 1\)
Question 3: Find the area bounded by \(y = x^3\), the x-axis, and the lines \(x = 1\) and \(x = 2\).
Answer:
Since \(y = x^3 > 0\) on [1, 2], area = \(\int_1^2 x^3 dx = \left[ \frac{x^4}{4} \right]_1^2\)
= \(\frac{16}{4} – \frac{1}{4} = 4 – 0.25 = 3.75\) or \(\frac{15}{4}\) square units.
Question 4: Solve the differential equation \(\frac{dy}{dx} = \frac{1}{x}\), given that \(y = 2\) when \(x = e\).
Answer:
\(y = \int \frac{1}{x} dx = \ln|x| + C\)
Using condition: When \(x = e\), \(y = 2\): \(2 = \ln e + C = 1 + C \Rightarrow C = 1\)
So \(y = \ln|x| + 1\) (or \(\ln x + 1\) for \(x > 0\))
Question 5: Evaluate \(\int (2x + 1)(x – 3) dx\)
Answer:
First expand: \((2x + 1)(x – 3) = 2x^2 – 6x + x – 3 = 2x^2 – 5x – 3\)
Then integrate: \(\frac{2x^3}{3} – \frac{5x^2}{2} – 3x + C\)

🎯 Key Concepts Summary

  • Indefinite Integral: \(\int f(x) dx = F(x) + C\) where \(F'(x) = f(x)\)
  • Definite Integral: \(\int_a^b f(x) dx = F(b) – F(a)\) = net signed area
  • Power Rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1\)
  • Common Integrals:
    • \(\int e^x dx = e^x + C\)
    • \(\int \frac{1}{x} dx = \ln|x| + C\)
    • \(\int \sin x dx = -\cos x + C\)
    • \(\int \cos x dx = \sin x + C\)
  • Area Calculations:
    • Under curve: \(\int_a^b |f(x)| dx\)
    • Between curves: \(\int_a^b [f(x) – g(x)] dx\) where \(f(x) \geq g(x)\)
  • Solving \(\frac{dy}{dx} = f(x)\): Integrate and use initial condition to find \(C\)
  • CSEC Exam Tips:
    • Always add \(+C\) for indefinite integrals
    • For definite integrals, show substitution of limits clearly
    • Rewrite functions in power form before integrating
    • For area problems, sketch if possible to identify regions
    • Check your answer by differentiating (when time permits)

CSEC Exam Strategy: Integration questions often appear in Paper 2 (structured questions). Common question types: (1) Find indefinite integrals, (2) Evaluate definite integrals, (3) Calculate areas under/between curves, (4) Solve simple differential equations. Always show your working: write the integral, show the antiderivative, substitute limits (for definite integrals), and simplify. For area questions, if the curve crosses the x-axis, split the integral at the intersection points.

Real-World Applications of Integration

Calculating area of irregular shapes (land surveying, architecture)
Finding distance from velocity-time graphs (physics applications)
Determining growth functions from rate of change (biology, economics)
Calculating work done by variable forces (engineering)
Finding accumulated quantities (total cost from marginal cost in business)
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