CIE AS/A Level Maths-1.8 Integration- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-1.8 Integration- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-1.8 Integration- Study Notes
Key Concepts:
- Integration as the Reverse Process of Differentiation
- Solving Problems Involving the Constant of Integration
- Evaluating Definite Integrals
- Finding Areas Using Definite Integration
- Volume of Revolution
Integration as the Reverse Process of Differentiation
Integration as the Reverse Process of Differentiation
Integration is the process of finding the original function from its derivative. If a function \( F(x) \) has a derivative \( f(x) \), then \( F(x) \) is called an antiderivative of \( f(x) \). The integral of \( f(x) \) with respect to \( x \) is written as:
\( \displaystyle \int f(x) \, dx = F(x) + C \)
The symbol \( \displaystyle \int \) denotes integration, \( dx \) indicates the variable of integration, and \( C \) is the constant of integration. The constant \( C \) appears because differentiation of any constant is zero, so the original function could have contained an arbitrary constant.
Basic Principle:
Integration essentially reverses the process of differentiation. For any function \( f(x) \), the integral \(\displaystyle \int f(x) dx \) represents a family of functions whose derivative is \( f(x) \).
Power Rule for Integration:
If \( f(x) = (ax + b)^n \), where \( n \) is any rational number except –1, the integral is given by:
\( \displaystyle \int (ax + b)^n dx = \dfrac{(ax + b)^{n+1}}{a(n+1)} + C \)
Here, \( n \neq -1 \) because when \( n = -1 \), the integral becomes \( \int \frac{1}{ax + b} dx \), which is a special case resulting in a logarithmic function.
The factor \( a \) in the denominator arises from the chain rule in differentiation. When differentiating \( (ax + b)^{n+1} \), the derivative of \( ax + b \) is \( a \), which must be compensated for during integration.
Integration of Constant Multiples:
If a function is multiplied by a constant \( k \), the constant can be factored out of the integral. This is expressed as:
\( \displaystyle \int k \cdot f(x) dx = k \int f(x) dx \)
This property allows simplification of integrals when constants are involved.
Integration of Sums and Differences:
Integration is a linear operation, which means it distributes over addition and subtraction. For any two functions \( f(x) \) and \( g(x) \), we have:
\( \displaystyle \int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx \)
\( \displaystyle \int [f(x) – g(x)] dx = \int f(x) dx – \int g(x) dx \)
This property is fundamental because it allows complex integrals to be broken down into simpler parts that can be integrated individually.
Example:
Evaluate the following integrals:
\( \displaystyle \int (2x^3 – 5x + 1) \, dx \), \( \displaystyle \int \frac{1}{(2x+3)^2} \, dx \)
▶️ Answer/Explanation
Step 1: Break the integral into simpler terms
\( \displaystyle \int (2x^3 – 5x + 1) \, dx = \int 2x^3 \, dx – \int 5x \, dx + \int 1 \, dx \)
Step 2: Apply the power rule for each term
\( \int 2x^3 \, dx = \dfrac{2x^4}{4} = \dfrac{x^4}{2} \)
\( \int 5x \, dx = \dfrac{5x^2}{2} \)
\( \int 1 \, dx = x \)
Step 3: Combine results and add constant of integration
\( \displaystyle \int (2x^3 – 5x + 1) \, dx = \dfrac{x^4}{2} – \dfrac{5x^2}{2} + x + C \)
Step 4: Second integral using the formula for \(\int (ax+b)^n dx\)
\( \displaystyle \int \frac{1}{(2x+3)^2} \, dx = \int (2x+3)^{-2} dx = \dfrac{(2x+3)^{-1}}{2(-1)} + C = -\dfrac{1}{2(2x+3)} + C \)
Example:
Evaluate the following integrals:
\( \displaystyle \int \left(4x^2 – \dfrac{3}{x} + 5\right) dx \), \(\displaystyle \int (3x+1)^{-3} dx \)
▶️ Answer/Explanation
Step 1: Separate the first integral into simpler terms
\( \displaystyle \int \left(4x^2 – \dfrac{3}{x} + 5\right) dx = \displaystyle \int 4x^2 dx – \displaystyle \int \dfrac{3}{x} dx + \displaystyle \int 5\, dx \).
