Calculus – OCR FSMQ Additional Maths

Calculus is the mathematics of change. It is almost certainly the most important new idea in this qualification, and it is the foundation of AS and A Level Mathematics.

The two halves, and how they relate

Differentiation finds a rate of change — the gradient of a curve
Integration reverses differentiation, and finds an area
Each undoes the other

What calculus adds to what you know. At GCSE you found the gradient of a straight line — the same everywhere. A curve's gradient changes from point to point, and differentiation gives you a formula for it: the gradient function.

The scope is deliberately limited. The specification restricts differentiation and integration to $kx^n$ where $n$ is a positive integer or zero, and sums of such terms. No products, quotients, chain rule, trigonometric or exponential calculus is required.

Notation for the derivative (CA1)

$\dfrac{\mathrm{d}y}{\mathrm{d}x}$ · $\mathrm{f}'(x)$ · $\dot{x}$ (for differentiation with respect to time)

Word	Meaning
Derivative	The result of differentiating — the gradient function.
Gradient function	A formula giving the gradient at any $x$.
Tangent	The line touching the curve, with the curve's gradient.
Normal	Perpendicular to the tangent at the same point.
Stationary point	Where $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ — the tangent is horizontal.
Second derivative	$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$: the rate of change of the gradient.
Indefinite integral	An integral with no limits, needing $+c$.
Definite integral	An integral between two limits — a number.
Ordinate	A vertical line at a given $x$, bounding an area.

The dot notation matters for kinematics. If $x$ is displacement and $t$ is time, $\dot{x}$ is velocity and $\ddot{x}$ is acceleration. The specification names this notation explicitly.

To differentiate (CA1)

$y = kx^n \;\Longrightarrow\; \dfrac{\mathrm{d}y}{\mathrm{d}x} = knx^{n-1}$
Multiply by the power, then reduce the power by one

To integrate (CA8)

$\displaystyle\int kx^n \,\mathrm{d}x = \dfrac{kx^{n+1}}{n+1} + c$
Raise the power by one, then divide by the new power

Notice that these are exact opposites. Differentiating multiplies by the old power and drops it; integrating raises the power and divides by the new one. That is what "integration is the reverse of differentiation" (CA9) means in practice.

Worked Example — There and back again

Differentiate $y = 4x^3 - 5x + 7$, then integrate the result.

①$4x^3 \to 4 \times 3 x^2 = 12x^2$

②$-5x \to -5$ (since $-5x^1 \to -5x^0 = -5$)

③$+7 \to 0$ (a constant has zero gradient)

④$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 - 5$

⑤Now integrate: $\displaystyle\int \left(12x^2 - 5\right)\mathrm{d}x = \dfrac{12x^3}{3} - 5x + c = 4x^3 - 5x + c$

⑥We recover the original — except the $7$ has become an unknown $c$.

This is exactly why $+c$ is needed. Differentiating destroyed the constant term, so integrating cannot recover it. Only extra information — a point on the curve — can pin $c$ down.

CA1
CA2
CA3DifferentiationThe gradient function, the rule for $kx^n$, and what the answer means.
CA4Tangents and NormalsFinding the equation of a tangent or normal at any point.
CA5
CA6
CA7Stationary PointsLocating them, determining their nature, and sketching the curve.
CA5
CA6Maximum and Minimum ProblemsUsing calculus to find the best design or the largest volume.
CA8
CA9
CA10
CA11IntegrationIndefinite and definite integrals, and finding a curve from its gradient.
CA12
CA13Areas Under and Between CurvesAreas bounded by a curve and the axis, and between two curves.
CA14
CA15Application to KinematicsDisplacement, velocity and acceleration, with variable acceleration.

Forgetting $+c$. Every indefinite integral needs it. It is usually a whole mark.

Differentiating a constant to itself. The derivative of $7$ is $0$, not $7$.

Mishandling $kx$. The derivative of $5x$ is $5$, not $5x$ and not $0$.

Stopping at the stationary point. If the question asks for the nature of the point, you must test it with the second derivative and state the conclusion.

Giving only the $x$-coordinate. "Find the stationary points" wants coordinates — substitute back to get $y$.

Using the calculator's integral function when detailed reasoning is demanded. The specification is explicit: writing down a definite integral's value straight from the calculator will not be awarded full marks.

Differentiate

$kx^n \to knx^{n-1}$.

Integrate

$kx^n \to \dfrac{kx^{n+1}}{n+1} + c$.

