1. Introduction: Welcome to Integration
Welcome to Integration! If you have already studied differentiation, you know how to find the rate of change or the gradient of a curve. Integration is simply the reverse process of differentiation. Think of differentiation as "unwinding" a function, and integration as "winding it back up".
In this chapter for AS 1 Pure Mathematics, you will discover how to reverse the differentiation process to find original functions, calculate unknown constants, and find the exact area under curves. Don't worry if this seems a bit abstract at first—once you learn the simple rules, integration becomes one of the most satisfying topics in A-Level Maths!
Real-World Analogy: Imagine driving a car. If your speedometer tells you your speed at every second (the rate of change of distance), integration is the mathematical tool that allows you to calculate the total distance you have traveled!
Did You Know? The elongated "S" symbol used for integration (\(\int\)) was introduced by Gottfried Wilhelm Leibniz in 1675. It stands for the Latin word summa (sum), because integration can be thought of as adding together infinitely many tiny slivers of area.
2. Indefinite Integration & The Power Rule
When we differentiate a term like \(x^3\), we multiply by the power and then subtract 1 from the power to get \(3x^2\).
Because integration is the exact opposite, we do the inverse operations in reverse order: we add 1 to the power and then divide by the new power.
The Fundamental Rule:
\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + c\) (where \(n \neq -1\))
Memory Trick: Remember the two-step chant: "Add one to the power, divide by the new power!"
Why do we need the \(+ c\)?
Consider differentiating the following functions:
• If \(y = x^2 + 5\), then \(\frac{dy}{dx} = 2x\)
• If \(y = x^2 - 100\), then \(\frac{dy}{dx} = 2x\)
• If \(y = x^2\), then \(\frac{dy}{dx} = 2x\)
Notice that the constant always disappears during differentiation! Therefore, when we reverse the process by integrating \(2x\), we cannot know what the original constant was just by looking at the derivative. To represent any possible constant number, we always add an arbitrary constant of integration, written as \(+ c\).
Key Rules for Integrating Expressions:
• Constant Multiple Rule: \(\int k f(x) \, dx = k \int f(x) \, dx\)
• Sum and Difference Rule: \(\int (f(x) \pm g(x)) \, dx = \int f(x) \, dx \pm \int g(x) \, dx\)
• Integrating a Constant: \(\int k \, dx = kx + c\)
Preparing Terms Before Integrating (Algebraic Preparation)
Just like in differentiation, before applying the integration rule, you must rewrite roots and fractions with \(x\) in the denominator as single powers of \(x\) using index laws:
• Fractions: \(\frac{1}{x^n} = x^{-n}\) (e.g. \(\frac{5}{x^3} = 5x^{-3}\))
• Roots: \(\sqrt[n]{x} = x^{\frac{1}{n}}\) (e.g. \(\sqrt{x} = x^{\frac{1}{2}}\) and \(\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\))
• Brackets & Quotients: Expand brackets and split fractions over a single denominator before integrating, as there is no simple "product rule" or "quotient rule" for basic integration!
Step-by-Step Example:
Problem: Find \(\int \left( 6x^2 - \frac{4}{x^2} + 3\sqrt{x} - 7 \right) \, dx\)
Step 1: Rewrite all terms in index form (\(x^n\)):
\(\int \left( 6x^2 - 4x^{-2} + 3x^{\frac{1}{2}} - 7 \right) \, dx\)
Step 2: Apply the integration rule to each term:
\(= \frac{6x^{2+1}}{2+1} - \frac{4x^{-2+1}}{-2+1} + \frac{3x^{\frac{1}{2}+1}}{\frac{1}{2}+1} - 7x + c\)
\(= \frac{6x^3}{3} - \frac{4x^{-1}}{-1} + \frac{3x^{\frac{3}{2}}}{\frac{3}{2}} - 7x + c\)
Step 3: Simplify coefficients and signs:
\(= 2x^3 + 4x^{-1} + 2x^{\frac{3}{2}} - 7x + c\)
\(= 2x^3 + \frac{4}{x} + 2x^{\frac{3}{2}} - 7x + c\)
Quick Review: For any indefinite integral (an integral without numbers on the top and bottom of the \(\int\) sign), always prepare terms into \(a x^n\) form, integrate each term individually, and never forget to write \(+ c\) at the end!
3. Finding the Constant of Integration (\(c\))
If an exam question provides extra information—such as a specific coordinate point \((x, y)\) that the curve passes through—you can determine the exact numerical value of \(c\). This is often called solving a differential equation or finding the equation of the curve given its gradient function \(\frac{dy}{dx}\).
Step-by-Step Method:
1. Integrate \(\frac{dy}{dx}\) with respect to \(x\) to get an expression for \(y\) including \(+ c\).
2. Substitute the given coordinates \((x, y)\) into your equation.
3. Solve the resulting linear equation to find the numerical value of \(c\).
4. Rewrite the complete equation for \(y\) with your calculated value of \(c\).
Step-by-Step Example:
Problem: A curve has gradient function \(\frac{dy}{dx} = 3x^2 - 4x + 1\) and passes through the point \((2, 7)\). Find the equation of the curve.
Step 1: Integrate the gradient function:
\(y = \int (3x^2 - 4x + 1) \, dx\)
\(y = \frac{3x^3}{3} - \frac{4x^2}{2} + x + c\)
\(y = x^3 - 2x^2 + x + c\)
Step 2: Substitute \(x = 2\) and \(y = 7\):
\(7 = (2)^3 - 2(2)^2 + (2) + c\)
\(7 = 8 - 8 + 2 + c\)
\(7 = 2 + c\)
Step 3: Solve for \(c\):
\(c = 7 - 2 = 5\)
Step 4: State the final equation:
\(y = x^3 - 2x^2 + x + 5\)
Key Takeaway: Having a known point allows you to nail down the exact curve out of the infinite family of parallel curves described by \(+ c\).