Step 2: Apply the power rule and the logarithmic rule
\( \displaystyle \int 4x^2 dx = 4 \cdot \dfrac{x^3}{3} = \dfrac{4x^3}{3} \).
\( \displaystyle \int \dfrac{3}{x} dx = 3 \displaystyle \int \dfrac{1}{x} dx = 3 \ln|x| \).
\( \displaystyle \int 5\, dx = 5x \).
Step 3: Combine results and add the constant of integration
\( \displaystyle \int \left(4x^2 – \dfrac{3}{x} + 5\right) dx = \dfrac{4x^3}{3} – 3 \ln|x| + 5x + C \).
Step 4: Second integral using the power rule for \(\displaystyle \int (ax+b)^n dx\)
\( \displaystyle \int (3x+1)^{-3} dx = \dfrac{(3x+1)^{-2}}{3(-2)} + C = -\dfrac{1}{6} (3x+1)^{-2} + C \).
Final Answers:
\( \displaystyle \int \left(4x^2 – \dfrac{3}{x} + 5\right) dx = \dfrac{4x^3}{3} – 3 \ln|x| + 5x + C \).
\( \displaystyle \int (3x+1)^{-3} dx = -\dfrac{1}{6} (3x+1)^{-2} + C \).
Solving Problems Involving the Constant of Integration
Solving Problems Involving the Constant of Integration
When we are given a differential equation of the form \(\dfrac{dy}{dx} = f(x)\), we can find the equation of the curve passing through a specific point \((x_0, y_0)\) by following these steps:
Step 1: Integrate the differential equation
The derivative \(\dfrac{dy}{dx}\) gives the slope of the curve. To find \(y\), integrate both sides with respect to \(x\):
\(y = \displaystyle \int f(x) \, dx\)
This introduces a constant of integration, usually denoted by \(C\), because indefinite integration only determines the curve up to a vertical shift.
Step 2: Apply substitution if needed
Sometimes the integral may require substitution, especially if it involves functions like \(\sqrt{ax+b}\), \((ax+b)^n\), or composite functions. Common substitutions:
- Let \(u = ax+b \implies du = a dx\)
- Then replace \(dx\) and the function in terms of \(u\)
- Integrate in terms of \(u\), then substitute back \(x\)
Step 3: Use the given point to find the constant of integration \(C\)
If the curve passes through a point \((x_0, y_0)\), substitute \(x_0\) and \(y_0\) into the integrated equation:
\(y_0 = \text{(integrated expression in } x_0) + C\)
Solve for \(C\) to determine the exact curve.
Step 4: Write the final equation of the curve
Substitute the value of \(C\) into the integrated equation to get the explicit equation of the curve:
\(y = \text{(integrated expression)} + C\)
Example:
Find the equation of the curve passing through the point \((1, -2)\) for which \(\dfrac{dy}{dx} = \sqrt{2x+1}\).
▶️ Answer/Explanation
Step 1: Integrate the differential equation
\(y = \int \sqrt{2x+1} \, dx\)
Step 2: Use substitution
Let \(u = 2x+1 \implies du = 2 dx \implies dx = \dfrac{du}{2}\)
\(y = \dfrac{1}{2} \int u^{1/2} du = \dfrac{1}{3} u^{3/2} + C = \dfrac{1}{3} (2x+1)^{3/2} + C\)
Step 3: Apply the given point to find \(C\)
Substitute \(x=1\), \(y=-2\):
\(-2 = \dfrac{1}{3} (2(1)+1)^{3/2} + C \implies -2 = \dfrac{1}{3} (3)^{3/2} + C\)
\(C = -2 – \sqrt{3}\)
Step 4: Write the final equation of the curve
\(y = \dfrac{1}{3} (2x+1)^{3/2} – 2 – \sqrt{3}\)
Example:
Find the equation of the curve passing through the point \((0, 1)\) for which \(\dfrac{dy}{dx} = 3x^2 + 2x\).
