AQA A-Level Maths: Calculus Applications — Complete Revision Guide (7357)

Calculus is the highest-yield content area on AQA A-Level Maths (7357). On a typical Paper 1 and Paper 2 sitting, the differentiation and integration sections together account for around a third of the available pure marks — more than algebra, more than trigonometry, more than coordinate geometry. The reason is straightforward. Once you have the basic toolkit (the chain, product and quotient rules; standard integrals; integration by substitution and by parts), AQA can build problems of almost any depth on top of that foundation: optimisation, kinematics, areas, volumes, differential equations, curve sketching. Calculus is where pure mathematics is at its most powerful, and AQA's papers reflect that.

This guide is the calculus applications depth course — it sits on top of the techniques you have already met and shows you how AQA examines them. It is the second of the three depth guides for 7357, designed to be read alongside the Pure 1 and Pure 2 overviews. The course it accompanies combines the application content of Section G (differentiation) with the integration techniques of Section H, because that is how AQA tests them — together, in the same long question, with marks distributed across both halves.

A note on what this guide is not. It is not a treatment of basic differentiation rules or standard integrals: those are assumed. If you are not yet fluent with $\frac{d}{dx}(\sin x) = \cos x$dxd​(sinx)=cosx or with substituting limits into a definite integral, work through Pure 1 first. What you will get here is the application layer — the question types AQA actually sets, the pitfalls candidates fall into, and a clear map of how the ten core topics connect. Use it as a checklist, a refresher, and a launchpad into focused practice.

What the AQA 7357 Specification Covers in Calculus Applications

AQA A-Level Maths (7357) is assessed through three two-hour papers, each worth 100 marks. Papers 1 and 2 cover pure mathematics; Paper 3 covers a mixture of pure, mechanics and statistics. Calculus content sits in Sections G (differentiation) and H (integration) of the specification, and can appear on any of the three papers, though long calculus questions tend to live on Papers 1 and 2.

The application content is what AQA examines most heavily once the basic techniques are in place. The table below shows the ten sub-topics covered in this guide, the part of the specification they sit under, and a realistic estimate of how many marks across a Paper 1 / Paper 2 sitting come from each.

Topic	Spec Section	Typical Paper 1/2 marks weight
Optimisation problems	G	6-10 marks
Connected rates of change	G	4-6 marks
Tangents and normals	G	4-6 marks
Stationary points and curve sketching	G	6-10 marks
Areas under curves	H	4-6 marks
Areas between curves	H	4-6 marks
Trapezium rule	H	3-5 marks
Integration by substitution	H	4-6 marks
Integration by parts	H	4-6 marks
Differential equations modelling	H	6-10 marks

These weights are estimates based on the spread of typical 7357 papers — not guarantees for any single year. What is reliable is that calculus applications are the single largest source of pure marks across Papers 1 and 2, and that long-tail questions (8 marks or more) are disproportionately drawn from this section. A candidate who is confident across all ten topics here is well placed for the top grade boundaries.

Optimisation Problems

Optimisation problems ask you to find the maximum or minimum of a quantity subject to a constraint. The classic AQA setup is a real-world scenario — a box of fixed surface area, a cylinder of fixed volume, a fenced region against a wall — where you must construct the function to be optimised, differentiate, set the derivative to zero, and verify the nature of the stationary point.

The workflow is the same every time. First, read the question carefully and identify the variable being optimised and the constraint that links the dimensions. Second, use the constraint to eliminate variables until the quantity to be optimised is a function of a single variable. Third, differentiate, set $\frac{dV}{dx} = 0$dxdV​=0 (or whichever variable applies), and solve. Fourth, verify the stationary point is the required type — usually by checking $\frac{d^2V}{dx^2}$dx2d2V​ — and substitute back to find the maximum or minimum value.

A short worked example. A closed cylinder of volume $1000\,\text{cm}^3$1000cm3 has radius $r$r and height $h$h. Find the radius that minimises the surface area. The constraint is $\pi r^2 h = 1000$πr2h=1000, so $h = \frac{1000}{\pi r^2}$h=πr21000​. The surface area is $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \frac{2000}{r}$S=2πr2+2πrh=2πr2+r2000​. Differentiating, $\frac{dS}{dr} = 4\pi r - \frac{2000}{r^2}$drdS​=4πr−r22000​. Setting this to zero gives $r^3 = \frac{500}{\pi}$r3=π500​, so $r = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42\,\text{cm}$r=(π500​)1/3≈5.42cm.

