👋 Welcome to the Integration Chapter!

Welcome, future mathematician! If you successfully navigated the world of differentiation (finding the gradient), you’re ready for the exciting challenge of Integration.

Integration is often called the reverse process of differentiation. While differentiation finds the instantaneous rate of change, integration finds the total accumulation or the total area under a curve.
Think of it this way: If differentiation breaks things down, integration builds them back up!

This chapter is essential because it moves mathematics from theory into real-world applications, helping us calculate things like area, volume, and displacement from velocity. Don't worry if this seems tricky at first; we will break down every concept into simple, step-by-step notes.

1. Introduction to Indefinite Integration (The Antiderivative)

1.1 What is the Antiderivative?

When we integrate a function, we are finding its antiderivative.

Consider a function \(F(x)\). If we differentiate it, we get \(f(x)\), where \(f(x) = F'(x)\).
Integration is the process of starting with \(f(x)\) and finding the original \(F(x)\).

The standard notation for integrating a function \(f(x)\) with respect to \(x\) is: $$ \int f(x) \,dx $$

The symbol \(\int\) is the integral sign (an elongated 'S'), and \(dx\) tells us which variable we are integrating with respect to (in this case, \(x\)).

1.2 The Power Rule for Integration

The most fundamental rule you need to master in P2 is how to integrate \(x\) raised to a power.

The Power Rule: $$ \int x^n \,dx = \frac{x^{n+1}}{n+1} + C $$

Step-by-Step Process for Integrating \(x^n\)
  1. Power Up: Add 1 to the existing power \(n\). (The new power is \(n+1\)).
  2. Divide: Divide the whole term by this new power, \((n+1)\).
  3. Add \(C\): Always remember to add the constant of integration, \(C\).

⚠️ Crucial P2 Restriction: This rule works for all rational values of \(n\), EXCEPT when \(n = -1\). If \(n = -1\), the denominator would be zero, which is undefined. The case \(\int x^{-1} \,dx\) is handled using logarithms (covered in P3).

Example: Integrate \(x^3\).
Solution: Power up (3+1=4). Divide by the new power (4). Add \(C\). $$ \int x^3 \,dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C $$

1.3 The Essential Constant of Integration, \(C\)

When finding an indefinite integral, we must always add the constant \(C\).

Why is \(C\) necessary?
Remember that when you differentiate a constant, the result is always zero.

If \(F_1(x) = x^2 + 5\), then \(F_1'(x) = 2x\).
If \(F_2(x) = x^2 - 100\), then \(F_2'(x) = 2x\).
If \(F_3(x) = x^2\), then \(F_3'(x) = 2x\).

Since integrating \(2x\) should take us back to the original function, and we don't know if the original function had a constant (like +5, -100, or 0), we use \(C\) to represent that unknown constant.

Memory Aid: Never forget the \(C\)! Forgetting \(C\) is the most common and easily avoidable mistake in indefinite integration.

Quick Review: Indefinite Integration
  • It is the process of finding the antiderivative.
  • Key Rule: \(\int x^n \,dx = \frac{x^{n+1}}{n+1} + C\) (\(n \ne -1\)).
  • You must include the Constant of Integration, \(C\).

2. Manipulating Functions Before Integrating

2.1 Integrating Polynomials and Sums

Integration is a linear operation. This means you can integrate terms separately and apply constants.

$$ \int (f(x) \pm g(x)) \,dx = \int f(x) \,dx \pm \int g(x) \,dx $$ $$ \int k \cdot f(x) \,dx = k \cdot \int f(x) \,dx $$

Example: Integrate \(4x^2 - 3x + 7\). $$ \int (4x^2 - 3x^1 + 7x^0) \,dx $$ Term 1: \(4x^2 \rightarrow 4 \cdot \frac{x^{2+1}}{3} = \frac{4}{3}x^3\)
Term 2: \(-3x^1 \rightarrow -3 \cdot \frac{x^{1+1}}{2} = -\frac{3}{2}x^2\)
Term 3: \(7\) (which is \(7x^0\)) \(\rightarrow 7 \cdot \frac{x^{0+1}}{1} = 7x\)
Result: \(\frac{4}{3}x^3 - \frac{3}{2}x^2 + 7x + C\)

2.2 Integrating Roots and Fractions

Just like in differentiation, you cannot integrate roots or fractions (with \(x\) in the denominator) until you rewrite them using index notation. This allows you to use the power rule.

Remember these conversions:

  • Roots: \(\sqrt[m]{x^n} = x^{n/m}\)
  • Fractions: \(\frac{1}{x^n} = x^{-n}\)

Example 1: Integrate \(\frac{1}{x^3}\). $$ \int \frac{1}{x^3} \,dx = \int x^{-3} \,dx $$ New Power: \(-3 + 1 = -2\)
Divide by New Power: \(\frac{x^{-2}}{-2} + C\)
Final Answer (simplified): \(-\frac{1}{2x^2} + C\)

Example 2: Integrate \(5\sqrt{x}\). $$ \int 5\sqrt{x} \,dx = \int 5x^{1/2} \,dx $$ New Power: \(1/2 + 1 = 3/2\)
Divide by New Power: \(\frac{5x^{3/2}}{3/2} + C = 5 \times \frac{2}{3} x^{3/2} + C\)
Final Answer: \(\frac{10}{3}x^{3/2} + C\)

Struggling Student Tip: Setup is Key!

