Welcome to Integration!
Welcome to one of the most powerful and exciting topics in CCEA GCSE Further Mathematics: Integration. In the previous chapters of Pure Mathematics, you learned about differentiation, which is all about finding the gradient or the rate of change. Integration is simply the reverse process!
Think of differentiation and integration like wrapping and unwrapping a present, or multiplying and dividing. If differentiation takes you from a distance to a speed, integration takes you from a speed back to a distance. In this unit, we will discover how to reverse differentiation, how to find the equations of curves, and how to calculate the exact area under a curve.
Did you know? Integration was developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century. It is used today by rocket scientists to calculate flight paths, by economists to predict market trends, and by game developers to simulate realistic physics!
Don't worry if this seems a bit abstract right now. We will break every single step down into easy, bite-sized pieces.
1. What is Indefinite Integration?
When we differentiate a term like \(x^3\), we multiply by the power and subtract \(1\) from the power to get \(3x^2\).
Because integration is the exact opposite (the inverse operation or anti-differentiation), we do the opposite steps in reverse order:
1. Add \(1\) to the power.
2. Divide by the new power.
The Fundamental Rule of Integration
For any term \(a x^n\) (where \(n \neq -1\)):
\(\int a x^n \, dx = \frac{a x^{n+1}}{n+1} + c\)
Let's look at the symbols:
\(\bullet\) The elongated 'S' symbol \(\int\) means "integrate with respect to...".
\(\bullet\) The \(dx\) at the end tells us that \(x\) is the variable we are integrating.
\(\bullet\) The \(+ c\) is the constant of integration.
Why do we need "+ c"?
Consider differentiating these three functions:
\(\bullet\) If \(y = x^2 + 5\), then \(\frac{dy}{dx} = 2x\)
\(\bullet\) If \(y = x^2 - 100\), then \(\frac{dy}{dx} = 2x\)
\(\bullet\) If \(y = x^2\), then \(\frac{dy}{dx} = 2x\)
Notice that whenever you differentiate a constant number, it disappears (becomes \(0\)). So, when we integrate \(2x\) backwards, we know the answer starts with \(x^2\), but we cannot know what constant number was originally there! It could have been \(+5\), \(-100\), or \(0\). To represent this unknown number, we always write \(+ c\).
Memory Trick: Whenever you do indefinite integration, remember: "Don't forget the \(+ c\)!" Losing the \(+ c\) is the easiest way to drop a mark in your exam.
Special Cases to Remember
\(\bullet\) Integrating a single constant: \(\int k \, dx = kx + c\). For example, \(\int 7 \, dx = 7x + c\).
\(\bullet\) Integrating a single variable: \(\int x \, dx = \int x^1 \, dx = \frac{x^2}{2} + c\).
Step-by-Step Example 1
Find \(\int (6x^2 - 8x + 5) \, dx\).
Step 1: Integrate each term separately by adding \(1\) to the power and dividing by the new power.
\(\bullet\) For \(6x^2\): \(\frac{6x^{2+1}}{2+1} = \frac{6x^3}{3} = 2x^3\)
\(\bullet\) For \(-8x\): \(\frac{-8x^{1+1}}{1+1} = \frac{-8x^2}{2} = -4x^2\)
\(\bullet\) For \(5\): \(5x\)
Step 2: Combine your terms and add the constant \(+ c\):
Answer: \(2x^3 - 4x^2 + 5x + c\)
Algebraic Preparation: Fractions and Brackets
Just like in differentiation, before integrating, you must rewrite expressions so that every term is in the form \(a x^n\):
\(\bullet\) Expand any brackets before integrating: \(\int x(x - 4) \, dx = \int (x^2 - 4x) \, dx = \frac{x^3}{3} - 2x^2 + c\).
\(\bullet\) Split single-denominator fractions: \(\int \frac{6x^3 + 4}{x^2} \, dx = \int (6x + 4x^{-2}) \, dx = 3x^2 - 4x^{-1} + c = 3x^2 - \frac{4}{x} + c\).
Key Takeaway for Section 1: Add \(1\) to the power, divide by the new power, and always add \(+ c\) for indefinite integrals!
2. Finding the Constant of Integration (\(c\))
Sometimes an exam question gives you extra information, such as the coordinates of a point \((x, y)\) through which the curve passes. This allows you to find the exact numerical value of \(c\)!
