Chapter: Integration (AS 1: Pure Mathematics)
Welcome to Integration! If you have already studied differentiation, you know how to find the gradient (rate of change) of a curve. Integration is simply the reverse process of differentiation. Think of differentiation like peeling an orange to see what's inside, and integration like putting the orange back together again.
Don't worry if this seems abstract right now. By breaking it down into simple, step-by-step rules, you will quickly see that integration is one of the most powerful and rewarding tools in AS Pure Mathematics.
1. Indefinite Integration: The Power Rule
The Concept of Anti-Differentiation
When you differentiate \(y = x^3\), you multiply by the power and subtract \(1\) from the power to get \(\frac{dy}{dx} = 3x^2\).
To reverse this process (integrate), we do the exact opposite in reverse order: add \(1\) to the power, then divide by the new power.
The Fundamental Rule
For any power of \(x\) (where \(n \neq -1\)):
\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + c\)
Here is what the symbols mean:
• \(\int\) is the integral sign (an elongated 'S' for sum).
• \(dx\) tells you that you are integrating with respect to the variable \(x\).
• \(c\) is the constant of integration.
Why do we need "\( + c\)"?
Consider differentiating the following functions:
• \(y = x^2 + 5 \implies \frac{dy}{dx} = 2x\)
• \(y = x^2 - 100 \implies \frac{dy}{dx} = 2x\)
• \(y = x^2 \implies \frac{dy}{dx} = 2x\)
Notice that all of them differentiate to give \(2x\) because the constant number disappears (its gradient is \(0\)). Therefore, when we go backwards and integrate \(2x\), we cannot know what the original constant was unless we have extra information. We write \(+ c\) to represent any possible constant number.
Integrating Constants and Multiples
• Constant multiple: \(\int k f(x) \, dx = k \int f(x) \, dx\)
• Constant number: \(\int k \, dx = kx + c\) (Think of \(k\) as \(k x^0\), so adding \(1\) to the power gives \(k x^1\))
• Sum or difference: \(\int \left( f(x) \pm g(x) \right) \, dx = \int f(x) \, dx \pm \int g(x) \, dx\)
Memory Aid: The Two Steps
To integrate any single power of \(x\):
Step 1: Power UP (Add \(1\) to the index).
Step 2: Divide DOWN (Divide by the new index).
Step 3: Don't forget \(+ c\)!
Preparing Expressions Before Integrating
Just like in differentiation, you must rewrite terms in index form \(a x^n\) before applying the rule:
• Fractions with powers in the denominator: \(\frac{1}{x^3} = x^{-3}\)
• Roots: \(\sqrt{x} = x^{\frac{1}{2}}\) and \(\sqrt[3]{x^2} = x^{\frac{2}{3}}\)
• Products (brackets): Expand them first, e.g., \((x + 2)(x - 3) = x^2 - x - 6\)
• Fractions with a single denominator: Split the fraction into separate terms, e.g., \(\frac{x^3 + 4x}{x^2} = \frac{x^3}{x^2} + \frac{4x}{x^2} = x + 4x^{-1}\)
Worked Example: Indefinite Integration
Find \(\int \left( 6x^2 - \frac{4}{\sqrt{x}} + 5 \right) \, dx\).
Step 1: Rewrite in index form:
\(= \int \left( 6x^2 - 4x^{-\frac{1}{2}} + 5 \right) \, dx\)
Step 2: Apply the power rule to each term:
\(= \frac{6x^{2+1}}{3} - \frac{4x^{-\frac{1}{2}+1}}{\frac{1}{2}} + 5x + c\)
Step 3: Simplify the coefficients:
\(= 2x^3 - 8x^{\frac{1}{2}} + 5x + c\) (or \(2x^3 - 8\sqrt{x} + 5x + c\))
Common Mistakes to Avoid
• Forgetting \(+ c\): Always write \(+ c\) for indefinite integrals.
• Integrating terms in a product separately: You cannot do \(\int (x)(x+1) \, dx = (\int x \, dx)(\int (x+1) \, dx)\). You MUST expand brackets first.
• Fractional powers error: Dividing by \(\frac{1}{2}\) means multiplying by \(2\), not dividing by \(2\).
Key Takeaway
For indefinite integrals, rewrite every term into \(a x^n\) form, add \(1\) to the power, divide by the new power, and always add the constant of integration \(+ c\).
