Calculus I: Differentiation and Integration
This domain tests technique: differentiating from first principles and by rule, differentiating implicit and parametric forms, and integrating standard forms by substitution, identity or partial fractions. Marks are lost through missing chain-rule factors and constants far more often than through ignorance of the rule itself.
Differentiation rules
From first principles, f′(x) = lim(h→0) [f(x + h) − f(x)]/h. Standard results: d/dx(xⁿ) = nxⁿ⁻¹; d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = sec²x; d/dx(eˣ) = eˣ; d/dx(ln x) = 1/x; d/dx(aˣ) = aˣ ln a. Product rule (uv)′ = u′v + uv′; quotient rule (u/v)′ = (u′v − uv′)/v²; chain rule dy/dx = (dy/du)(du/dx).
For implicit equations, differentiate every term with respect to x, applying the chain rule to y-terms, then collect dy/dx. For parametric curves, dy/dx = (dy/dt)/(dx/dt).
Integration
Standard integrals: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C for n ≠ −1; ∫(1/x) dx = ln|x| + C; ∫eˣ dx = eˣ + C; ∫sin x dx = −cos x + C; ∫cos x dx = sin x + C; ∫sec²x dx = tan x + C. For ∫f′(x)/f(x) dx use ln|f(x)|. Use the substitution u = g(x), du = g′(x) dx, and for definite integrals change the limits too. Integrate sin²x and cos²x with the double-angle identities.
Definite integrals
∫ₐᵇ f(x) dx = F(b) − F(a). Regions below the x-axis give negative values, so a total area needs splitting at the roots.
Worked example
Evaluate ∫₀^(π/2) sin²x dx. Use sin²x = (1 − cos 2x)/2, so ∫sin²x dx = x/2 − (sin 2x)/4 + C. Then [x/2 − (sin 2x)/4] from 0 to π/2 = (π/4 − 0) − (0 − 0) = π/4.
Traps
- Omitting the inner derivative in the chain rule.
- Forgetting + C in an indefinite integral.
- Writing ln x instead of ln|x|.
- Not changing limits when substituting into a definite integral.
Calculus II: Applications of Calculus
Applications questions ask you to turn a calculus result into geometry or motion: stationary points and curve shape, tangents and normals, connected rates of change, area, volume of revolution, the trapezium rule and kinematics. The calculus is usually routine; the marks lie in setting up the correct expression and interpreting the answer.
Stationary points and shape
Solve f′(x) = 0. Classify with f″: f″ > 0 gives a minimum, f″ < 0 a maximum, and f″ = 0 is inconclusive, so check for a sign change in f′. f is increasing where f′ > 0 and decreasing where f′ < 0. A point of inflection needs f″ = 0 together with a change of concavity. On a closed interval, compare stationary values with endpoint values.
Tangents, normals and rates
The tangent at x = a has gradient f′(a); the normal gradient is −1/f′(a). For connected rates, dy/dt = (dy/dx)(dx/dt). Small changes satisfy δy ≈ f′(x)δx.
Area and volume
Area between a curve and the x-axis = ∫y dx; between two curves = ∫(upper − lower) dx. Volume about the x-axis: V = π∫y² dx. Trapezium rule with n strips of width h = (b − a)/n: ∫ ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)].
Kinematics
v = ds/dt and a = dv/dt = v(dv/ds). Displacement = ∫v dt; total distance = ∫|v| dt, splitting where v changes sign.
Worked example
The region bounded by y = √x, the x-axis and x = 4 is rotated through 360° about the x-axis. Find the volume. V = π∫₀⁴ y² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π(8 − 0) = 8π cubic units. Note that y is squared before integrating; integrating y and then squaring gives the wrong value.
Traps
- Omitting π or forgetting to square y in a volume of revolution.
- Treating displacement and distance as the same quantity.
- Stating a stationary point without classifying it when the question asks for the nature.
- Using h = n rather than (b − a)/n in the trapezium rule.