Common pitfalls include: forgetting the constraint and trying to differentiate a function of two variables; setting up the wrong objective (minimising volume instead of surface area); and skipping the second-derivative check, which AQA usually awards a method mark for. Always verify and always substitute back to give the requested optimum value, not just the value of $x$x at which it occurs.

For full coverage with worked optimisation problems and a clean construction workflow, see the Optimisation Problems lesson.

Connected Rates of Change

Connected rates of change problems use the chain rule to relate the rate at which one quantity is changing to the rate at which another is changing. The textbook scenario is a balloon being inflated: you are told the rate at which the radius is increasing and asked for the rate at which the volume is increasing.

The setup is almost mechanical. You have two variables linked by a known formula — for the balloon, $V = \frac{4}{3}\pi r^3$V=34​πr3. You are told one rate, say $\frac{dr}{dt} = 0.5\,\text{cm/s}$dtdr​=0.5cm/s, and asked for another, $\frac{dV}{dt}$dtdV​. The chain rule gives $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$dtdV​=drdV​⋅dtdr​. Differentiate $V$V with respect to $r$r, multiply by the given rate, and substitute the value of $r$r at the moment in question.

A short worked example. A spherical balloon is being inflated so that its radius increases at $0.5\,\text{cm/s}$0.5cm/s. Find the rate at which the volume is increasing when $r = 4\,\text{cm}$r=4cm. We have $\frac{dV}{dr} = 4\pi r^2$drdV​=4πr2, so $\frac{dV}{dt} = 4\pi r^2 \cdot 0.5 = 2\pi r^2$dtdV​=4πr2⋅0.5=2πr2. At $r = 4$r=4, $\frac{dV}{dt} = 32\pi \approx 100.5\,\text{cm}^3/\text{s}$dtdV​=32π≈100.5cm3/s.

Common pitfalls are using the wrong chain — for instance, computing $\frac{dr}{dV}$dVdr​ when $\frac{dV}{dr}$drdV​ was needed — and substituting the value of $r$r before differentiating, which produces a constant and a derivative of zero. Always differentiate first, then substitute. Another pitfall is forgetting to include units in the final answer, which AQA usually requires for the final mark.

For chain-rule practice on connected-rates problems, see the Rates of Change lesson.

Tangents and Normals

The tangent to a curve at a point is the straight line that touches the curve at that point and has the same gradient as the curve there. The normal is the straight line through the same point that is perpendicular to the tangent. Both are tested constantly at A-Level, often as the first part of a longer question.

The workflow is short. Differentiate to get $\frac{dy}{dx}$dxdy​, substitute the $x$x-coordinate of the point to get the tangent gradient $m$m, and use the point-slope form $y - y_1 = m(x - x_1)$y−y1​=m(x−x1​). For the normal, the gradient is $-\frac{1}{m}$−m1​ (provided $m \neq 0$m=0). The point through which both lines pass is the same — the point of tangency.

A short worked example. Find the equation of the tangent to $y = x^3 - 2x + 1$y=x3−2x+1 at the point where $x = 2$x=2. The $y$y-coordinate is $y = 8 - 4 + 1 = 5$y=8−4+1=5, so the point is $(2, 5)$(2,5). Differentiating, $\frac{dy}{dx} = 3x^2 - 2$dxdy​=3x2−2, which at $x = 2$x=2 equals $10$10. The tangent equation is $y - 5 = 10(x - 2)$y−5=10(x−2), or $y = 10x - 15$y=10x−15. The normal has gradient $-\frac{1}{10}$−101​ and equation $y - 5 = -\frac{1}{10}(x - 2)$y−5=−101​(x−2).

Common pitfalls include differentiating but forgetting to substitute the $x$x-value before quoting a gradient, leaving the answer in terms of $x$x. Another is reciprocating the wrong sign for the normal — the rule is negative reciprocal, both operations together. A subtler error is finding the gradient of the curve at a point but using a different point's coordinates in the line equation, often because the point was given as $x = 2$x=2 but the candidate copied $(0, 1)$(0,1) from earlier working.

For step-by-step worked tangent and normal problems, see the Tangents and Normals lesson.

Stationary Points and Curve Sketching

A stationary point of a curve $y = f(x)$y=f(x) is a point where $\frac{dy}{dx} = 0$dxdy​=0. The three types AQA expects you to distinguish are local maxima, local minima, and points of inflection. Identifying the type is part of almost every stationary-point question, and there are two standard tests.