▶️ Answer/Explanation
Step 1: Integrate the differential equation
\(y = \int (3x^2 + 2x) \, dx = \int 3x^2 dx + \int 2x dx\)
\(y = x^3 + x^2 + C\)
Step 2: Apply the given point to find \(C\)
Substitute \(x=0\), \(y=1\):
\(1 = 0^3 + 0^2 + C \implies C = 1\)
Step 3: Write the final equation of the curve
\(y = x^3 + x^2 + 1\)
Evaluating Definite Integrals
Evaluating Definite Integrals
A definite integral calculates the area under a curve between two limits \(a\) and \(b\):
\(\displaystyle \int_a^b f(x) \, dx\)
The steps to evaluate a definite integral are:
- Find the indefinite integral \(F(x) =\displaystyle \int f(x) dx\).
- Apply the Fundamental Theorem of Calculus:
Step 1: Identify the function and limits
Check the function \(f(x)\) and the limits of integration \(a\) and \(b\).
Step 2: Integrate the function
Use standard formulas, substitution, or integration by parts as needed.
Step 3: Apply limits
Substitute \(b\) and \(a\) into the antiderivative and subtract: \(F(b) – F(a)\).
Improper Integrals (Simple Cases)
An improper integral occurs when:
- The interval of integration is infinite, e.g., \(\displaystyle \int_a^\infty f(x) dx\).
- The integrand becomes infinite at a limit, e.g., \(\displaystyle \int_0^1 \dfrac{1}{\sqrt{x}} dx\).
To evaluate, replace the problematic limit with a variable and take a limit:
\(\displaystyle \int_a^\infty f(x) dx = \lim_{t \to \infty} \int_a^t f(x) dx\)
\(\displaystyle \int_0^b f(x) dx = \lim_{t \to 0^+} \int_t^b f(x) dx\)
Example:
Evaluate \(\displaystyle \int_1^3 (2x + 1) dx\).
▶️ Answer/Explanation
Step 1: Integrate the function
\(\int (2x + 1) dx = x^2 + x + C\)
Step 2: Apply the limits
\(F(3) – F(1) = (3^2 + 3) – (1^2 + 1) = (9 + 3) – (1 + 1) = 12 – 2 = 10\)
Final Answer: 10
Example:
Evaluate \(\displaystyle \int_1^\infty \frac{1}{x^2} dx\).
▶️ Answer/Explanation
Step 1: Replace the infinite limit with a variable
\(\displaystyle \int_1^\infty \frac{1}{x^2} dx = \lim_{t \to \infty} \int_1^t x^{-2} dx\)
Step 2: Integrate
\(\int x^{-2} dx = -x^{-1} + C = -\dfrac{1}{x} + C\)
Step 3: Apply limits and take the limit
\(\lim_{t \to \infty} \left[ -\dfrac{1}{t} + \dfrac{1}{1} \right] = 0 + 1 = 1\)
Final Answer: 1
Example:
Evaluate the improper integral \(\displaystyle \int_0^1 x^{-1/2} dx\).
▶️ Answer/Explanation
Step 1: Recognize the singularity at \(x=0\)
Since \(x^{-1/2}\) becomes infinite at \(x=0\), we rewrite the integral as a limit:
\(\displaystyle \int_0^1 x^{-1/2} dx = \lim_{t \to 0^+} \int_t^1 x^{-1/2} dx\)
Step 2: Integrate
\(\int x^{-1/2} dx = 2 x^{1/2} + C\)
Step 3: Apply the limits
\(\lim_{t \to 0^+} \left[ 2(1)^{1/2} – 2(t)^{1/2} \right] = 2 – 0 = 2\)
Final Answer: 2
Finding Areas Using Definite Integration
Finding Areas Using Definite Integration
Definite integrals can be used to calculate the area of a region bounded by curves, lines, or axes. The general idea is:
\(\text{Area} = \displaystyle\int_a^b f(x) \, dx\)
where \(f(x) \ge 0\) in the interval \([a, b]\).