If the function looks complicated (like \(\frac{x^2+4x}{\sqrt{x}}\)), always simplify and rewrite every term as \(Ax^n\) first. Don't even start integrating until you have used algebra to prepare the expression.
Preparation example: $$\frac{x^2+4x}{x^{1/2}} = \frac{x^2}{x^{1/2}} + \frac{4x}{x^{1/2}} = x^{2-1/2} + 4x^{1-1/2} = x^{3/2} + 4x^{1/2}$$ Now you can integrate this term by term easily!

3. Definite Integration and Area

3.1 Defining Definite Integrals

Unlike indefinite integration, which results in a function containing \(C\), definite integration results in a single numerical value. This value usually represents the area under the curve between two specific points.

Definite integrals are written with limits of integration (\(a\) and \(b\)): $$ \int_a^b f(x) \,dx $$ Where \(a\) is the lower limit and \(b\) is the upper limit.

3.2 The Fundamental Theorem of Calculus (Evaluation)

The definite integral is evaluated using the antiderivative \(F(x)\): $$ \int_a^b f(x) \,dx = [F(x)]_a^b = F(b) - F(a) $$

Step-by-Step: Evaluating a Definite Integral
  1. Integrate: Find the indefinite integral \(F(x)\). (Crucially, you do not need to include the \(+C\) here, as it cancels out: \((F(b)+C) - (F(a)+C) = F(b) - F(a)\)).
  2. Bracket and Label: Place \(F(x)\) in square brackets with the limits: \([F(x)]_a^b\).
  3. Substitute \(b\): Calculate \(F(b)\) (substitute the upper limit).
  4. Substitute \(a\): Calculate \(F(a)\) (substitute the lower limit).
  5. Subtract: Find the final value \(F(b) - F(a)\).

Example: Evaluate \(\int_1^3 (3x^2 - 2) \,dx\).
1. Integrate: \(F(x) = \frac{3x^3}{3} - 2x = x^3 - 2x\).
2. Bracket: \([x^3 - 2x]_1^3\)
3. Substitute \(b=3\): \(F(3) = (3)^3 - 2(3) = 27 - 6 = 21\)
4. Substitute \(a=1\): \(F(1) = (1)^3 - 2(1) = 1 - 2 = -1\)
5. Subtract: \(F(3) - F(1) = 21 - (-1) = 22\)

The value of the definite integral is 22.

3.3 Calculating the Area Under a Curve

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

Did you know? This connection between differentiation and finding area (integration) was a revolutionary discovery in mathematics, formalised by Newton and Leibniz, and is the foundation of calculus!

Dealing with Negative Areas

This is one of the trickiest parts of P2 area problems.

When you integrate, any area that lies above the x-axis will yield a positive result. Any area that lies below the x-axis will yield a negative result.

If the question asks for the Total Area bounded by the curve and the x-axis, you must treat negative areas as positive.

Procedure for finding Total Area:

  1. Find the x-intercepts (where \(f(x)=0\)) to see where the curve crosses the x-axis.
  2. Split the integral into separate sections at each intercept.
  3. Integrate each section separately.
  4. For any section that yields a negative value (area below the axis), take the absolute value (make it positive).
  5. Add all the positive area values together to find the total area.

Example Scenario: If you integrate from \(x=0\) to \(x=4\), but the curve crosses the axis at \(x=2\): $$ \text{Total Area} = \left| \int_0^2 f(x) \,dx \right| + \left| \int_2^4 f(x) \,dx \right| $$

Common Mistake Alert! Never integrate straight across the x-axis if part of the curve is below it. If you integrate \(\int_a^b f(x) \,dx\) and the area above cancels out the area below, you might get an answer of zero, which is clearly not the true area!

3.4 Area Between a Curve and a Line

Sometimes you need to find the area bounded by a curve \(y=f(x)\) and a straight line \(y=g(x)\).

The area between two functions, \(f(x)\) (the upper function) and \(g(x)\) (the lower function), between points \(a\) and \(b\) is found by integrating the difference between the two functions: $$ \text{Area} = \int_a^b (y_{\text{upper}} - y_{\text{lower}}) \,dx = \int_a^b (f(x) - g(x)) \,dx $$

Step-by-Step: Area Between Two Curves
  1. Find Intersections (\(a\) and \(b\)): Set \(f(x) = g(x)\) and solve to find the limits of integration.
  2. Identify Upper/Lower: Determine which function is on top within the required region (you can test a point between \(a\) and \(b\)).
  3. Set up Integral: Write the integral of the difference: \(\int_a^b (f(x) - g(x)) \,dx\).
  4. Integrate and Evaluate: Use definite integration rules to find the numerical area.
Key Takeaway for P2 Integration

The core skills required for P2 are:
1. Algebraic preparation (converting roots/fractions to \(x^n\)).
2. Applying the Power Rule correctly (including dividing by the new power).
3. Understanding when to use \(+C\) (indefinite) versus limits (definite).
4. Dealing with negative areas separately when calculating total bounded area.