The second-derivative test is fastest. Compute $\frac{d^2y}{dx^2}$dx2d2y​ at the stationary point. If it is negative, the point is a local maximum; if positive, a local minimum; if zero, the test is inconclusive and you must use the first-derivative test instead. The first-derivative test examines the sign of $\frac{dy}{dx}$dxdy​ just before and just after the stationary point. A sign change from positive to negative indicates a maximum; negative to positive indicates a minimum; no sign change indicates a point of inflection (specifically, a stationary point of inflection).

For curve sketching, the routine is: find intercepts (set $x = 0$x=0 and $y = 0$y=0), find stationary points (set $\frac{dy}{dx} = 0$dxdy​=0), classify them, identify any asymptotes (especially for rational functions), and consider end behaviour as $x \to \pm\infty$x→±∞. A clean sketch shows all of these features, labelled, with the curve drawn smoothly through them. Examiners want shape and key features correct, not a graph drawn to scale.

A common pitfall is concluding from $\frac{d^2y}{dx^2} = 0$dx2d2y​=0 that the point is automatically a point of inflection. It might be — but it might also be a maximum or minimum where the second derivative happens to vanish. Always fall back on the first-derivative test when the second-derivative test is inconclusive. Another pitfall is missing intercepts on a sketch, particularly the $y$y-intercept, which is often a single mark that students forget to claim.

For sketching practice and a workflow for any cubic or quartic, see the Stationary Points and Curve Sketching lesson.

Areas Under Curves

The area under a curve $y = f(x)$y=f(x) between $x = a$x=a and $x = b$x=b, where $f(x) \geq 0$f(x)≥0 on $[a, b]$[a,b], is given by the definite integral

$\text{Area} = \int_a^b f(x)\,dx.$Area=∫ab​f(x)dx.

This is the most basic application of integration on the spec, but it is examined every series and there are several traps. The first is that the formula assumes $f(x) \geq 0$f(x)≥0 throughout the interval. If the curve dips below the $x$x-axis, the integral counts that region as negative and the answer is no longer the geometric area.

For a curve that crosses the $x$x-axis between $a$a and $b$b, you must split the integral at the crossing points and take the absolute value of each piece. So if $f(x) = x^3 - x$f(x)=x3−x between $x = -1$x=−1 and $x = 2$x=2, the curve crosses at $x = 0$x=0 and $x = 1$x=1, and the geometric area is $\left|\int_{-1}^{0} (x^3 - x)\,dx\right| + \left|\int_{0}^{1} (x^3 - x)\,dx\right| + \left|\int_{1}^{2} (x^3 - x)\,dx\right|$​∫−10​(x3−x)dx​+​∫01​(x3−x)dx​+​∫12​(x3−x)dx​. Computing one definite integral from $-1$−1 to $2$2 gives a different (and wrong) number.

A short worked example. Find the area enclosed between $y = 4 - x^2$y=4−x2 and the $x$x-axis. The curve crosses the $x$x-axis at $x = \pm 2$x=±2 and is positive in between, so $\text{Area} = \int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{32}{3}$Area=∫−22​(4−x2)dx=[4x−3x3​]−22​=(8−38​)−(−8+38​)=332​.

Common pitfalls include integrating across an axis crossing without splitting; computing $\int f(x)\,dx$∫f(x)dx but forgetting to evaluate at the limits; and substitution sign errors when the lower limit is negative. Always sketch the curve before integrating — five seconds of drawing saves most of these errors.

For worked area problems with split integrals and limit-substitution practice, see the Areas Under Curves lesson.

Areas Between Curves

The area between two curves $y = f(x)$y=f(x) and $y = g(x)$y=g(x) between their intersection points is given by

$\text{Area} = \int_a^b [f(x) - g(x)]\,dx,$Area=∫ab​[f(x)−g(x)]dx,

where $f(x) \geq g(x)$f(x)≥g(x) on $[a, b]$[a,b] and $a, b$a,b are the $x$x-coordinates of the intersection points. The formula is the area under the upper curve minus the area under the lower curve, which leaves the enclosed region.

