Calculus is the largest and most heavily examined section of the edexcel igcse further pure mathematics specification, and everything you have learned in the other sections feeds into it
Calculus gives you two fundamental tools: differentiation (finding the rate at which a quantity changes) and integration (finding the total accumulated effect of that rate of change). In the Pearson Edexcel IGCSE Further Pure Mathematics course, these tools are applied to polynomials, trigonometric functions and exponential functions. You need to differentiate products, quotients and composites using the product rule, quotient rule and chain rule. You need to integrate standard functions and apply integration to find areas, volumes and solutions to kinematics problems. And you need to combine these skills with coordinate geometry to find tangents, normals and stationary points.
These edexcel igcse further pure mathematics calculus revision notes cover all five sub-topics. The treatment is thorough because the exam rewards depth here more than in any other section. If you master this material, you have the single strongest preparation advantage available to you.
Differentiation
Standard results
The specification requires you to differentiate the following function types:
| Function f(x) | Derivative f'(x) |
|---|---|
| xn | nxn-1 |
| sin(ax) | a cos(ax) |
| cos(ax) | -a sin(ax) |
| eax | aeax |
These results extend to sums and constant multiples. If y = 3x4 - 2sin(5x) + e2x, then dy/dx = 12x3 - 10cos(5x) + 2e2x.
The product rule
If y = uv, then dy/dx = u(dv/dx) + v(du/dx).
Worked example: Differentiate y = x3sin(2x).
Let u = x3, v = sin(2x). Then du/dx = 3x2, dv/dx = 2cos(2x).
dy/dx = x3(2cos(2x)) + sin(2x)(3x2) = 2x3cos(2x) + 3x2sin(2x).
The quotient rule
If y = u/v, then dy/dx = (v(du/dx) - u(dv/dx)) / v2.
Worked example: Differentiate y = e3x / (x + 1).
Let u = e3x, v = x + 1. Then du/dx = 3e3x, dv/dx = 1.
dy/dx = ((x + 1)(3e3x) - e3x(1)) / (x + 1)2 = e3x(3x + 3 - 1) / (x + 1)2 = e3x(3x + 2) / (x + 1)2.
The chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) . g'(x).
Worked example: Differentiate y = (3x + 1)5.
Let u = 3x + 1, so y = u5. dy/du = 5u4, du/dx = 3.
dy/dx = 5(3x + 1)4 x 3 = 15(3x + 1)4.
Integration
Standard results
| Function f(x) | Integral (+ constant c) |
|---|---|
| xn (n is not -1) | xn+1/(n + 1) + c |
| sin(ax) | -cos(ax)/a + c |
| cos(ax) | sin(ax)/a + c |
| eax | eax/a + c |
Worked example: Find the integral of 4x3 - 3cos(2x) + e-x with respect to x.
Integral = x4 - (3/2)sin(2x) - e-x + c.
Applications of calculus
Kinematics: displacement, velocity and acceleration
If s is displacement, v is velocity and a is acceleration, then:
- v = ds/dt (velocity is the derivative of displacement with respect to time)
- a = dv/dt (acceleration is the derivative of velocity with respect to time)
- s = integral of v dt (displacement is the integral of velocity)
- v = integral of a dt (velocity is the integral of acceleration)
Worked example: A particle moves along a straight line with displacement s = 2t3 - 9t2 + 12t metres at time t seconds. Find the velocity and acceleration at t = 2.
v = ds/dt = 6t2 - 18t + 12. At t = 2: v = 24 - 36 + 12 = 0 m/s.
a = dv/dt = 12t - 18. At t = 2: a = 24 - 18 = 6 m/s2.
The particle is momentarily at rest (v = 0) and accelerating at 6 m/s2 at t = 2.
Areas under and between curves
The area under y = f(x) between x = a and x = b is the definite integral from a to b of f(x) dx. When the curve dips below the x-axis, the integral gives a negative value, so you must split the integral at the roots and take absolute values.
Worked example: Find the area enclosed between y = x2 and y = 2x.
Find the intersection points: x2 = 2x, so x2 - 2x = 0, x(x - 2) = 0, giving x = 0 and x = 2.
Between x = 0 and x = 2, y = 2x is above y = x2.
Area = integral from 0 to 2 of (2x - x2) dx = [x2 - x3/3] from 0 to 2 = (4 - 8/3) - 0 = 4/3 square units.
Volumes of revolution
The volume generated when the region under y = f(x) from x = a to x = b is rotated through 360 degrees about the x-axis is:
V = pi x integral from a to b of y2 dx.
For rotation about the y-axis: V = pi x integral from y = c to y = d of x2 dy.
Worked example: Find the volume of revolution when the area under y = x2 from x = 0 to x = 3 is rotated 360 degrees about the x-axis.
V = pi x integral from 0 to 3 of (x2)2 dx = pi x integral from 0 to 3 of x4 dx = pi [x5/5] from 0 to 3 = pi (243/5) = 243pi/5 cubic units.
Stationary points and optimisation
A stationary point occurs where dy/dx = 0. To classify it:
| Second derivative d2y/dx2 | Nature of stationary point |
|---|---|
| Positive | Minimum |
| Negative | Maximum |
| Zero | Test is inconclusive; use a sign-change test on dy/dx |
Worked example: Find and classify the stationary points of y = x3 - 6x2 + 9x + 1.
dy/dx = 3x2 - 12x + 9 = 3(x2 - 4x + 3) = 3(x - 1)(x - 3).
dy/dx = 0 when x = 1 or x = 3.
d2y/dx2 = 6x - 12.
