AQA A-Level Further Maths: Further Calculus — In-Depth Revision Guide (7367)

Further calculus is where the integration and differentiation you met in A-Level Maths grow up. AQA A-Level Further Mathematics (7367) takes the familiar machinery and points it at genuinely new questions: integrals that run to infinity, the volume of a solid swept out by a curve, the length of a curve and the area of the surface it sweeps, and differential equations that describe how quantities change over time. These are the tools that physics, engineering and applied mathematics actually use, and they are some of the most satisfying techniques in the whole qualification.

This guide is the in-depth companion for AQA's further-calculus course. It runs from improper integrals and the mean value of a function, through volumes of revolution about both axes, repeated integration by parts and reduction formulae, into arc length and surface area of revolution, and finishes with first- and second-order differential equations. Each section gives the core skills, a worked example, the common pitfalls, and a link to the matching lesson.

The aim is not to replace working through problems. Calculus fluency comes only from doing many integrals and differential equations until the standard patterns are automatic. The aim of this guide is to give you a clear map of what AQA can examine, in roughly the order they teach it, so your revision is targeted rather than scattered.

What the AQA 7367 Specification Covers

AQA A-Level Further Mathematics (7367) is a linear qualification assessed through three written papers, each 2 hours long, each worth 100 marks, and each carrying 33⅓% of the marks. Papers 1 and 2 are compulsory and cover the Pure content — which is where further calculus, complex numbers and matrices all live. Paper 3 is the optional applications paper, on which you choose two of Mechanics, Statistics and Discrete Mathematics. Further calculus is compulsory Pure, so it can appear on either Paper 1 or Paper 2, often as a multi-step question that combines, say, a reduction formula with a volume of revolution.

Further Mathematics is one of the most highly valued A-Levels for STEM study, and is effectively expected for mathematics and many physics and engineering courses at Oxbridge and other competitive universities. The further-calculus content also rewards exactly the sustained, multi-step reasoning that the STEP and MAT admissions tests look for.

The table below shows the further-calculus sub-topics, what each involves, and why it matters across the course.

Topic	What it involves	Why it matters
Improper integrals	Integrals to infinity or over a singularity, via limits	Foundation for many advanced results
Mean value of a function	The average height of a curve over an interval	A clean, examinable application of integration
Volumes of revolution (x-axis)	Rotating a curve about $Ox$Ox, integrating $\pi y^2$πy2	Classic solid-of-revolution questions
Volumes of revolution (y-axis)	Rotating about $Oy$Oy, integrating $\pi x^2$πx2	The companion case; requires care with variables
Repeated integration by parts	Applying parts more than once	Integrals of products like $x^2 e^x$x2ex
Reduction formulae	A recurrence relating $I_n$In​ to $I_{n-1}$In−1​	Powerful for high-power integrands
Arc length	The length of a curve, $\int\sqrt{1 + (\mathrm{d}y/\mathrm{d}x)^2}\,\mathrm{d}x$∫1+(dy/dx)2​dx	Measuring curves, not just areas
Surface area of revolution	The area swept by rotating a curve	Builds on arc length
First-order differential equations	Separable equations and integrating factors	Modelling rates of change
Second-order differential equations	The auxiliary-equation method	Oscillations and growth/decay

These topics build on one another. Improper integrals and integration by parts underpin reduction formulae; arc length leads to surface area; and the differential-equation methods bring the whole toolkit together to model real systems. Revising them in order pays off.

Improper Integrals

A definite integral is improper if either a limit of integration is infinite, or the integrand becomes infinite (has a singularity) somewhere on the interval. You cannot simply substitute "infinity" into an antiderivative, so AQA's method is to replace the awkward limit with a variable and take a limit at the end:

$\int_a^{\infty} f(x)\,\mathrm{d}x = \lim_{t \to \infty} \int_a^{t} f(x)\,\mathrm{d}x.$∫a∞​f(x)dx=limt→∞​∫at​f(x)dx.