The standard workflow is: solve $f(x) = g(x)$f(x)=g(x) to find the intersection points; identify which curve is above on the interval (a quick sketch or testing a midpoint value works); set up the integral with upper minus lower; integrate and substitute the limits. The order of subtraction matters — if you put the lower curve first you get a negative answer. AQA generally accepts the absolute value, but you should aim to get the order right first time.

A short worked example. Find the area enclosed between $y = x$y=x and $y = x^2$y=x2. Setting $x = x^2$x=x2 gives intersection at $x = 0$x=0 and $x = 1$x=1. On $[0, 1]$[0,1], $x \geq x^2$x≥x2, so the line is above. $\text{Area} = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$Area=∫01​(x−x2)dx=[2x2​−3x3​]01​=21​−31​=61​.

Common pitfalls include using $f(x) + g(x)$f(x)+g(x) instead of $f(x) - g(x)$f(x)−g(x), getting the upper-lower order wrong, and missing intersection points by not solving the simultaneous equations carefully. Cubics and higher-degree intersections can have more than two crossing points; check for them by factorising the difference $f(x) - g(x)$f(x)−g(x) fully.

For worked between-curves problems and intersection finding, see the Areas Between Curves lesson.

The Trapezium Rule

The trapezium rule is a numerical method for approximating a definite integral when an exact antiderivative is unavailable or impractical. The rule divides the interval $[a, b]$[a,b] into $n$n equal-width strips, each of width $h = \frac{b - a}{n}$h=nb−a​, and approximates the area under the curve as a sum of trapezium areas:

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\right],$∫ab​f(x)dx≈2h​[y0​+yn​+2(y1​+y2​+⋯+yn−1​)],

where $y_i = f(a + ih)$yi​=f(a+ih) are the function values at the strip boundaries.

The setup is mechanical. Compute $h$h, tabulate the $y$y-values at $x = a, a + h, a + 2h, \ldots, b$x=a,a+h,a+2h,…,b, plug them into the formula, and evaluate. The standard table format is essential — laying out the values in a clean column with $i$i and $y_i$yi​ headings prevents arithmetic slips and earns method marks even if the final number is wrong.

A short worked example. Approximate $\int_0^1 \sqrt{1 + x^3}\,dx$∫01​1+x3​dx using the trapezium rule with four strips. Here $h = 0.25$h=0.25 and the $y$y-values at $x = 0, 0.25, 0.5, 0.75, 1$x=0,0.25,0.5,0.75,1 are approximately $1, 1.0078, 1.0607, 1.1924, 1.4142$1,1.0078,1.0607,1.1924,1.4142. The estimate is $\frac{0.25}{2}\left[1 + 1.4142 + 2(1.0078 + 1.0607 + 1.1924)\right] \approx 1.111$20.25​[1+1.4142+2(1.0078+1.0607+1.1924)]≈1.111.

Common pitfalls include using $n$n ordinates when $n$n strips were asked for (or vice versa) — they differ by one — and forgetting to double the interior $y$y-values. AQA also tests the direction of the error: for a curve that is concave up, the trapezium rule overestimates the true area; for a curve that is concave down, it underestimates. A sketch reveals which case you are in.

For worked trapezium-rule problems and a clean tabulation method, see the Trapezium Rule lesson.

Integration by Substitution

Integration by substitution is the integration analogue of the chain rule. To integrate $\int f(g(x)) \cdot g'(x)\,dx$∫f(g(x))⋅g′(x)dx, the substitution $u = g(x)$u=g(x) converts it into the simpler integral $\int f(u)\,du$∫f(u)du. Recognising when a substitution will work is the core skill — once you have the substitution, the mechanics are straightforward.

The standard signal is an integrand that contains a function and its derivative (or a constant multiple of its derivative). For instance, $\int 2x \cos(x^2)\,dx$∫2xcos(x2)dx contains $x^2$x2 inside the cosine and $2x$2x as a multiplier — exactly the derivative of $x^2$x2. Substituting $u = x^2$u=x2, $du = 2x\,dx$du=2xdx, gives $\int \cos u\,du = \sin u + c = \sin(x^2) + c$∫cosudu=sinu+c=sin(x2)+c.

For a definite integral, change the limits when you change the variable. If you substitute $u = g(x)$u=g(x), then when $x = a$x=a, $u = g(a)$u=g(a), and when $x = b$x=b, $u = g(b)$u=g(b). Evaluate the new integral with the new limits. The alternative is to do the integration in $u$u, change back to $x$x, and use the original limits — both approaches are valid but the limit-changing route is faster and less error-prone.

