Unit P2: Pure Mathematics 2 – Chapter 6: Integration
Hello future calculus master! Welcome to the world of Integration. If differentiation was about finding the rate of change (the slope of a curve), integration is like putting the pieces back together. It’s the essential reverse process, and it allows us to calculate things like the area under a curve, which is incredibly useful!
Don't worry if this seems tricky at first. We will break down this powerful tool step by step. By the end of this chapter, you will be able to 'undo' differentiation and solve crucial area problems!
1. Introduction to Anti-Differentiation (Indefinite Integration)
The Idea: Reversing the Process
Think of integration as anti-differentiation. If differentiation is like tying your shoelaces, integration is like untying them. You are going backwards to find the original function.
If you have a function, say \(F(x)\), and you differentiate it to get \(f(x)\):
$$F(x) \xrightarrow{\text{Differentiate}} f(x)$$
Then, integrating \(f(x)\) takes you back to \(F(x)\):
$$f(x) \xrightarrow{\text{Integrate}} F(x)$$
The symbol we use for integration is the elongated 'S': \(\int\). This symbol represents the process of summation, which is fundamentally what integration is doing (summing up tiny changes).
- Integrand: The function being integrated (e.g., \(f(x)\) in \(\int f(x) dx\)).
- Variable of Integration: Indicated by \(dx\) (meaning we are integrating with respect to \(x\)).
Did You Know?
The symbol \(\int\) is called the integral sign and was introduced by the German mathematician Gottfried Wilhelm Leibniz in the late 17th century. It is a stylised letter 'S', standing for "summa" (sum).
2. The Power Rule for Integration
The most important rule in P2 integration is the Power Rule. It reverses the power rule you used in differentiation.
Step-by-Step Power Rule
To integrate a term \(x^n\):
- Increase the power (exponent) by 1.
- Divide the entire term by the new power.
- Add the constant of integration, \(+C\). (More on this crucial step next!)
The Rule:
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$
(This rule holds for all values of \(n\), except when \(n=-1\). Integration for \(x^{-1}\) is reserved for P3.)
Examples of the Power Rule
Example 1: Integrate \(x^3\)
$$\int x^3 dx = \frac{x^{3+1}}{3+1} + C = \mathbf{\frac{1}{4}x^4 + C}$$
Example 2: Integrate \(\sqrt{x}\)
First, rewrite the term using indices: \(\sqrt{x} = x^{1/2}\)
$$\int x^{1/2} dx = \frac{x^{1/2 + 1}}{1/2 + 1} + C = \frac{x^{3/2}}{3/2} + C = \mathbf{\frac{2}{3}x^{3/2} + C}$$
Example 3: Integrate \(\frac{1}{x^2}\)
First, rewrite the term using indices: \(\frac{1}{x^2} = x^{-2}\)
$$\int x^{-2} dx = \frac{x^{-2 + 1}}{-2 + 1} + C = \frac{x^{-1}}{-1} + C = \mathbf{-\frac{1}{x} + C}$$
Integrating Multiple Terms
Just like differentiation, integration is linear. This means you can integrate each term in a polynomial expression separately.
Rule: If \(k\) is a constant, \(\int k f(x) dx = k \int f(x) dx\)
Example: Integrate \(f(x) = 4x^3 - 6x + 5\)
$$\int (4x^3 - 6x + 5) dx = 4 \left( \frac{x^4}{4} \right) - 6 \left( \frac{x^2}{2} \right) + 5x + C$$
$$= \mathbf{x^4 - 3x^2 + 5x + C}$$
Quick Review: Always prepare your terms (move everything to the numerator, convert roots to powers) before you apply the power rule!
3. The Constant of Integration (+C)
The constant \(+C\) is arguably the most important part of indefinite integration and a common place where students lose marks!
Why is +C Necessary?
When you differentiate a constant, the result is always zero. Look at these three functions:
- \(y = x^2 + 5 \quad \implies \frac{dy}{dx} = 2x\)
- \(y = x^2 - 100 \quad \implies \frac{dy}{dx} = 2x\)
- \(y = x^2 \quad \implies \frac{dy}{dx} = 2x\)
If we start with \(\frac{dy}{dx} = 2x\) and integrate it, how do we know which original function we started with? We don't! The \(+C\) acts as a placeholder for that unknown constant that disappeared during differentiation.
Analogy: If you know a car's speed (\(2x\)) but not its starting position, you know its velocity function, but you don't know its exact position function until someone gives you a starting point.
Finding the Value of C
In many exam questions, you will be given an additional piece of information—a boundary condition or a specific point \((x, y)\) that the original curve passes through. This point allows you to solve for \(C\).
Process for finding C:
- Integrate the function \(\frac{dy}{dx}\), including the \(+C\).
- Substitute the given \(x\) and \(y\) values into the integrated equation.
- Solve the resulting equation for \(C\).
- Write out the final, complete function \(y = F(x)\) with the calculated value of \(C\).
Example: A curve has gradient function \(\frac{dy}{dx} = 3x^2 - 4\). The curve passes through the point \((2, 1)\). Find the equation of the curve.
