Mathematical Methods

Calculus: Differentiation, Integration and Applications

Year 11 · Year 12

  • ✓Understand the concepts of instantaneous rate of change and accumulation as the foundations of differentiation and integration.
  • ✓Accurately apply differentiation rules (power, chain, product, quotient) to a range of algebraic, exponential, logarithmic, and trigonometric functions.
  • ✓Accurately apply integration rules (power, reverse chain) to find indefinite and definite integrals of a range of functions.
  • ✓Solve practical problems involving rates of change, tangents, stationary points, optimisation, and areas under curves using calculus.

Key concepts

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative, denoted as f'(x) or dy/dx, represents the instantaneous rate of change of a function with respect to its independent variable. Geometrically, it gives the gradient of the tangent to the curve at any given point.

Basic Differentiation Rules

These are fundamental rules for differentiating common function types:\n* Power Rule: If f(x) = ax^n, then f'(x) = nax^(n-1).\n* Constant Multiple Rule: If g(x) = cf(x), then g'(x) = cf'(x).\n* Sum/Difference Rule: If h(x) = f(x) ± g(x), then h'(x) = f'(x) ± g'(x).

Chain Rule

Used to differentiate composite functions. If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). Alternatively, if y = f(u) and u = g(x), then dy/dx = dy/du * du/dx.

dy/dx = dy/du * du/dx
Product Rule

Used to differentiate the product of two functions. If y = u(x)v(x), then dy/dx = u'(x)v(x) + u(x)v'(x).

dy/dx = u'v + uv'
Quotient Rule

Used to differentiate the quotient of two functions. If y = u(x)/v(x), then dy/dx = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2.

dy/dx = (u'v - uv') / v^2
Derivatives of Standard Functions

Key derivatives to memorise:\n* d/dx (e^x) = e^x\n* d/dx (ln(x)) = 1/x (for x > 0)\n* d/dx (sin(x)) = cos(x)\n* d/dx (cos(x)) = -sin(x)\n* d/dx (tan(x)) = sec^2(x)

Second Derivative

The second derivative, f''(x) or d^2y/dx^2, is the derivative of the first derivative. It describes the rate of change of the gradient and is used to determine concavity and classify stationary points (local maxima/minima, points of inflection).

Integration (Antidifferentiation)

Integration is the reverse process of differentiation. It is used to find a function given its derivative. An indefinite integral represents a family of functions, differing by a constant, and is denoted by ∫f(x) dx. The '+ C' (constant of integration) is crucial.

Basic Integration Rules

Fundamental rules for integrating common function types:\n* Power Rule: ∫x^n dx = (x^(n+1))/(n+1) + C (for n ≠ -1)\n* Constant Multiple Rule: ∫cf(x) dx = c∫f(x) dx\n* Sum/Difference Rule: ∫(f(x) ± g(x)) dx = ∫f(x) dx ± ∫g(x) dx

Integrals of Standard Functions

Key integrals to memorise:\n* ∫e^x dx = e^x + C\n* ∫(1/x) dx = ln|x| + C (for x ≠ 0)\n* ∫sin(x) dx = -cos(x) + C\n* ∫cos(x) dx = sin(x) + C

Definite Integrals

A definite integral ∫_a^b f(x) dx evaluates the accumulation of a function over a specific interval [a, b]. It represents the signed area between the curve y = f(x) and the x-axis from x = a to x = b. The Fundamental Theorem of Calculus states ∫_a^b f(x) dx = F(b) - F(a), where F(x) is an antiderivative of f(x).

∫_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)
Applications of Differentiation

Differentiation is used in various practical contexts:\n* Rates of Change: The derivative dy/dx gives the instantaneous rate of change of y with respect to x.\n* Tangents and Normals: The gradient of the tangent to y = f(x) at x=a is f'(a). The normal is perpendicular to the tangent, so its gradient is -1/f'(a).\n* Stationary Points: Points where f'(x) = 0. These can be local maxima, local minima, or horizontal points of inflection. The second derivative test (f''(x)) helps classify them.\n* Optimisation: Finding maximum or minimum values of a quantity by setting the first derivative to zero.

Applications of Integration

Integration is used in various practical contexts:\n* Area Under a Curve: The definite integral ∫_a^b f(x) dx calculates the signed area between y = f(x) and the x-axis from x = a to x = b. For geometric area, if f(x) is below the x-axis, the integral will be negative, so absolute value or splitting the integral may be needed.\n* Displacement from Velocity: If v(t) is the velocity function, then ∫v(t) dt gives the displacement function. ∫_a^b v(t) dt gives the net displacement from time a to b.

Key facts to remember

  • 1The derivative f'(x) represents the instantaneous rate of change of f(x) and the gradient of the tangent to the curve.
  • 2The power rule for differentiation is d/dx (ax^n) = nax^(n-1).
  • 3The power rule for integration is ∫x^n dx = (x^(n+1))/(n+1) + C (for n ≠ -1).
  • 4Remember the '+ C' (constant of integration) for all indefinite integrals.
  • 5The Chain Rule, Product Rule, and Quotient Rule are essential for differentiating complex functions.
  • 6Stationary points occur where f'(x) = 0. Use the second derivative test (f''(x)) to classify them.
  • 7A definite integral ∫_a^b f(x) dx calculates the signed area between the curve and the x-axis.
  • 8For geometric area, if the curve is below the x-axis, the integral will be negative; take its absolute value.

Worked examples

Example 1

Given f(x) = 3x^4 - 2x^2 + 5e^x, find f'(x) and ∫f(x) dx.

