🚀 Chapter Study Notes: Integration (P2: Pure Maths)
Welcome to the world of Integration! If differentiation was about finding the rate of change (like speed), integration is the reverse process—it's about finding the original function, often used to calculate total accumulated amounts, like distance traveled, or the areas of complex shapes.
Don't worry if this seems tricky at first; integration is just reversing the steps you learned in differentiation. By breaking it down, you'll see it's a powerful and fundamental tool in Mathematics (9660).
Section 1: Indefinite Integration (The Anti-Derivative)
Indefinite Integration is the official term for reversing differentiation. It finds the original function, often called the antiderivative.
The Golden Rule: Never Forget +c
When you differentiate a constant, it disappears (its derivative is zero). This means when you integrate, you have no way of knowing what constant was there originally.
We solve this by always adding the Constant of Integration, \(+c\), at the end of every indefinite integral.
Analogy: Imagine differentiation is finding the speed of a car. Integration is finding the distance it travelled. If you only know the speed, you don't know where the journey started (was the initial distance 0 km, 5 km, or 100 km?). The \(+c\) represents that unknown starting position.
Key Takeaway:
If \( \frac{dy}{dx} = f(x) \), then \( y = \int f(x) dx = F(x) + c \).
To find the value of 'c', you need a known point \((x, y)\) on the original curve.
Basic Integration of Powers (P1 Core)
This is the fundamental rule for integrating polynomial terms.
The Power Rule:
$$ \int ax^n dx = \frac{a}{n+1}x^{n+1} + c $$
You increase the power by 1, and then divide by the new power.
- This rule applies for all rational numbers \(n\), EXCEPT \(n = -1\).
- Always rewrite terms like \(\frac{1}{x^2}\) as \(x^{-2}\) before integrating.
- Constants: \(\int a dx = ax + c\).
Example:
If we want to integrate \( 6x^2 + 5x^{-1/2} \):
$$ \int (6x^2 + 5x^{-1/2}) dx = \frac{6}{3}x^3 + \frac{5}{(-1/2) + 1}x^{(-1/2) + 1} + c $$
$$ = 2x^3 + \frac{5}{1/2}x^{1/2} + c = 2x^3 + 10\sqrt{x} + c $$
Quick Review: The Power Rule
\( \rightarrow \) Increase power by 1.
\( \rightarrow \) Divide by the new power.
\( \rightarrow \) Add \(+c\).
Section 2: Standard Integrals (P2 Extended)
In A Level Pure Maths, you extend the standard functions you can integrate.
1. The Special Case: \(n = -1\)
Since the power rule fails when \(n=-1\), we use the integral of \(\ln x\):
$$ \int \frac{1}{x} dx = \int x^{-1} dx = \ln |x| + c $$
Important: You must use the modulus sign, \(|x|\), because although the derivative of \(\ln x\) is only defined for \(x>0\), the integral of \(1/x\) exists for all \(x \ne 0\). The constant \(c\) must still be included.
2. Integrating \(e^x\) (Exponential Functions)
Exponential functions are very friendly when integrating:
- $$ \int e^{x} dx = e^{x} + c $$
- For a function with a linear factor in the power:
- $$ \int e^{kx} dx = \frac{1}{k} e^{kx} + c $$
Did you know? \(e^x\) is its own derivative and its own integral! It's one of the few functions that remains unchanged under these operations.
3. Integrating Trigonometric Functions
Remember that integration reverses differentiation, so be extra careful with signs!
If \( \frac{d}{dx}(\sin x) = \cos x \), then \(\int \cos x dx = \sin x + c\).
If \( \frac{d}{dx}(\cos x) = -\sin x \), then \(\int \sin x dx = -\cos x + c\).
Standard Trigonometric Integrals:
- $$ \int \cos x dx = \sin x + c $$
- $$ \int \sin x dx = -\cos x + c $$
- With a linear factor \(kx\) inside:
- $$ \int \cos(kx) dx = \frac{1}{k} \sin(kx) + c $$
- $$ \int \sin(kx) dx = -\frac{1}{k} \cos(kx) + c $$
Memory Aid (Sign Trick): When differentiating C gets a negative sign (\(\cos \rightarrow -\sin\)). When integrating, if the answer starts with C, it gets a negative sign (\(\sin \rightarrow -\cos\)).
Section 3: Definite Integration and Area (P1 Core)
Definite Integration involves finding the value of an integral between two specific limits, \(a\) (lower limit) and \(b\) (upper limit).
$$ \int^b_a f(x) dx = [F(x)]^b_a = F(b) - F(a) $$
Notice that the constant of integration \(c\) is not needed for definite integrals, because it cancels out: \((F(b) + c) - (F(a) + c) = F(b) - F(a)\).
Geometric Interpretation: Area Under a Curve
The definite integral \(\int^b_a y dx\) represents the area between the curve \(y=f(x)\), the x-axis, and the vertical lines \(x=a\) and \(x=b\).
Crucial Point about Area:
If the region is entirely above the x-axis (\(y \ge 0\)), the definite integral gives the area directly (a positive value).
If the region is entirely below the x-axis (\(y < 0\)), the definite integral will give a negative value. The actual area is the absolute (positive) value of this result.
Common Mistake to Avoid: If a curve crosses the x-axis between \(a\) and \(b\), you must split the integral into two parts (one for the positive area, one for the negative area), integrate them separately, and then add their absolute values together to find the total area.
Key Takeaway: Integration finds the signed area (area above x-axis minus area below x-axis). To find the total area, integrate parts separately and take the magnitude of any negative result.