1. Integrate:
$$y = \int (3x^2 - 4) dx = \frac{3x^3}{3} - 4x + C = x^3 - 4x + C$$2. Substitute \((x=2, y=1)\):
$$1 = (2)^3 - 4(2) + C$$ $$1 = 8 - 8 + C$$ $$1 = C$$3. Final Equation:
$$y = \mathbf{x^3 - 4x + 1}$$Key Takeaway: If the question asks you to find the equation of the curve or the original function, you MUST find the value of \(C\). If it's just an indefinite integral, leave the \(+C\).
4. Definite Integration
When you integrate a function between two specified limits, it is called a definite integral. This gives you a numerical value, usually representing an area.
Understanding the Limits
A definite integral looks like this:
$$\int_a^b f(x) dx$$- \(b\) is the upper limit.
- \(a\) is the lower limit.
The result of a definite integral is calculated using the Fundamental Theorem of Calculus.
The Process of Definite Integration
1. Find the indefinite integral \(F(x)\) of \(f(x)\). (You can omit the \(+C\) here, as it always cancels out.)
2. Evaluate the integral at the upper limit \(b\), giving \(F(b)\).
3. Evaluate the integral at the lower limit \(a\), giving \(F(a)\).
4. Subtract the results: \(F(b) - F(a)\).
Notation: We write \([F(x)]_a^b\):
$$\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$$
Example: Evaluate \(\int_1^3 (2x + 1) dx\)
1. Integrate:
$$F(x) = x^2 + x$$2. Set up the evaluation:
$$[x^2 + x]_1^3$$3. Substitute upper limit (3):
$$F(3) = (3)^2 + 3 = 9 + 3 = 12$$4. Substitute lower limit (1):
$$F(1) = (1)^2 + 1 = 1 + 1 = 2$$5. Subtract:
$$F(3) - F(1) = 12 - 2 = \mathbf{10}$$Common Mistake to Avoid!
When substituting the limits, be extremely careful with negative numbers and brackets. A simple sign error in \(F(a)\) can ruin the whole answer.
Always calculate \(F(b)\) and \(F(a)\) separately before performing the final subtraction.
5. Area Under a Curve
The most common geometric application of definite integration in P2 is finding the area bounded by a curve \(y = f(x)\), the x-axis, and two vertical lines \(x=a\) and \(x=b\).
The Simple Case: Area Above the x-axis
If the function \(f(x)\) is completely above the x-axis between \(x=a\) and \(x=b\), the area \(A\) is simply the value of the definite integral:
$$A = \int_a^b f(x) dx$$The Tricky Case: Area Below the x-axis
If the curve lies below the x-axis, the value of the definite integral will be negative.
Since area is a physical quantity, it must always be positive. If your integral yields a negative result, it means the area lies below the axis.
Rule: If the area is below the x-axis, take the modulus (absolute value) of the result to make it positive.
$$A = \left| \int_a^b f(x) dx \right|$$Combining Areas (When the Curve Crosses the Axis)
If the curve crosses the x-axis between \(a\) and \(b\), you cannot integrate in one step! The positive and negative areas will cancel each other out, giving you an incorrect total area.
Process for Crossing the Axis:
- Find the x-intercepts (where \(f(x) = 0\)) between \(a\) and \(b\). This point divides the total area into separate regions.
- Calculate the definite integral for each region separately.
- Take the absolute value of any region that falls below the x-axis (negative result).
- Sum the positive results to find the total area.
Example Analogy: Imagine walking 5km forward, then 3km backward. Your total distance travelled is 8km, not 2km. Similarly, we must add the magnitudes of the separate integrated sections.
Example Scenario: Find the area between \(x=0\) and \(x=4\) for a function that crosses the x-axis at \(x=2\).
Total Area \(A = \left| \int_0^2 f(x) dx \right| + \left| \int_2^4 f(x) dx \right|\)
Summary of Area Rules
Area calculation must yield a positive number. Always graph or check intercepts if the function is unfamiliar to ensure you don't accidentally subtract areas!
6. Integrating Simple Functions of (ax + b)
While P3 introduces the full chain rule for integration, P2 requires you to handle simple linear terms raised to a power, such as \((3x + 2)^5\).
This is essentially the reverse of the chain rule used for differentiation.
The Rule for \( (ax+b)^n \)
To integrate \( (ax+b)^n \):
- Apply the power rule as usual: increase the power by 1 and divide by the new power.
- Additionally, divide by the coefficient of \(x\) (which is \(a\)).
- Don't forget the \(+C\) if indefinite!
The Formula:
$$\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$$
Example: Integrate \( (2x - 1)^3 \)
Here, \(a=2\) and \(n=3\).
$$\int (2x - 1)^3 dx = \frac{(2x - 1)^{3+1}}{2(3+1)} + C = \frac{(2x - 1)^4}{8} + C$$Tip: Check your answer by differentiating the result! Differentiating \(\frac{(2x - 1)^4}{8}\) should bring you back to \((2x - 1)^3\).
Chapter Key Takeaways
- Indefinite Integral: Uses the Power Rule: Increase power by 1, divide by new power, and add \(+C\).
- Finding C: Requires a given point \((x, y)\) to substitute into the integrated equation.
- Definite Integral: Calculates a numerical value, representing the net area, using \([F(b) - F(a)]\).
- Area Calculation: If any part of the area is below the x-axis, you must integrate in separate parts and use the modulus (absolute value) for the negative region before summing the areas.
Keep practising the setup and algebra—Integration relies heavily on your skills with indices and fractions! You've got this!