Mathematics (9709) | Integration

Pure Mathematics 1 (Paper 1) Study Notes: Chapter 1.8 Integration

Welcome to Integration! If Differentiation was about finding the rate of change (like speed or gradient), Integration is about reversing that process and finding the total accumulation (like distance or area). Think of it as the mathematical rewind button!

This chapter is fundamental. Mastering it will allow you to calculate areas under complicated curves and find volumes of unusual shapes. Don't worry if it feels tricky; practice is the key to mastering the 'reverse' thinking required here.

1. Indefinite Integration: The Reverse Process

1.1 The Idea of Anti-Differentiation

Integration is officially known as anti-differentiation. If we differentiate a function \(y = F(x)\) to get \(f(x)\), then integrating \(f(x)\) takes us back to \(F(x)\).

• If \( \frac{dy}{dx} = f(x) \), then \( \int f(x) dx = y = F(x) \).

The symbol \( \int \) is the integral sign, and \( dx \) tells us we are integrating with respect to x.

1.2 The Integration Power Rule (The Key Formula)

This is the core rule for integrating simple polynomial terms \(x^n\):

If \( n \) is any rational number (except -1):

$$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $$

Step-by-Step Power Rule:

1. Increase the power by 1 (from \(n\) to \(n+1\)).
2. Divide the whole term by the new power (\(n+1\)).
3. Always add the constant of integration, \(C\).

Example: To integrate \( \int x^3 dx \):
New power is \( 3 + 1 = 4 \). Divide by 4.
$$ \int x^3 dx = \frac{x^4}{4} + C $$

WARNING: The special case n = -1
Remember, for P1, you cannot use the power rule when \(n = -1\) (i.e., for integrating \( \frac{1}{x} \)). This specific case is covered in Paper 2/3. For P1, the syllabus explicitly states the rule is for any rational \(n\) except -1.

1.3 The Constant of Integration, C

When you differentiate any constant, the result is zero. Therefore, when you integrate, you lose information about the original constant. To account for this unknown value, we must always add C.

Analogy: Imagine tracking a moving object. Differentiation gives you the speed. Integration gives you the position. If you know the speed, you know *how* the position is changing, but you don't know the object's *starting location*. That unknown starting location is \(C\).

Finding C (Solving Problems)

To find the specific value of \(C\), you need an initial condition—a point \((x, y)\) that the original curve passes through. This is often necessary when solving problems like:

"Given that \( \frac{dy}{dx} = 2x - 3 \) and the curve passes through \((1, -2)\), find the equation of the curve."

Step-by-Step:

1. Integrate \( \frac{dy}{dx} \) to get \(y\) in terms of \(x\) and \(C\). $$ y = \int (2x - 3) dx = x^2 - 3x + C $$
2. Substitute the given point \((1, -2)\) into the equation. $$ -2 = (1)^2 - 3(1) + C $$ $$ -2 = 1 - 3 + C \implies -2 = -2 + C $$
3. Solve for \(C\). Here, \(C = 0\).
4. State the particular solution (the equation of the curve). $$ y = x^2 - 3x $$

1.4 Integrating Composite Functions: The Reverse Chain Rule

The P1 syllabus requires you to integrate linear composite functions of the form \((ax+b)^n\).

$$ \int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C $$

This is just the standard power rule, but you also have to divide by a (the derivative of the expression inside the bracket).

Example: Integrate \( \int (2x + 3)^4 dx \):
Here, \(n=4\) and \(a=2\).
$$ \int (2x + 3)^4 dx = \frac{(2x + 3)^5}{2(5)} + C = \frac{1}{10}(2x + 3)^5 + C $$

2. Definite Integration: Finding Net Change

2.1 What are Definite Integrals?

A definite integral is an integral with specific upper and lower limits (or bounds), \(a\) and \(b\). It produces a numerical value, usually representing the area or net change between \(x=a\) and \(x=b\).

$$ \int_a^b f(x) dx $$

2.2 The Fundamental Theorem of Calculus

This theorem tells us how to calculate a definite integral:

$$ \int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a) $$

where \(F(x)\) is the antiderivative of \(f(x)\).

Step-by-Step Evaluation:

1. Find the indefinite integral, \(F(x)\). (You do not need to include \(+C\) here, as it cancels out when calculating \(F(b) - F(a)\)).
2. Substitute the upper limit (\(b\)) into \(F(x)\) to get \(F(b)\).
3. Substitute the lower limit (\(a\)) into \(F(x)\) to get \(F(a)\).
4. Calculate the difference: \(F(b) - F(a)\).