If this limit exists as a finite number, the integral converges to that value; if the limit is infinite or does not exist, the integral diverges and has no finite value. The same idea handles a singularity: if $f$f blows up at the lower limit $a$a, you integrate from $a + \varepsilon$a+ε and let $\varepsilon \to 0$ε→0.

A worked example. Evaluate $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,\mathrm{d}x$∫1∞​x21​dx. Replace the upper limit: $\displaystyle\int_1^{t} x^{-2}\,\mathrm{d}x = \left[-\frac{1}{x}\right]_1^{t} = -\frac{1}{t} + 1$∫1t​x−2dx=[−x1​]1t​=−t1​+1. As $t \to \infty$t→∞, $-\tfrac{1}{t} \to 0$−t1​→0, so the integral converges to $1$1. By contrast, $\int_1^{\infty} \tfrac{1}{x}\,\mathrm{d}x$∫1∞​x1​dx gives $\ln t \to \infty$lnt→∞, which diverges — a famous and frequently examined contrast.

A common pitfall is treating $\infty$∞ as a number you can substitute directly, rather than taking a proper limit. Another is failing to notice a singularity inside the interval, which makes an innocent-looking integral improper. For convergence-versus-divergence practice, see the Improper Integrals lesson.

Mean Value of a Function

The mean value of a continuous function $f$f over the interval $[a, b]$[a,b] is the average height of its graph, defined as

$\bar{f} = \frac{1}{b - a}\int_a^{b} f(x)\,\mathrm{d}x.$fˉ​=b−a1​∫ab​f(x)dx.

Geometrically, this is the height of the rectangle on base $[a, b]$[a,b] that has the same area as the region under the curve. It is a clean, self-contained application of integration that AQA likes to set, sometimes combined with a model (the average value of a current or displacement over a cycle, for instance).

A worked example. Find the mean value of $f(x) = x^2$f(x)=x2 over the interval $[0, 3]$[0,3]. Compute $\displaystyle\int_0^{3} x^2\,\mathrm{d}x = \left[\frac{x^3}{3}\right]_0^{3} = 9$∫03​x2dx=[3x3​]03​=9. The interval width is $3 - 0 = 3$3−0=3, so the mean value is $\bar{f} = \tfrac{1}{3}\times 9 = 3$fˉ​=31​×9=3. As a sanity check, this lies between the minimum value 0 and the maximum value 9 on the interval, as a mean must.

A common pitfall is forgetting the $\frac{1}{b - a}$b−a1​ factor and quoting the integral itself as the mean. Another is dividing by the wrong interval width, especially when limits are negative. For worked mean-value problems, see the Mean Value of a Function lesson.

Volumes of Revolution about the x-axis

When the region under a curve $y = f(x)$y=f(x) between $x = a$x=a and $x = b$x=b is rotated a full turn about the $x$x-axis, it sweeps out a solid whose volume is

$V = \pi \int_a^{b} y^2 \,\mathrm{d}x.$V=π∫ab​y2dx.

The reasoning is worth holding onto: slice the solid into thin discs perpendicular to the axis. Each disc at position $x$x is a circle of radius $y$y, so its area is $\pi y^2$πy2 and its volume is $\pi y^2\,\mathrm{d}x$πy2dx; summing (integrating) the discs gives the total. The single most common error is squaring incorrectly — you integrate $\pi y^2$πy2, so you must square the whole expression for $y$y before integrating, never integrate first and square afterwards.

A worked example. Find the volume when the region under $y = x^2$y=x2 from $x = 0$x=0 to $x = 2$x=2 is rotated about the $x$x-axis. Here $y^2 = x^4$y2=x4, so $V = \pi \displaystyle\int_0^{2} x^4\,\mathrm{d}x = \pi\left[\frac{x^5}{5}\right]_0^{2} = \pi\cdot\frac{32}{5} = \frac{32\pi}{5}$V=π∫02​x4dx=π[5x5​]02​=π⋅532​=532π​.

A common pitfall is dropping the $\pi$π, or integrating $y$y and then squaring (which is wrong — square first). Another is using the curve's equation without isolating $y$y cleanly. For disc-method practice, see the Volumes of Revolution about the x-axis lesson.