Case 1: Area under a curve and lines parallel to the x-axis
If a curve \(y = f(x)\) is bounded between \(x = a\) and \(x = b\) and above the x-axis:
\(\text{Area} = \displaystyle\int_a^b f(x) dx\)
If part of the curve is below the x-axis, the integral gives a signed area. To get the actual area, take the absolute value where necessary.
Case 2: Area between a curve and a line
If a curve \(y = f(x)\) is bounded between \(x = a\) and \(x = b\) and a horizontal line \(y = c\), the area is:
\(\text{Area} = \displaystyle\int_a^b |f(x) – c| dx\)
Take absolute value to ensure the area is positive.
Case 3: Area between two curves
If a region is bounded by two curves \(y = f(x)\) and \(y = g(x)\) with \(f(x) \ge g(x)\) for \(x \in [a, b]\):
\(\text{Area} = \displaystyle\int_a^b [f(x) – g(x)] dx\)
Steps:
- Find points of intersection to determine \(a\) and \(b\).
- Identify which curve is on top (\(f(x)\)) and which is below (\(g(x)\)).
- Integrate the difference \(f(x) – g(x)\) over \([a,b]\).
Case 4: Area with vertical and horizontal boundaries
Sometimes the region is bounded by \(x = a\), \(x = b\), \(y = f(x)\) and \(y = g(x)\) or curves with vertical lines as boundaries. The area formula remains:
\(\text{Area} = \displaystyle\int_a^b |f(x) – g(x)| dx\)
Choose \(f(x)\) as the function on top and \(g(x)\) as the function below.
Important Notes:
- Always check which function is above the other over the interval to avoid negative area.
- If the curves cross, break the integral at the intersection points and sum the areas of sub-intervals.
- Definite integration automatically accounts for the width of the interval and accumulation of area.
Example
Find the area under the curve \(y = x^2 + 1\) from \(x = 0\) to \(x = 2\).
▶️ Answer/Explanation
Step 1: Integrate the function
\(\int_0^2 (x^2 + 1) dx = \left[ \dfrac{x^3}{3} + x \right]_0^2\)
Step 2: Apply limits
\(\dfrac{8}{3} + 2 – (0 + 0) = \dfrac{14}{3}\)
Final Answer: \(\dfrac{14}{3}\)
Example
Find the area between \(y = x^2\) and the line \(y = 4\) from \(x = 0\) to \(x = 2\).
▶️ Answer/Explanation
Step 1: Express the area as an integral
\(\int_0^2 |4 – x^2| dx = \int_0^2 (4 – x^2) dx\)
Step 2: Integrate
\(\int_0^2 4 dx – \int_0^2 x^2 dx = [4x]_0^2 – [\dfrac{x^3}{3}]_0^2 = 8 – \dfrac{8}{3} = \dfrac{16}{3}\)
Final Answer: \(\dfrac{16}{3}\)
Example
Find the area between \(y = x^2\) and \(y = x\) from \(x = 0\) to \(x = 1\).
▶️ Answer/Explanation
Step 1: Identify top and bottom curves
For \(0 \le x \le 1\), \(x \ge x^2 \implies f(x) = x\), \(g(x) = x^2\)
Step 2: Integrate the difference
\(\int_0^1 (x – x^2) dx = \left[ \dfrac{x^2}{2} – \dfrac{x^3}{3} \right]_0^1 = \dfrac{1}{2} – \dfrac{1}{3} = \dfrac{1}{6}\)
Final Answer: \(\dfrac{1}{6}\)
Example
Find the area under the curve \(y = x^3 – x\) from \(x = -1\) to \(x = 1\).