Example: Evaluate \( \int_1^2 (3x^2 - 4) dx \)

1. Integrate: \( F(x) = x^3 - 4x \)

2. Substitute limits: $$ [x^3 - 4x]_1^2 = [(2)^3 - 4(2)] - [(1)^3 - 4(1)] $$ $$ = [8 - 8] - [1 - 4] $$ $$ = 0 - (-3) = 3 $$

The definite integral is 3.

3. Applications of Integration: Area

The primary geometrical application of definite integration in P1 is finding the area of a region bounded by curves and lines.

3.1 Area Bounded by a Curve and the x-axis

The area \(A\) bounded by the curve \(y = f(x)\), the x-axis, and the lines \(x=a\) and \(x=b\) is:

$$ A = \int_a^b f(x) dx $$

Crucial Consideration: Areas Below the Axis

• If \(f(x) \ge 0\) over the interval \([a, b]\), the integral result is the positive area.
• If \(f(x) < 0\) (the curve is below the x-axis), the integral result will be negative. Since area must be positive, you must take the absolute value of this integral result.

Common Mistake Alert!
If the curve crosses the x-axis between \(a\) and \(b\), you must split the integral into separate sections and calculate the absolute value of the area below the axis before summing the results.

If \(f(x)\) crosses the axis at \(c\) (\(a < c < b\)):
$$ A = \left| \int_a^c f(x) dx \right| + \int_c^b f(x) dx $$

3.2 Area Between a Curve and a Straight Line

If a region is bounded by a curve \(y_C\) and a straight line \(y_L\), the area between them is found by integrating the difference between the top function and the bottom function.

$$ A = \int_a^b (y_{top} - y_{bottom}) dx $$

Steps:

1. Find the intersection points (\(a\) and \(b\)) by solving \(y_C = y_L\). These are your limits.
2. Determine which function is on top (\(y_{top}\)) over the interval \([a, b]\). Sketching a graph helps greatly!
3. Set up and evaluate the definite integral.

3.3 Area Between Two Curves

The principle remains the same as 3.2. If \(y_1\) is the top curve and \(y_2\) is the bottom curve between limits \(a\) and \(b\):

$$ A = \int_a^b (y_1 - y_2) dx $$

4. Applications of Integration: Volume of Revolution

Integration can be used to calculate the volume of a solid formed when a two-dimensional region is rotated (revolved) 360° about an axis (usually the x-axis or y-axis).

Did you know? This method works by summing the volumes of infinitely many thin discs (like tiny coins) stacked up along the axis of rotation.

4.1 Revolution About the x-axis

When the area bounded by \(y = f(x)\), the x-axis, and \(x=a\) and \(x=b\) is rotated 360° about the x-axis, the volume \(V\) is:

$$ V = \int_a^b \pi y^2 dx $$

Note: You must express \(y^2\) purely in terms of \(x\) before integrating.

4.2 Revolution About the y-axis

When the area bounded by the curve, the y-axis, and \(y=c\) and \(y=d\) is rotated 360° about the y-axis, the volume \(V\) is:

$$ V = \int_c^d \pi x^2 dy $$

Note: You must express \(x^2\) purely in terms of \(y\). If you start with \(y = f(x)\), you need to rearrange it to find \(x^2\) in terms of \(y\).

Example: If \(y = x^2\), then \(x^2 = y\). If rotating about the y-axis, you integrate \( V = \int_c^d \pi (y) dy \).

4.3 Volume Between Two Curves (The Washer Method)

If the region rotated lies between two curves, $y_{outer}$ and $y_{inner}$, the resulting volume is the difference between the volume generated by the outer curve and the volume generated by the inner curve.

When rotating about the x-axis, between $x=a$ and $x=b$:

$$ V = \pi \int_a^b (y_{outer}^2 - y_{inner}^2) dx $$

This technique is used whenever the boundary of the region being rotated is not the axis of rotation itself (for example, revolving the area between a curve and the line $y=5$ about the x-axis).

You have now covered the essentials of Integration for Pure Mathematics 1! Keep practicing your algebraic manipulation, especially rearranging equations for $x^2$ or $y^2$ in volume problems, and remember to always check for zero-crossings when calculating area.