ITo find f'(x), differentiate each term using the power rule and the derivative of e^x:
IIf'(x) = d/dx (3x^4) - d/dx (2x^2) + d/dx (5e^x)
IIIf'(x) = 3 * 4x^(4-1) - 2 * 2x^(2-1) + 5e^x
IVf'(x) = 12x^3 - 4x + 5e^x
VTo find ∫f(x) dx, integrate each term using the power rule for integration and the integral of e^x:
VI∫f(x) dx = ∫(3x^4 - 2x^2 + 5e^x) dx
VII= ∫3x^4 dx - ∫2x^2 dx + ∫5e^x dx
VIII= 3 * (x^(4+1))/(4+1) - 2 * (x^(2+1))/(2+1) + 5e^x + C
9= (3/5)x^5 - (2/3)x^3 + 5e^x + C

Answer

f'(x) = 12x^3 - 4x + 5e^x and ∫f(x) dx = (3/5)x^5 - (2/3)x^3 + 5e^x + C

Remember the constant of integration '+ C' for indefinite integrals.

Example 2

Find the coordinates and nature of the stationary points of the curve y = x^3 - 6x^2 + 9x - 2.

I1. Find the first derivative, dy/dx:
IIdy/dx = 3x^2 - 12x + 9
III2. Set dy/dx = 0 to find the x-coordinates of stationary points:
IV3x^2 - 12x + 9 = 0
VDivide by 3: x^2 - 4x + 3 = 0
VIFactorise: (x - 1)(x - 3) = 0
VIISo, x = 1 or x = 3.
VIII3. Find the corresponding y-coordinates by substituting x values into the original equation:
9If x = 1, y = (1)^3 - 6(1)^2 + 9(1) - 2 = 1 - 6 + 9 - 2 = 2. Point: (1, 2).
10If x = 3, y = (3)^3 - 6(3)^2 + 9(3) - 2 = 27 - 54 + 27 - 2 = -2. Point: (3, -2).
114. Find the second derivative, d^2y/dx^2:
12d^2y/dx^2 = d/dx (3x^2 - 12x + 9) = 6x - 12
135. Test the nature of stationary points using the second derivative:
14At x = 1: d^2y/dx^2 = 6(1) - 12 = -6. Since -6 < 0, (1, 2) is a local maximum.
15At x = 3: d^2y/dx^2 = 6(3) - 12 = 18 - 12 = 6. Since 6 > 0, (3, -2) is a local minimum.

Answer

Local maximum at (1, 2) and local minimum at (3, -2).

Always state both the coordinates and the nature of the stationary points.

Example 3

Find the area enclosed by the curve y = x^2 - 4x, the x-axis, and the lines x = 0 and x = 5.

I1. Find the x-intercepts of the curve by setting y = 0:
IIx^2 - 4x = 0
IIIx(x - 4) = 0
IVSo, x = 0 and x = 4. These are where the curve crosses the x-axis.
V2. Observe the interval [0, 5]. The curve is below the x-axis between x = 0 and x = 4 (e.g., at x=1, y = 1-4 = -3). It is above the x-axis between x = 4 and x = 5 (e.g., at x=5, y = 25-20 = 5).
VI3. To find the total geometric area, we must split the integral and take the absolute value for the part below the x-axis:
VIIArea = |∫_0^4 (x^2 - 4x) dx| + ∫_4^5 (x^2 - 4x) dx
VIII4. Find the antiderivative of x^2 - 4x:
9∫(x^2 - 4x) dx = x^3/3 - 2x^2
105. Calculate the first part of the area (A1):
11A1 = |[(x^3/3 - 2x^2)]_0^4|
12A1 = |((4^3/3 - 2(4^2)) - (0^3/3 - 2(0^2)))|
13A1 = |(64/3 - 32) - 0| = |64/3 - 96/3| = |-32/3| = 32/3 square units.
146. Calculate the second part of the area (A2):
15A2 = [(x^3/3 - 2x^2)]_4^5
16A2 = (5^3/3 - 2(5^2)) - (4^3/3 - 2(4^2))
17A2 = (125/3 - 50) - (64/3 - 32)
18A2 = (125/3 - 150/3) - (64/3 - 96/3)
19A2 = (-25/3) - (-32/3) = -25/3 + 32/3 = 7/3 square units.
207. Add the areas together for the total area:
21Total Area = A1 + A2 = 32/3 + 7/3 = 39/3 = 13 square units.

Answer

The total area enclosed is 13 square units.

When calculating geometric area, always consider if the curve goes below the x-axis and take the absolute value of those integral parts.

Common mistakes

  • ✗Forgetting the '+ C' when performing indefinite integration.
  • ✗Confusing differentiation rules (e.g., applying the product rule when the chain rule is needed).
  • ✗Incorrectly applying the quotient rule (numerator order u'v - uv').
  • ✗Not checking for x-intercepts when calculating area under a curve, leading to incorrect total geometric area.
  • ✗Algebraic errors when simplifying derivatives or evaluating definite integrals.

Exam tips

  • ★Clearly show all steps in your working, especially for differentiation and integration, as partial marks are often awarded.
  • ★Double-check your algebraic manipulations and arithmetic, particularly when substituting values into functions.
  • ★For optimisation problems, always define your variables, set up the function to be optimised, differentiate, find stationary points, and verify the nature of the extremum.
  • ★When calculating areas, sketch the graph to identify any regions where the curve is below the x-axis, and split the integral accordingly.

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