At x = 1: d2y/dx2 = 6 - 12 = -6 < 0, so (1, 5) is a maximum.
At x = 3: d2y/dx2 = 18 - 12 = 6 > 0, so (3, 1) is a minimum.
Optimisation in context
Worked example: An open-top box is made from a square sheet of metal by cutting squares of side x cm from each corner and folding up. The sheet is 24 cm square. Find the value of x that maximises the volume.
The base has side length (24 - 2x) and the height is x. V = x(24 - 2x)2.
Expand: V = x(576 - 96x + 4x2) = 4x3 - 96x2 + 576x.
dV/dx = 12x2 - 192x + 576 = 12(x2 - 16x + 48) = 12(x - 4)(x - 12).
dV/dx = 0 when x = 4 or x = 12. Since x must be less than 12 (otherwise no box), x = 4.
d2V/dx2 = 24x - 192. At x = 4: 96 - 192 = -96 < 0, confirming a maximum.
Maximum volume = 4(24 - 8)2 = 4(16)2 = 4 x 256 = 1024 cm3.
Tangents, normals and rates of change
Equations of tangents and normals
The tangent to y = f(x) at the point (a, f(a)) has gradient f'(a). Its equation is y - f(a) = f'(a)(x - a).
The normal at the same point has gradient -1/f'(a). Its equation is y - f(a) = (-1/f'(a))(x - a).
Worked example: Find the equation of the tangent and the normal to y = x2 - 3x + 5 at the point where x = 2.
At x = 2: y = 4 - 6 + 5 = 3. The point is (2, 3).
dy/dx = 2x - 3. At x = 2: dy/dx = 1.
Tangent: y - 3 = 1(x - 2), so y = x + 1.
Normal: y - 3 = -1(x - 2), so y = -x + 5.
Connected rates of change
When two quantities are related by a formula and both change with time, the chain rule connects their rates of change.
Worked example: The radius of a circle increases at 0.5 cm/s. Find the rate at which the area increases when the radius is 10 cm.
A = pi r2. dA/dr = 2pi r. dr/dt = 0.5.
dA/dt = dA/dr x dr/dt = 2pi(10)(0.5) = 10pi cm2/s.
Small changes approximation
The specification includes the approximation: delta y is approximately equal to (dy/dx) x delta x, for small delta x.
Worked example: Use calculus to estimate the change in y = x3 + 2x when x increases from 3 to 3.02.
dy/dx = 3x2 + 2. At x = 3: dy/dx = 29. delta x = 0.02.
delta y is approximately 29 x 0.02 = 0.58.
Exact change: f(3.02) - f(3) = (27.543608 + 6.04) - (27 + 6) = 33.583608 - 33 = 0.583608. The approximation 0.58 is close.
Practice questions
These edexcel igcse further pure mathematics practice questions cover the full range of calculus skills. Work through each one completely before checking the solutions.
- Differentiate y = (2x + 1)4 sin x.
- Find the integral of (3e2x - 4cos(3x) + x-2) dx.
- A particle has velocity v = 6t - t2 m/s. Find the displacement between t = 0 and t = 6 seconds.
- Find the volume generated when the area under y = ex from x = 0 to x = 1 is rotated 360 degrees about the x-axis.
- The curve y = 2x3 - 9x2 + 12x has two stationary points. Find them and determine which is the maximum and which is the minimum.
Solutions
Question 1: Product rule with u = (2x + 1)4 and v = sin x. du/dx = 4(2x + 1)3(2) = 8(2x + 1)3. dv/dx = cos x. dy/dx = (2x + 1)4 cos x + 8(2x + 1)3 sin x.
Question 2: (3/2)e2x - (4/3)sin(3x) - x-1 + c = (3/2)e2x - (4/3)sin(3x) - 1/x + c.
Question 3: s = integral from 0 to 6 of (6t - t2) dt = [3t2 - t3/3] from 0 to 6 = (108 - 72) - 0 = 36 m.
Question 4: V = pi x integral from 0 to 1 of (ex)2 dx = pi x integral from 0 to 1 of e2x dx = pi [e2x/2] from 0 to 1 = pi(e2/2 - 1/2) = pi(e2 - 1)/2 cubic units.
Question 5: dy/dx = 6x2 - 18x + 12 = 6(x2 - 3x + 2) = 6(x - 1)(x - 2). Stationary at x = 1 and x = 2. d2y/dx2 = 12x - 18. At x = 1: 12 - 18 = -6 < 0, maximum at (1, 5). At x = 2: 24 - 18 = 6 > 0, minimum at (2, 4).
The igcse 4PM1 calculus section is where the most marks are available and where thorough preparation has the greatest impact. These edexcel igcse further pure mathematics notes cover the essential methods; for more edexcel igcse further pure mathematics explained worked examples and additional practice, use edexcel igcse further pure mathematics revision notes on the Green Bridge CBT platform to build the exam-ready fluency that distinguishes top-grade candidates.
Edexcel IGCSE Further Pure Mathematics revision notes on calculus: differentiation, integration, stationary points, areas, volumes and rates of change.
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