▶️ Answer/Explanation
Step 1: Split the integral where curve changes sign
Curve crosses x-axis at \(x = -1, 0, 1\). Use absolute value:
\(\text{Area} = \int_{-1}^{0} |x^3 – x| dx + \int_0^1 |x^3 – x| dx\)
Step 2: Simplify absolute value
For \(-1 \le x \le 0\), \(x^3 – x = x(x^2 -1) < 0 \implies |x^3 – x| = -(x^3 – x) = -x^3 + x\)
For \(0 \le x \le 1\), \(x^3 – x < 0 \implies |x^3 – x| = -x^3 + x\)
Step 3: Integrate
\(\int_{-1}^0 (-x^3 + x) dx = \left[-\dfrac{x^4}{4} + \dfrac{x^2}{2} \right]_{-1}^0 = 0 – \left(-\dfrac{1}{4} + \dfrac{1}{2}\right) = \dfrac{1}{4}\)
\(\int_0^1 (-x^3 + x) dx = \left[-\dfrac{x^4}{4} + \dfrac{x^2}{2} \right]_0^1 = -\dfrac{1}{4} + \dfrac{1}{2} = \dfrac{1}{4}\)
Step 4: Total area
\(\text{Area} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}\)
Volume of Revolution
Volume of Revolution
A solid of revolution is formed when a plane region is rotated about an axis. The volume can be calculated using definite integrals with the following formulas.
Case 1: Revolution about the x-axis
If a curve \(y = f(x)\) is rotated about the x-axis from \(x = a\) to \(x = b\), the volume is given by the disk method:
\(\text{Volume} = \pi \displaystyle\int_a^b [f(x)]^2 dx\)
Steps:
- Square the function \(f(x)\) to get the radius of the disk.
- Integrate over the interval \([a, b]\).
- Multiply by \(\pi\) to account for the circular cross-section.
Case 2: Revolution about the y-axis
If a curve \(x = g(y)\) is rotated about the y-axis from \(y = c\) to \(y = d\), the volume is:
\(\text{Volume} = \pi \displaystyle\int_c^d [g(y)]^2 dy\)
Steps:
- Square the function \(g(y)\) to get the radius of the disk.
- Integrate with respect to \(y\) over \([c, d]\).
- Multiply by \(\pi\).
Case 3: Using shells (optional for y-axis if curve in x)
If a curve \(y = f(x)\) is rotated about the y-axis, an alternative is the cylindrical shells method:
\(\text{Volume} = 2 \pi \displaystyle\int_a^b x \, f(x) dx\)
This method is particularly useful when the function is expressed as \(y = f(x)\) and rotation is around the y-axis.
Example:
Find the volume of the solid formed when \(y = \sqrt{x}\) is rotated about the x-axis from \(x = 0\) to \(x = 4\).
▶️ Answer/Explanation
Step 1: Use the disk method formula
\(\text{Volume} = \pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx\)
Step 2: Integrate
\(\pi \int_0^4 x dx = \pi \left[ \dfrac{x^2}{2} \right]_0^4 = \pi \cdot \dfrac{16}{2} = 8 \pi\)
Final Answer: \(8 \pi\)
Example:
Find the volume of the solid formed when \(x = \sqrt{y}\) is rotated about the y-axis from \(y = 0\) to \(y = 4\).
▶️ Answer/Explanation
Step 1: Use the disk method formula about y-axis
\(\text{Volume} = \pi \int_0^4 (\sqrt{y})^2 dy = \pi \int_0^4 y dy\)
Step 2: Integrate
\(\pi \int_0^4 y dy = \pi \left[ \dfrac{y^2}{2} \right]_0^4 = \pi \cdot \dfrac{16}{2} = 8 \pi\)
Final Answer: \(8 \pi\)
Example:
Find the volume of the solid formed when \(y = x^2\) is rotated about the y-axis from \(x = 0\) to \(x = 2\) using cylindrical shells.
▶️ Answer/Explanation
Step 1: Use the shells method formula
\(\text{Volume} = 2 \pi \int_0^2 x (x^2) dx = 2 \pi \int_0^2 x^3 dx\)
Step 2: Integrate
\(2 \pi \int_0^2 x^3 dx = 2 \pi \left[ \dfrac{x^4}{4} \right]_0^2 = 2 \pi \cdot \dfrac{16}{4} = 8 \pi\)
Final Answer: \(8 \pi\)