4. Definite Integrals
A definite integral has numbers written at the top and bottom of the integral sign, called limits of integration:
\(\int_{a}^{b} f(x) \, dx\)
Here, \(a\) is the lower limit and \(b\) is the upper limit. A definite integral evaluates to a single numerical value rather than a function with \(+ c\).
The Fundamental Theorem of Calculus (Evaluation Rule):
\(\int_{a}^{b} f(x) \, dx = \left[ F(x) \right]_{a}^{b} = F(b) - F(a)\)
where \(F(x)\) is the integrated function.
Why does \(+ c\) disappear?
If you included \(+ c\), you would compute \((F(b) + c) - (F(a) + c) = F(b) - F(a) + c - c = F(b) - F(a)\). Because the \(c\) terms always cancel out, we do not need to write \(+ c\) when evaluating definite integrals!
Step-by-Step Example:
Problem: Evaluate \(\int_{1}^{3} (3x^2 - 2x + 4) \, dx\)
Step 1: Integrate and place inside square brackets with limits outside:
\(\left[ \frac{3x^3}{3} - \frac{2x^2}{2} + 4x \right]_{1}^{3} = \left[ x^3 - x^2 + 4x \right]_{1}^{3}\)
Step 2: Substitute the upper limit (\(x = 3\)):
\(F(3) = (3)^3 - (3)^2 + 4(3) = 27 - 9 + 12 = 30\)
Step 3: Substitute the lower limit (\(x = 1\)):
\(F(1) = (1)^3 - (1)^2 + 4(1) = 1 - 1 + 4 = 4\)
Step 4: Subtract lower limit value from upper limit value (\(F(3) - F(1)\)):
\(30 - 4 = 26\)
Top Tip on Calculations: Always use brackets around the substituted expressions: \([ \dots ] - [ \dots ]\). Negative signs are the most common source of lost marks in definite integration!
5. Finding Areas Under Curves
One of the most important geometric applications of integration is finding the exact area bounded by a curve \(y = f(x)\), the \(x\)-axis, and the vertical lines \(x = a\) and \(x = b\).
Case 1: Area Entirely Above the \(x\)-Axis
When the curve lies completely on or above the \(x\)-axis between \(x = a\) and \(x = b\) (i.e. \(y \ge 0\)), the area \(A\) is given directly by the definite integral:
\(\text{Area} = \int_{a}^{b} y \, dx\)
Example: Find the area enclosed by the curve \(y = 4 - x^2\) and the \(x\)-axis.
• Find the limits: The curve crosses the \(x\)-axis when \(y = 0\):
\(4 - x^2 = 0 \implies x^2 = 4 \implies x = -2 \text{ and } x = 2\).
• Set up and evaluate the integral:
\(\text{Area} = \int_{-2}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}\)
\(\text{Upper limit } (x=2): 4(2) - \frac{2^3}{3} = 8 - \frac{8}{3} = \frac{16}{3}\)
\(\text{Lower limit } (x=-2): 4(-2) - \frac{(-2)^3}{3} = -8 - \left(-\frac{8}{3}\right) = -\frac{16}{3}\)
\(\text{Area} = \left(\frac{16}{3}\right) - \left(-\frac{16}{3}\right) = \frac{32}{3} \text{ square units}\)
Case 2: Area Below the \(x\)-Axis
When a curve lies entirely below the \(x\)-axis (where \(y < 0\)), integrating will give a negative numerical answer. Since geometric area can never be negative, you take the absolute (positive) magnitude of the value:
\(\text{Area} = \left| \int_{a}^{b} y \, dx \right|\)
Case 3: Regions Spanning Above and Below the \(x\)-Axis
Danger Alert! If a curve crosses the \(x\)-axis between your starting and ending values, you cannot integrate across the entire interval in a single step. The negative area below the axis will cancel out part of the positive area above the axis, giving an incorrect total area.
Strategy for Combined Areas:
1. Set \(y = 0\) to find any \(x\)-intercepts between your limits.
2. Split the integration into separate sections (e.g., from \(a\) to the root, and from the root to \(b\)).
3. Evaluate each integral separately.
4. Make the negative result positive, and add the separate positive areas together to get the total area.
Key Takeaway: Sketching the curve (or at least finding all \(x\)-intercepts) is essential before calculating areas, ensuring you know if and where the curve drops below the \(x\)-axis!
6. Common Mistakes to Avoid & Exam Checklist
• Forgetting \(+ c\): Always include \(+ c\) for indefinite integrals. You will lose an easy mark if it is missing!
• Dividing by zero: Remember that \(\int x^{-1} \, dx \neq \frac{x^0}{0}\). (The power rule is valid for all powers \(n\) except \(n = -1\)).
• Multiplying/Dividing incorrectly: Never integrate the numerator and denominator of a fraction separately. Split the fraction into individual terms first.
• Expanding brackets before integrating: For expressions like \(\int (2x + 1)(x - 3) \, dx\), always expand to \(\int (2x^2 - 5x - 3) \, dx\) before applying the power rule.
• Bracket errors with negative limits: Be meticulous when subtracting the lower limit in definite integration, especially with negative numbers: \(F(b) - (F(a))\).
Quick Chapter Summary:
• Indefinite Integral: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + c\)
• Finding \(c\): Integrate, substitute a given point \((x, y)\), solve for \(c\).
• Definite Integral: \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\)
• Area Under a Curve: \(\int_{a}^{b} y \, dx\); always split integrals at \(x\)-intercepts if the curve goes below the \(x\)-axis.