Step-by-Step Method:
1. Integrate the derivative \(\frac{dy}{dx}\) to get an expression for \(y\) that includes \(+ c\).
2. Substitute the given values of \(x\) and \(y\) into your equation.
3. Solve the resulting equation to find the value of \(c\).
4. Rewrite the full equation of the curve with your calculated value of \(c\).
Step-by-Step Example 2
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 \(\frac{dy}{dx}\) to find \(y\):
\(y = \int (3x^2 - 4x + 1) \, dx\)
\(y = \frac{3x^3}{3} - \frac{4x^2}{2} + x + c\)
\(y = x^3 - 2x^2 + x + c\)
Step 2: Substitute \(x = 2\) and \(y = 7\) into the equation:
\(7 = (2)^3 - 2(2)^2 + (2) + c\)
\(7 = 8 - 8 + 2 + c\)
\(7 = 2 + c\)
Step 3: Solve for \(c\):
\(c = 7 - 2 = 5\)
Step 4: State the final equation:
Answer: \(y = x^3 - 2x^2 + x + 5\)
Key Takeaway for Section 2: If you are given a coordinate point \((x, y)\), integrate first with \(+ c\), substitute the numbers in, and solve for \(c\).
3. Definite Integration
A definite integral has numbers at the top and bottom of the integral sign. These numbers are called limits.
We write it like this: \(\int_{a}^{b} f(x) \, dx\), where \(b\) is the upper limit and \(a\) is the lower limit.
How to Evaluate a Definite Integral
1. Integrate the function as normal, putting square brackets around the result: \(\left[ F(x) \right]_{a}^{b}\).
2. Notice: We do not need to write \(+ c\) for definite integrals because it cancels out during subtraction!
3. Substitute the upper limit \(b\) into the integrated expression: \(F(b)\).
4. Substitute the lower limit \(a\) into the integrated expression: \(F(a)\).
5. Subtract the lower limit result from the upper limit result: \(F(b) - F(a)\).
Step-by-Step Example 3
Evaluate \(\int_{1}^{3} (3x^2 + 2x) \, dx\).
Step 1: Integrate the terms:
\(\int (3x^2 + 2x) \, dx = \left[ x^3 + x^2 \right]_{1}^{3}\)
Step 2: Substitute the upper limit (\(x = 3\)):
\((3)^3 + (3)^2 = 27 + 9 = 36\)
Step 3: Substitute the lower limit (\(x = 1\)):
\((1)^3 + (1)^2 = 1 + 1 = 2\)
Step 4: Subtract the lower evaluation from the upper evaluation:
\(36 - 2 = 34\)
Answer: \(34\)
Key Takeaway for Section 3: Definite integration produces a single numerical value. Calculate \(\text{Top Limit} - \text{Bottom Limit}\) using square brackets.
4. Finding the Area Under a Curve
One of the primary uses of integration is finding the exact area bounded by a curve, the \(x\)-axis, and the vertical lines \(x = a\) and \(x = b\).
\(\text{Area} = \int_{a}^{b} y \, dx\)
Analogy: How Integration Finds Area
Imagine slicing the region under a curve into thousands of extremely thin vertical rectangles. Each rectangle has height \(y\) and a tiny width \(dx\). The area of each tiny rectangle is \(y \times dx\). Integration adds all these infinitely thin rectangles together to give the exact total area under the curve!
Step-by-Step Example 4
Find the area enclosed by the curve \(y = 6x - x^2\) and the \(x\)-axis.
Step 1: Find the limits (where the curve crosses the \(x\)-axis):
Set \(y = 0\):
\(6x - x^2 = 0\)
\(x(6 - x) = 0 \implies x = 0 \text{ or } x = 6\)
So our limits are \(a = 0\) (lower) and \(b = 6\) (upper).