2. Finding the Constant of Integration (\(c\))
If you are given the gradient function \(\frac{dy}{dx}\) and the coordinates of a point \((x, y)\) through which the curve passes, you can find the exact value of \(c\).
Step-by-Step Method
Step 1: Integrate the gradient function \(\frac{dy}{dx}\) to obtain \(y = F(x) + c\).
Step 2: Substitute the given coordinates \((x, y)\) into the integrated equation.
Step 3: Solve the resulting equation for \(c\).
Step 4: Rewrite the final equation of the curve with your calculated value of \(c\).
Worked Example
A curve has gradient \(\frac{dy}{dx} = 3x^2 - 4x + 1\) and passes through the point \((2, 7)\). Find the equation of the curve.
Step 1: Integrate \(\frac{dy}{dx}\):
\(y = \int (3x^2 - 4x + 1) \, dx\)
\(y = \frac{3x^3}{3} - \frac{4x^2}{2} + x + c\)
\(y = x^3 - 2x^2 + x + c\)
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 \(c\):
\(c = 5\)
Step 4: State the final equation:
\(y = x^3 - 2x^2 + x + 5\)
Key Takeaway
To find the equation of a curve from its derivative, integrate to find the general equation, substitute the known point \((x, y)\), and solve for \(c\).
3. Definite Integrals
What is a Definite Integral?
A definite integral has numbers at the top and bottom of the integral sign, called the limits of integration:
\(\int_a^b f(x) \, dx\)
Here, \(b\) is the upper limit and \(a\) is the lower limit.
Definite integrals give a numerical value, not a function of \(x\). You do not need to include \(+ c\) because it cancels out when you subtract!
How to Evaluate a Definite Integral
By the Fundamental Theorem of Calculus:
\(\int_a^b f(x) \, dx = \left[ F(x) \right]_a^b = F(b) - F(a)\)
where \(F(x)\) is the integrated function.
Step-by-Step Procedure
Step 1: Integrate the function inside square brackets \([ \; ]\), writing the limits on the right bracket.
Step 2: Substitute the upper limit \(b\) into the integrated expression.
Step 3: Substitute the lower limit \(a\) into the integrated expression.
Step 4: Subtract the second bracket from the first: \(( \text{Top} ) - ( \text{Bottom} )\).
Worked Example
Evaluate \(\int_1^3 (3x^2 - 2x) \, dx\).
Step 1: Integrate into square brackets:
\(\left[ x^3 - x^2 \right]_1^3\)
Step 2 & 3: Substitute upper limit \(3\) and lower limit \(1\):
\(= \left( (3)^3 - (3)^2 \right) - \left( (1)^3 - (1)^2 \right)\)
\(= (27 - 9) - (1 - 1)\)
\(= 18 - 0 = 18\)
Quick Review: Handling Negative Limits
Be extremely careful with minus signs when substituting negative limits. Always write full brackets around terms, for example: \((-2)^3 = -8\), whereas \(-(-2)^2 = -4\).
Key Takeaway
Definite integration produces a numerical value using \(\left[ F(x) \right]_a^b = F(b) - F(a)\). No \(+ c\) is needed.
4. Finding Areas Under Curves
Geometric Meaning of the Integral
One of the primary applications of definite integration is finding the area bounded by a curve \(y = f(x)\), the \(x\)-axis, and the vertical lines \(x = a\) and \(x = b\).
\(\text{Area} = \int_a^b y \, dx = \int_a^b f(x) \, dx\)
Case 1: Area Above the \(x\)-axis
When the curve lies entirely above the \(x\)-axis between \(x = a\) and \(x = b\), the value of the integral is positive and represents the exact area.
Case 2: Area Below the \(x\)-axis
When a curve lies entirely below the \(x\)-axis, the definite integral gives a negative value. Since physical area must always be positive, take the absolute value (magnitude) of the result:
\(\text{Area} = \left| \int_a^b y \, dx \right|\)
Case 3: Area Crossing the \(x\)-axis
Did you know? If a curve has parts both above and below the \(x\)-axis, evaluating a single integral from start to end will cause the negative area below the axis to cancel out some of the positive area above!