Volumes of Revolution about the y-axis

When the region is rotated about the $y$y-axis instead, the discs are perpendicular to the $y$y-axis, each of radius $x$x, so the volume is

$V = \pi \int_c^{d} x^2 \,\mathrm{d}y.$V=π∫cd​x2dy.

The crucial difference is that everything must be expressed in terms of $y$y: you rearrange the curve to make $x^2$x2 the subject, and the limits $c$c and $d$d are now $y$y-values, not $x$x-values. Forgetting to convert the limits — leaving them as $x$x-values — is the classic mistake here, and it is worth a deliberate pause in every such question.

A worked example. Find the volume when the region bounded by $y = x^2$y=x2, the $y$y-axis and the line $y = 4$y=4 is rotated about the $y$y-axis. Rearrange to $x^2 = y$x2=y. The $y$y-limits run from $0$0 to $4$4, so $V = \pi\displaystyle\int_0^{4} y\,\mathrm{d}y = \pi\left[\frac{y^2}{2}\right]_0^{4} = \pi\cdot 8 = 8\pi$V=π∫04​ydy=π[2y2​]04​=π⋅8=8π.

A common pitfall is integrating with respect to $x$x out of habit, or keeping the original $x$x-limits instead of converting to $y$y. Another is failing to rearrange the equation to give $x^2$x2 explicitly. For the companion method and mixed-axis problems, see the Volumes of Revolution about the y-axis lesson.

Integration by Parts: Repeated Application

Integration by parts reverses the product rule:

$\int u\,\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\,\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x.$∫udxdv​dx=uv−∫vdxdu​dx.

For integrands like $x^2 e^x$x2ex or $x^2 \sin x$x2sinx, one application is not enough — you have to apply parts repeatedly, each pass lowering the power of the polynomial factor by one until it disappears. Choosing $u$u wisely is the whole game: pick $u$u to be the factor that simplifies when differentiated (the polynomial), and $\tfrac{\mathrm{d}v}{\mathrm{d}x}$dxdv​ to be the factor you can integrate (the exponential or trig). The acronym LIATE is a useful guide to which factor to call $u$u.

A second, elegant technique is the "loop" or recovery method, used for integrals like $\int e^x \cos x\,\mathrm{d}x$∫excosxdx. Applying parts twice reproduces the original integral on the right-hand side; you then treat the integral as an unknown, collect it on one side, and solve algebraically.

A worked example. Evaluate $\displaystyle\int x^2 e^x\,\mathrm{d}x$∫x2exdx. Take $u = x^2$u=x2, $\tfrac{\mathrm{d}v}{\mathrm{d}x} = e^x$dxdv​=ex: the integral becomes $x^2 e^x - \int 2x e^x\,\mathrm{d}x$x2ex−∫2xexdx. Apply parts again to $\int 2x e^x\,\mathrm{d}x$∫2xexdx with $u = 2x$u=2x: this gives $2x e^x - \int 2 e^x\,\mathrm{d}x = 2x e^x - 2e^x$2xex−∫2exdx=2xex−2ex. Combining, $\int x^2 e^x\,\mathrm{d}x = x^2 e^x - 2x e^x + 2e^x + C = e^x(x^2 - 2x + 2) + C$∫x2exdx=x2ex−2xex+2ex+C=ex(x2−2x+2)+C.

A common pitfall is choosing $u$u and $\tfrac{\mathrm{d}v}{\mathrm{d}x}$dxdv​ the wrong way round, which makes the new integral harder, not easier. Another is sign-slips across the repeated minus signs. For the loop method and repeated-parts practice, see the Integration by Parts: Repeated Application lesson.

Reduction Formulae

A reduction formula is a recurrence that expresses an integral $I_n$In​ (depending on a positive integer $n$n) in terms of an integral $I_{n-1}$In−1​ or $I_{n-2}$In−2​ with a smaller index. They are the efficient way to handle integrands with a high power, such as $\int_0^{\pi/2} \sin^n x\,\mathrm{d}x$∫0π/2​sinnxdx: rather than integrating a tenth power directly, you derive a formula once and then step down.