A short worked example. Evaluate $\int_0^2 x e^{x^2}\,dx$∫02​xex2dx. Let $u = x^2$u=x2, so $du = 2x\,dx$du=2xdx and $x\,dx = \frac{1}{2}\,du$xdx=21​du. When $x = 0$x=0, $u = 0$u=0; when $x = 2$x=2, $u = 4$u=4. The integral becomes $\int_0^4 \frac{1}{2}e^u\,du = \frac{1}{2}\left[e^u\right]_0^4 = \frac{1}{2}(e^4 - 1)$∫04​21​eudu=21​[eu]04​=21​(e4−1).

Common pitfalls include forgetting to change the limits for definite integrals; substituting $u$u but leaving an $x$x in the integrand; and choosing a substitution that does not simplify the integral. If the substitution makes the problem harder, it is the wrong substitution — try a different one.

For worked substitution problems with limit-changing practice, see the Integration by Substitution lesson.

Integration by Parts

Integration by parts is the integration analogue of the product rule. The formula is

$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx,$∫udxdv​dx=uv−∫vdxdu​dx,

and it converts an integral of a product into a different integral that is (hopefully) easier. The skill is choosing the right $u$u and $\frac{dv}{dx}$dxdv​.

A reliable mnemonic for the choice of $u$u is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Whichever type appears earliest in the list should be chosen as $u$u. So in $\int x \ln x\,dx$∫xlnxdx, the logarithm comes before the algebraic term, so $u = \ln x$u=lnx and $\frac{dv}{dx} = x$dxdv​=x. In $\int x e^x\,dx$∫xexdx, the algebraic term comes before the exponential, so $u = x$u=x and $\frac{dv}{dx} = e^x$dxdv​=ex.

A short worked example. Evaluate $\int x e^x\,dx$∫xexdx. Take $u = x$u=x, so $\frac{du}{dx} = 1$dxdu​=1, and $\frac{dv}{dx} = e^x$dxdv​=ex, so $v = e^x$v=ex. The formula gives $\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + c = e^x(x - 1) + c$∫xexdx=xex−∫exdx=xex−ex+c=ex(x−1)+c.

Some integrals require integration by parts twice, particularly products of polynomials with exponentials or trigonometric functions where the polynomial reduces in degree each application. A famous case is $\int e^x \cos x\,dx$∫excosxdx, where applying parts twice produces the original integral on the right-hand side, which can then be solved algebraically.

Common pitfalls include choosing $u$u and $\frac{dv}{dx}$dxdv​ the wrong way round (which usually makes the integral harder); errors in finding $v$v by integrating $\frac{dv}{dx}$dxdv​; and sign errors in the formula, particularly the minus sign before the second integral. The LIATE rule is reliable — use it as a default and only deviate when you have a specific reason.

For worked parts problems including the double-application case, see the Integration by Parts lesson.

Differential Equations and Modelling

A differential equation is an equation involving a function and its derivatives. The 7357 specification covers first-order separable differential equations of the form

$\frac{dy}{dx} = f(x)g(y),$dxdy​=f(x)g(y),

solved by separating variables — moving all the $y$y terms to one side, all the $x$x terms to the other, and integrating both sides:

$\int \frac{1}{g(y)}\,dy = \int f(x)\,dx.$∫g(y)1​dy=∫f(x)dx.

After integrating, you have an implicit relationship between $y$y and $x$x involving an arbitrary constant. If a boundary condition is given (a specific point that the solution passes through), substitute it to find the constant; this gives the particular solution.

The big modelling application is exponential growth and decay. Many real-world quantities — population, radioactive material, heated objects — change at a rate proportional to their current amount, giving $\frac{dy}{dt} = ky$dtdy​=ky. Separating and integrating gives $y = Ae^{kt}$y=Aekt, the exponential growth (or decay, if $k$k is negative) law. AQA tests this in physical contexts: cooling, drug clearance, capacitor discharge, population dynamics.

A short worked example. A population $P$P grows at a rate proportional to its size, with $P = 100$P=100 at $t = 0$t=0 and $P = 200$P=200 at $t = 5$t=5. Find $P$P at $t = 10$t=10. From $\frac{dP}{dt} = kP$dtdP​=kP, separation gives $\ln P = kt + c$lnP=kt+c, so $P = Ae^{kt}$P=Aekt. From $P(0) = 100$P(0)=100, $A = 100$A=100. From $P(5) = 200$P(5)=200, $200 = 100 e^{5k}$200=100e5k, so $k = \frac{\ln 2}{5}$k=5ln2​. Then $P(10) = 100 e^{10 \cdot \ln 2 / 5} = 100 \cdot 4 = 400$P(10)=100e10⋅ln2/5=100⋅4=400.