How to handle regions above and below the \(x\)-axis:
1. Find where the curve crosses the \(x\)-axis by solving \(y = 0\).
2. Split the integral into separate regions at each \(x\)-intercept.
3. Evaluate each integral individually.
4. Make any negative area positive, then add all the separate areas together.
Worked Example: Area Crossing the \(x\)-axis
Find the total area enclosed between the curve \(y = x^2 - 4\) and the \(x\)-axis from \(x = 0\) to \(x = 3\).
Step 1: Find intercepts by setting \(y = 0\):
\(x^2 - 4 = 0 \implies x = 2\) (within the interval \(0 \le x \le 3\)).
Between \(x = 0\) and \(x = 2\), the curve is below the \(x\)-axis.
Between \(x = 2\) and \(x = 3\), the curve is above the \(x\)-axis.
Step 2: Integrate Region 1 (\(x = 0\) to \(x = 2\)):
\(\int_0^2 (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_0^2 = \left( \frac{8}{3} - 8 \right) - (0) = -\frac{16}{3}\)
Therefore, \(\text{Area}_1 = \frac{16}{3}\).
Step 3: Integrate Region 2 (\(x = 2\) to \(x = 3\)):
\(\int_2^3 (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_2^3 = \left( \frac{27}{3} - 12 \right) - \left( \frac{8}{3} - 8 \right) = (-3) - \left(-\frac{16}{3}\right) = \frac{7}{3}\)
Therefore, \(\text{Area}_2 = \frac{7}{3}\).
Step 4: Add the positive areas together:
\(\text{Total Area} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} = 7\frac{2}{3} \text{ square units}\).
Key Takeaway
Never integrate straight across an \(x\)-intercept when finding total area. Always split the integral at every point where \(y = 0\) and calculate the separate areas as positive values.
5. Area Between a Curve and a Line (or Two Curves)
To find the area trapped between two functions \(y_1\) and \(y_2\):
\(\text{Area} = \int_a^b \left( y_{\text{top}} - y_{\text{bottom}} \right) \, dx\)
Standard Method
Step 1: Find the limits \(a\) and \(b\) by equating the two functions (\(y_1 = y_2\)) and solving for \(x\).
Step 2: Identify which curve/line is on top in the interval (you can sketch the graphs or test a value of \(x\) between \(a\) and \(b\)).
Step 3: Subtract the bottom equation from the top equation: \((y_{\text{top}} - y_{\text{bottom}})\).
Step 4: Integrate the simplified expression between the limits \(a\) and \(b\).
Worked Example
Find the area enclosed between the curve \(y = 4x - x^2\) and the line \(y = x\).
Step 1: Find intersection points:
\(4x - x^2 = x\)
\(3x - x^2 = 0\)
\(x(3 - x) = 0 \implies x = 0 \text{ and } x = 3\)
So the limits are \(a = 0\) and \(b = 3\).
Step 2: Identify top and bottom curves:
For \(x = 1\): curve \(y = 4(1) - 1^2 = 3\), line \(y = 1\).
The curve is on top (\(y_{\text{top}} = 4x - x^2\)) and the line is on the bottom (\(y_{\text{bottom}} = x\)).
Step 3 & 4: Set up and evaluate the integral:
\(\text{Area} = \int_0^3 \left( (4x - x^2) - x \right) \, dx\)
\(\text{Area} = \int_0^3 (3x - x^2) \, dx\)
\(\text{Area} = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_0^3\)
\(\text{Area} = \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - (0)\)
\(\text{Area} = \left( \frac{27}{2} - 9 \right) = \frac{9}{2} = 4.5 \text{ square units}\).
Key Takeaway
The area between two curves is found using \(\int_a^b (y_{\text{top}} - y_{\text{bottom}}) \, dx\). Always find the \(x\)-coordinates of intersection first to determine your integration limits.
Quick Reference Summary & Revision Checklist
• Indefinite Integral: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + c\) (Power UP, Divide DOWN, add \(c\)).
• Constant Rule: \(\int k \, dx = kx + c\).
• Definite Integral: \(\int_a^b f(x) \, dx = \left[ F(x) \right]_a^b = F(b) - F(a)\).
• Area under curve: Split at \(x\)-intercepts if the graph dips below the \(x\)-axis.
• Area between two curves: \(\int_a^b (y_{\text{top}} - y_{\text{bottom}}) \, dx\).