Constants

Differentiate to $0$; integrate to $kx$.

Gradient at a point

Substitute $x$ into the derivative.

Tangent

$y - y_1 = m(x-x_1)$ with $m$ from the derivative.

Normal

Gradient $-\dfrac1m$.

Stationary

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$.

Nature

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$ minimum; $< 0$ maximum.

Definite integral

$\left[F(x)\right]_a^b = F(b) - F(a)$.

Kinematics

Differentiate to go $s \to v \to a$; integrate to come back.

Question 1

Differentiate $y = x^4$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3$

Question 2

Differentiate $y = 3x^2 - 7x + 4$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x - 7$

The constant $4$ differentiates to $0$.

Question 3

Find $\displaystyle\int \left(6x^2 + 2\right)\mathrm{d}x$.

▶ Show solution

$\dfrac{6x^3}{3} + 2x + c = 2x^3 + 2x + c$

Question 4

Find the gradient of $y = x^3 - 2x$ at $x = 2$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2$

At $x = 2$: $12 - 2 = 10$

Question 5

Evaluate $\displaystyle\int_0^2 3x^2 \,\mathrm{d}x$.

▶ Show solution

$\left[x^3\right]_0^2 = 8 - 0 = 8$

Question 6

Find the $x$-coordinates of the stationary points of $y = x^3 - 3x$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 3 = 0$

$x^2 = 1$, so $x = 1$ or $x = -1$.

Question 7

Determine the nature of the stationary point of $y = x^2 - 6x + 1$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 6 = 0$ gives $x = 3$, and then $y = 9 - 18 + 1 = -8$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0$, so $(3, -8)$ is a minimum.

Question 8

A curve has gradient function $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x$ and passes through $(1, 5)$. Find its equation.

▶ Show solution

$y = 2x^2 + c$

Substituting $(1,5)$: $5 = 2 + c$, so $c = 3$.

$y = 2x^2 + 3$

Question 9

A particle has displacement $s = t^3 - 4t^2$ metres at time $t$ seconds. Find its velocity and acceleration at $t = 3$.

▶ Show solution

$v = \dfrac{\mathrm{d}s}{\mathrm{d}t} = 3t^2 - 8t$

At $t = 3$: $v = 27 - 24 = 3$ m s⁻¹

$a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 6t - 8$

At $t = 3$: $a = 18 - 8 = 10$ m s⁻²

Question 10

The curve $C$ has equation $y = x^3 - 6x^2 + 9x$.

(a) Find the coordinates of the stationary points and determine their nature. (b) Find where $C$ crosses the $x$-axis. (c) Sketch $C$. (d) Find the area enclosed between $C$ and the $x$-axis between $x=0$ and $x=3$.

▶ Show solution

(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x + 9 = 3\left(x^2 - 4x + 3\right) = 3(x-1)(x-3)$

Stationary where this is zero: $x = 1$ or $x = 3$.

At $x=1$: $y = 1 - 6 + 9 = 4$. At $x=3$: $y = 27 - 54 + 27 = 0$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 12$

At $x=1$: $6 - 12 = -6 < 0$, so $(1, 4)$ is a maximum.

At $x=3$: $18 - 12 = 6 > 0$, so $(3, 0)$ is a minimum.

(b) $x^3 - 6x^2 + 9x = x\left(x^2-6x+9\right) = x(x-3)^2$

So $C$ crosses at $x = 0$ and touches at $x = 3$ (a repeated factor) — consistent with $(3,0)$ being a stationary point.

(c) A positive cubic: rising from the bottom left, through the origin, up to the maximum $(1,4)$, down to touch the axis at the minimum $(3,0)$, then rising again.

(d) The curve is above the axis throughout $0 < x < 3$ (it only meets the axis at the two ends), so a single integral gives the area.

$\displaystyle\int_0^3 \left(x^3 - 6x^2 + 9x\right)\mathrm{d}x = \left[\dfrac{x^4}{4} - 2x^3 + \dfrac{9x^2}{2}\right]_0^3$

At $x=3$: $\dfrac{81}{4} - 54 + \dfrac{81}{2} = 20.25 - 54 + 40.5 = 6.75$

At $x=0$: $0$

Area $= \dfrac{27}{4} = 6.75$ square units.

Sense check: the region sits inside a $3 \times 4$ rectangle of area $12$, and the curve dips well below the top of it, so $6.75$ is plausible.