Common pitfalls include forgetting the constant of integration; applying the boundary condition before integrating; and arithmetic slips when working with $\ln$ln to find $k$k. AQA modelling questions also expect you to comment on the suitability of the model — for instance, that exponential growth cannot continue indefinitely because resources are finite. A short evaluative sentence is often worth a mark.

For worked separable-equation problems and modelling applications, see the Differential Equations and Modelling lesson.

Common Mark-Loss Patterns Across Calculus Applications

Across the whole calculus applications section, a small set of habits accounts for a disproportionate share of lost marks. None of these are about content you do not know. They are all about content you do know, applied carelessly.

• Forgetting the constant of integration in an indefinite integral. This is the single most common slip, and AQA penalises it consistently.
• Not changing the limits when substituting in a definite integral, or changing them but forgetting to change back if you also change variable back. Pick one approach and stick with it.
• Skipping the second-derivative check in optimisation and stationary-point problems. Even when the answer is obvious from context, AQA mark schemes usually award an explicit-verification method mark.
• Integrating across an axis crossing without splitting the integral. For geometric area, take absolute values of each piece; for signed area, the single integral is fine.
• Choosing $u$u and $\frac{dv}{dx}$dxdv​ the wrong way round in integration by parts. Apply LIATE as a default.
• Wrong direction of error in the trapezium rule. Sketch the curve and check concavity before stating overestimate or underestimate.
• Sign errors with negative reciprocals when finding normals. The rule is negative and reciprocal — both operations.
• Differentiating before substituting in connected-rates problems is correct; substituting before differentiating turns the function into a constant. Make this routine.
• Not commenting on model suitability in differential-equation modelling questions. A one-sentence evaluative comment often scores a mark.

A revision plan that explicitly drills these habits — not just the content — will move your grade more than another pass through the textbook.

Recommended Six-Week Revision Plan

This plan is designed for a candidate who has covered the calculus content in lessons but wants to revise it cleanly before the exam. It assumes about 5-6 hours per week on this section. Adjust pace if you are starting earlier or later.

Week	Topics	Practice
1	Optimisation problems; connected rates of change	10 optimisation problems with full second-derivative checks; 10 chain-rule rates problems
2	Tangents and normals; stationary points and curve sketching	15 tangent/normal problems; 10 stationary-point classifications; 8 full curve sketches
3	Areas under curves; areas between curves	12 area-under problems including 3 with axis crossings; 10 area-between problems
4	Trapezium rule; integration by substitution	6 trapezium-rule problems with concavity-direction checks; 15 substitution problems including 8 with definite limits
5	Integration by parts; differential equations modelling	12 parts problems including 3 double-application; 10 separable equations including 5 modelling contexts
6	Mixed practice; full long-question calculus papers; targeted review of weakest topics	One full mixed problem set per day; review marking-scheme working for any question scoring below 60%

The point of the plan is to keep moving forward while maintaining contact with earlier topics. Do not spend three weeks on optimisation and run out of time before differential equations. By the end of week 5, every topic in the section should have had focused contact and a practice round. Week 6 is consolidation and weakness-targeting.

A useful discipline through the whole plan is to treat any question you got wrong not as a mistake but as a diagnostic. Was it a content gap? A method error? A careless arithmetic slip? Logging the cause means your next review session targets the right thing.

How LearningBro's AQA A-Level Maths Calculus Applications Course Helps

LearningBro's AQA A-Level Maths: Calculus Applications course is built around the structure of this guide. Each of the ten lessons covers one application area of Sections G and H, in the order AQA teaches them, with worked examples, practice questions and full mark-scheme-style solutions. Lessons end with a short review and quick-recall questions designed for spaced revisits.

The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building each topic on the last — optimisation builds on stationary points, areas between curves builds on areas under curves, integration by parts and substitution feed into differential equations. As a revision tool, you can drop into any lesson and work the practice independently — for example, drilling integration by parts for a week before mocks. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.

If you want one place to revise this section of the spec well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the AQA A-Level Maths: Calculus Applications course.

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