Section 4: Advanced Integration Techniques (P2 Extended)
These methods are essential when the function is not a simple polynomial or standard trig/exp function. They are the reverse of the Chain Rule, Product Rule, and algebraic processes.
1. Integration by Inspection (Reverse Chain Rule)
We look for functions of the form \(\int f'(x) [g(x)]^n dx\) or \(\int \frac{f'(x)}{f(x)} dx\).
Type 1: Reverse Power Rule
If you have a function multiplied by its derivative, you can often integrate quickly.
Example: \(\int x(x^2+1)^3 dx\). The derivative of the inner function (\(x^2+1\)) is \(2x\). We have \(x\), so we need to adjust by a factor of 1/2.
$$ \int x(x^2+1)^3 dx = \frac{1}{2} \int 2x(x^2+1)^3 dx $$
$$ = \frac{1}{2} \frac{(x^2+1)^4}{4} + c = \frac{1}{8}(x^2+1)^4 + c $$
Type 2: Reverse Log Rule
If the integrand is a fraction where the numerator is the derivative of the denominator:
$$ \int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + c $$
Example: \(\int \frac{2x}{x^2+5} dx\). The derivative of \(x^2+5\) is \(2x\).
$$ \int \frac{2x}{x^2+5} dx = \ln |x^2+5| + c $$
2. Integration by Substitution
This is the formal way to handle the reverse chain rule, especially useful for complicated integrals.
Step-by-Step Substitution:
- Choose a substitution, usually \(u = g(x)\) (often the inner function or the expression under a square root).
- Find \( \frac{du}{dx} \) and rearrange to express \( dx \) in terms of \( du \) and \( dx \): \( dx = \frac{du}{g'(x)} \).
- Substitute everything into the integral, ensuring all terms involving \(x\) are eliminated.
- If it is a definite integral, change the limits from \(x\) values to \(u\) values.
- Integrate with respect to \(u\).
- If indefinite, substitute \(u\) back with \(g(x)\).
3. Integration by Parts
This method reverses the Product Rule for differentiation. It is used to integrate products of two functions (e.g., \(x \ln x\), \(x e^x\)).
The Formula:
$$ \int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx $$
The Trick: Choosing \(u\) and \(\frac{dv}{dx}\)
The goal is to choose \(u\) such that \(\frac{du}{dx}\) simplifies the integral, and \(\frac{dv}{dx}\) is easy to integrate.
- Choose \(u\) as the term that becomes simpler when differentiated (e.g., \(x^n\), \(\ln x\)).
- Choose \(\frac{dv}{dx}\) as the term that is easy to integrate (e.g., \(e^x\), \(\cos x\)).
- For functions like \(\ln x\), we treat it as \(u = \ln x\) and \(\frac{dv}{dx} = 1\).
4. Integration using Partial Fractions
If the integrand is a rational function (a fraction where numerator and denominator are polynomials), we often decompose it into simpler fractions using partial fractions (P2.1 content).
This works because simple partial fractions lead to integrals of the form \(\int \frac{A}{ax+b} dx\), which integrates easily into \(\ln\) terms (using the reverse log rule).
Syllabus Note: You will only encounter simple cases, generally involving distinct linear factors in the denominator (no repeated linear factors or irreducible quadratics).
Section 5: Applications - Volumes of Revolution (P2 Extended)
Integration can find the volume created when an area under a curve is rotated (revolved) 360 degrees around an axis. The resulting shape is called a Solid of Revolution.
Think of slicing the solid into thin discs (like coins). The volume of one disc is \(\pi r^2 \times (\text{thickness})\). We use integration to sum up the volume of these infinite, thin discs.
Rotation about the x-axis
The radius of the disc is \(y\), and the thickness is \(dx\):
$$ V_x = \int^b_a \pi y^2 dx $$
Rotation about the y-axis
The radius of the disc is \(x\), and the thickness is \(dy\). You must express \(x\) in terms of \(y\), and use the limits \(c\) and \(d\) on the y-axis:
$$ V_y = \int^d_c \pi x^2 dy $$
Encouragement: Remember to always square the function before integrating, not after! This is a very common slip.
Section 6: Numerical Integration - The Trapezium Rule (P1 Core)
Sometimes, integration is algebraically impossible, or you only have data points (not a function). Here, we use Numerical Methods to approximate the area.
The Trapezium Rule approximates the area under a curve by dividing the region into vertical strips (trapezia) and summing their areas.
The Trapezium Rule Formula
For an area between \(x=a\) and \(x=b\), divided into \(n\) strips (or \(n+1\) ordinates \(y_0, y_1, \ldots, y_n\)), the width of each strip is \(h = \frac{b-a}{n}\).
The approximation \(I\) is given by:
$$ I \approx \frac{h}{2} \left[ (y_0 + y_n) + 2(y_1 + y_2 + \ldots + y_{n-1}) \right] $$
Memory Aid: This reads as: Half the width, multiplied by (First ordinate + Last ordinate + 2 times the sum of the Middle ordinates).
Over- or Under-Estimation
The question may ask if your estimate is too high (over-estimate) or too low (under-estimate).
- If the curve is concave (shaped like a cave or a frowning face \(\cap\)), the trapezoids lie below the curve, giving an under-estimate.
- If the curve is convex (shaped like a smile \(\cup\)), the trapezoids lie above the curve, giving an over-estimate.
To improve accuracy, simply increase the number of strips, \(n\) (making \(h\) smaller).