Step 2: Set up and evaluate the definite integral:
\(\text{Area} = \int_{0}^{6} (6x - x^2) \, dx\)
\(\text{Area} = \left[ 3x^2 - \frac{x^3}{3} \right]_{0}^{6}\)
Step 3: Substitute the limits:
Upper limit (\(x = 6\)): \(3(6)^2 - \frac{(6)^3}{3} = 3(36) - \frac{216}{3} = 108 - 72 = 36\)
Lower limit (\(x = 0\)): \(3(0)^2 - \frac{(0)^3}{3} = 0 - 0 = 0\)
Step 4: Subtract:
\(\text{Area} = 36 - 0 = 36 \text{ units}^2\)
Areas Below the \(x\)-Axis
If a section of a curve drops below the \(x\)-axis, the value of \(y\) is negative, so the definite integral will give a negative value.
Because physical area is always positive, we simply take the positive value (magnitude) of the result:
\(\text{Physical Area} = \left| \int_{a}^{b} y \, dx \right|\)
Important Exam Warning: If a region has parts both above and below the \(x\)-axis, you must integrate each section separately and add their positive areas together! If you integrate across the entire range in one single step, the negative area will cancel out part of the positive area, giving an incorrect total.
Key Takeaway for Section 4: Area under a curve between \(x = a\) and \(x = b\) is given by \(\int_{a}^{b} y \, dx\). If the area lies below the \(x\)-axis, change the negative result to a positive value.
5. Area Between a Curve and a Line
In Further Mathematics, you will often need to calculate the area bounded between a curve and a straight line.
The General Method
1. Find the points of intersection between the line and curve by setting \(y_{\text{curve}} = y_{\text{line}}\) and solving for \(x\). These \(x\)-values become your limits \(a\) and \(b\).
2. Determine which graph is on top over that interval.
3. Subtract the bottom equation from the top equation:
\(\text{Area} = \int_{a}^{b} \left( y_{\text{top}} - y_{\text{bottom}} \right) \, dx\)
4. Evaluate the definite integral.
Step-by-Step Example 5
Find the area enclosed between the line \(y = x + 2\) and the curve \(y = x^2\).
Step 1: Find limits by finding points of intersection:
\(x^2 = x + 2\)
\(x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0 \implies x = -1 \text{ and } x = 2\)
Step 2: Set up \(\int (\text{top} - \text{bottom}) \, dx\):
Between \(x = -1\) and \(x = 2\), the straight line is above the curve.
\(\text{Area} = \int_{-1}^{2} \left[ (x + 2) - x^2 \right] \, dx = \int_{-1}^{2} (2 + x - x^2) \, dx\)
Step 3: Integrate:
\(\left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}\)
Step 4: Evaluate with limits:
Upper limit (\(x = 2\)): \(2(2) + \frac{(2)^2}{2} - \frac{(2)^3}{3} = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}\)
Lower limit (\(x = -1\)): \(2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3} = -2 + \frac{1}{2} - \left(-\frac{1}{3}\right) = -2 + \frac{1}{2} + \frac{1}{3} = -\frac{7}{6}\)
Step 5: Subtract:
\(\text{Area} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = 4.5 \text{ units}^2\)
Key Takeaway for Section 5: To find the area between two graphs, identify the intersection points for limits, and integrate \(\int (\text{Upper Graph} - \text{Lower Graph}) \, dx\).
Quick Review: Common Mistakes to Avoid
\(\bullet\) Forgetting \(+ c\): Always add \(+ c\) on indefinite integrals.
\(\bullet\) Mixing up Differentiation and Integration: Remember: in integration, the power gets bigger (\(+1\)). In differentiation, it gets smaller (\(-1\)).
\(\bullet\) Brackets and Negative Signs: When substituting negative numbers into limits, always use brackets on your calculator (e.g. write \((-2)^2\) instead of \(-2^2\)).
\(\bullet\) Integrating terms in denominators: Never integrate \(\frac{1}{x^2}\) as \(\frac{1}{\frac{x^3}{3}}\). Always rewrite it as \(x^{-2}\) first, then integrate to get \(-x^{-1} = -\frac{1}{x}\).
Summary Checklist
\(\bullet\) Indefinite Integral Formula: \(\int a x^n \, dx = \frac{a x^{n+1}}{n+1} + c\)
\(\bullet\) Finding \(c\): Integrate, substitute \((x, y)\), solve for \(c\).
\(\bullet\) Definite Integral Formula: \(\int_{a}^{b} f(x) \, dx = \left[ F(x) \right]_{a}^{b} = F(b) - F(a)\)
\(\bullet\) Area Under a Curve: Area \(= \int_{a}^{b} y \, dx\)