Calculus: Differentiation and Integration with Applications

The big idea

Differentiate → gradient and rate of change. Integrate → area and total amount. They undo each other.

Differentiate when asked for a gradient, tangent, normal, rate of change, velocity from displacement, acceleration, or a maximum, minimum or turning point. Integrate when asked for the equation of a curve from dy/dx, an area under or between curves, distance from velocity, velocity from acceleration, or a total amount.

PART A: DIFFERENTIATION

What dy/dx means

A car moves so that its distance is s = t² metres. Its average speed between t = 2 and t = 2 + h is ((2 + h)² − 4) ÷ h = 4 + h. As h shrinks to zero (1, 0.1, 0.01 …), the average speed settles at 4 m/s. That limiting value is the derivative.

dy/dx ("dee-y by dee-x") is the rate of change of y with respect to x. At any point it equals the gradient of the tangent to the curve there. Other notations: f′(x), y′.

Differentiation from first principles

dy/dx = lim (h → 0) [f(x + h) − f(x)] ÷ h

The four steps examiners mark: write f(x + h) → subtract f(x) → divide by h → let h → 0.

The rules

• Power rule: y = axⁿ → dy/dx = nax^{n − 1} (bring the power down, then reduce it by 1)
• Constant: d/dx(7) = 0; linear: d/dx(5x) = 5
• Sums and differences: differentiate term by term
• Rewrite first: 1/x² = x⁻², √x = x^½

Product rule: y = uv → dy/dx = v·(du/dx) + u·(dv/dx)

Quotient rule: y = u/v → dy/dx = (v·du/dx − u·dv/dx) ÷ v² ("bottom first" in the numerator)

Chain rule (function of a function): y = f(u), u = g(x) → dy/dx = (dy/du) × (du/dx)

Sine and cosine (x in radians; mainly Further Maths and UTME): d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(sin ax) = a cos ax.

Tangents and normals

1. Find the point (x₁, y₁) on the curve.
2. Find the gradient m = dy/dx at that point.
3. Tangent: y − y₁ = m(x − x₁). Normal (at right angles to the tangent): gradient −1/m.

Rates of change and motion

If s is displacement at time t: velocity v = ds/dt and acceleration a = dv/dt.

Maximum and minimum (stationary) points

At a turning point the gradient is zero: set dy/dx = 0 and solve. Then use the second derivative d²y/dx²:

• negative → maximum (∩)
• positive → minimum (∪)

Optimisation word problems

Write the quantity to be maximised or minimised in terms of one variable, differentiate, set to zero, and check it is a maximum or minimum.

Differentiation traps

1. A constant differentiates to 0, not itself.
2. Rewrite 1/x³ as x⁻³ before differentiating.
3. Negative powers go down: x⁻² → −2x⁻³.
4. Don't forget the chain rule: (3x + 2)⁵ needs × 3 from the inside.
5. d/dx(uv) is not u′ × v′.
6. Quotient rule: v·u′ − u·v′ (bottom first).

PART B: INTEGRATION

What is integration?

Integration is the reverse of differentiation. Differentiation finds the gradient (rate of change) of a function; integration starts from the gradient and finds the original function.

Notation and the constant of integration

∫ f(x) dx = F(x) + C

• ∫ is the integral sign; f(x) is the integrand; dx means "with respect to x".
• C is the constant of integration. x³ + 5, x³ − 10 and x³ + 0.5 all differentiate to 3x², so when we integrate 3x² we must write x³ + C.

The basic rules

• Power rule: ∫ xⁿ dx = x^{n + 1}/(n + 1) + C (for n ≠ −1). In words: add 1 to the power, then divide by the new power.
• Constant multiple: ∫ k f(x) dx = k ∫ f(x) dx
• Sum and difference: integrate term by term.
• A constant on its own: ∫ k dx = kx + C

Check by differentiating

Differentiate your answer: you should get back the integrand.

Special integrals to memorise

• ∫ 1/x dx = ln|x| + C (the power rule fails when n = −1)
• ∫ e^x dx = e^x + C
• ∫ cos x dx = sin x + C
• ∫ sin x dx = −cos x + C

Integrating (ax + b)ⁿ: a linear "inside"

For a bracket with a linear inside, integrate as normal and then divide by the coefficient of x:

∫ (ax + b)ⁿ dx = (ax + b)^{n + 1}/(a(n + 1)) + C

Finding the equation of a curve from its gradient

If you know dy/dx and one point on the curve, integrate and then use the point to find C.

Definite integrals

A definite integral has limits. Integrate (no C needed), then substitute the upper limit minus the lower limit:

∫_a^b f(x) dx = [F(x)]_a^b = F(b) − F(a)

Area under a curve

The area between a curve y = f(x), the x-axis and the lines x = a and x = b is ∫_a^b f(x) dx (when the curve is above the x-axis).

Area below the x-axis and between curves

Area between two curves = ∫ (top curve − bottom curve) dx, between the points where they meet.

Motion by integration

Distance s = ∫ v dt (the area under a velocity–time graph) and velocity v = ∫ a dt.

A full NECO Paper 2 question

Going further (Further Maths and A-Level)

Harder integrals use extra techniques:

• Substitution for a function and its derivative together, e.g. ∫ x e^x² dx = ½e^x² + C.
• Integration by parts for products: ∫ u dv = uv − ∫ v du. For example, ∫ x e^x dx = x e^x − e^x + C.
• Partial fractions for rational functions, e.g. ∫ 1/((x + 1)(x + 2)) dx = ln|x + 1| − ln|x + 2| + C. (See the Algebra class for how to split the fraction.)

Integration traps

1. Forgetting + C on an indefinite integral.
2. Multiplying instead of dividing: ∫ x³ dx = x⁴/4, not 4x⁴.
3. Integrating brackets directly: expand (x + 1)² first.
4. Subtracting the limits the wrong way round: always F(upper) − F(lower).
5. Leaving an area below the axis negative: take the positive value.
6. Mixing up the operations: area or distance → integrate; gradient or rate → differentiate.

Practice

1. Differentiate y = 5x⁴ − 3x² + 8.
2. Differentiate y = (4x − 1)³.
3. Differentiate y = 2x/(x + 3).
4. Find the equation of the tangent to y = x² + 1 at x = 1.
5. Find and classify the turning point of y = x² − 6x + 5.
6. s = 2t³ − 9t² + 12t. When is the particle at rest?
7. Two positive numbers add up to 20. Find the largest possible product.
8. Find ∫ (4x³ + 2x) dx and ∫ (1/x³) dx.
9. Evaluate ∫ from 0 to 2 of (x² + 2) dx.
10. Find the area enclosed by y = 4 − x² and the x-axis.
11. dy/dx = 2x + 3 and the curve passes through (1, 6). Find y.
12. Find the maximum value of y = 12x − x³ for x > 0.

Answers: 1. 20x³ − 6x · 2. 12(4x − 1)² · 3. 6/(x + 3)² · 4. y = 2x · 5. (3, −4), a minimum · 6. t = 1 s and t = 2 s · 7. 10 × 10 = 100 · 8. x⁴ + x² + C; −1/(2x²) + C · 9. 20/3 = 6⅔ · 10. 32/3 = 10⅔ square units · 11. y = x² + 3x + 2 · 12. 16 (at x = 2)

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