The formula is almost always derived using integration by parts: split off one factor, apply parts, and manipulate the result until the original integral reappears with a reduced index. Once you have the recurrence, you apply it repeatedly down to a base case — typically $I_0$I0​ or $I_1$I1​ — which you evaluate directly.

A worked example. Given the reduction formula $I_n = \dfrac{n-1}{n}\,I_{n-2}$In​=nn−1​In−2​ for $I_n = \displaystyle\int_0^{\pi/2} \sin^n x\,\mathrm{d}x$In​=∫0π/2​sinnxdx, find $I_4$I4​. The base case is $I_0 = \int_0^{\pi/2} 1\,\mathrm{d}x = \tfrac{\pi}{2}$I0​=∫0π/2​1dx=2π​. Step up: $I_2 = \tfrac{1}{2}I_0 = \tfrac{\pi}{4}$I2​=21​I0​=4π​, then $I_4 = \tfrac{3}{4}I_2 = \tfrac{3}{4}\cdot\tfrac{\pi}{4} = \tfrac{3\pi}{16}$I4​=43​I2​=43​⋅4π​=163π​.

A common pitfall is making an algebra error in the derivation of the formula by parts, which then corrupts every subsequent value. Another is using the wrong base case or stopping the recurrence one step early. For derivations and step-down practice, see the Reduction Formulae lesson.

Arc Length

Calculus can measure the length of a curve, not just the area beneath it. For a curve $y = f(x)$y=f(x) from $x = a$x=a to $x = b$x=b, the arc length is

$s = \int_a^{b} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\;\mathrm{d}x.$s=∫ab​1+(dxdy​)2​dx.

The idea behind the formula is Pythagoras applied to an infinitesimal step: a tiny piece of the curve has horizontal extent $\mathrm{d}x$dx and vertical extent $\mathrm{d}y$dy, so its length is $\sqrt{(\mathrm{d}x)^2 + (\mathrm{d}y)^2}$(dx)2+(dy)2​; factoring out $\mathrm{d}x$dx gives the integrand. There is a companion parametric version, $s = \int \sqrt{(\mathrm{d}x/\mathrm{d}t)^2 + (\mathrm{d}y/\mathrm{d}t)^2}\,\mathrm{d}t$s=∫(dx/dt)2+(dy/dt)2​dt, for curves given as $x(t), y(t)$x(t),y(t).

A worked example. Find the arc length of $y = \tfrac{2}{3}x^{3/2}$y=32​x3/2 from $x = 0$x=0 to $x = 3$x=3. Differentiate: $\tfrac{\mathrm{d}y}{\mathrm{d}x} = x^{1/2}$dxdy​=x1/2, so $\left(\tfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = x$(dxdy​)2=x. The integrand becomes $\sqrt{1 + x}$1+x​, giving $s = \displaystyle\int_0^{3}\sqrt{1 + x}\,\mathrm{d}x = \left[\tfrac{2}{3}(1+x)^{3/2}\right]_0^{3} = \tfrac{2}{3}(8 - 1) = \tfrac{14}{3}$s=∫03​1+x​dx=[32​(1+x)3/2]03​=32​(8−1)=314​.

A common pitfall is forgetting to square the derivative inside the root, or dropping the $+1$+1. Another is choosing examples where the integrand does not simplify — AQA questions are designed so the expression under the root becomes a perfect square or a standard integral, so if yours does not, recheck the differentiation. For both Cartesian and parametric practice, see the Arc Length lesson.

Surface Area of Revolution

Rotating a curve about an axis sweeps out a surface, and its area builds directly on the arc-length idea. For a curve $y = f(x)$y=f(x) rotated about the $x$x-axis, the surface area is

$S = 2\pi \int_a^{b} y\,\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\;\mathrm{d}x.$S=2π∫ab​y1+(dxdy​)2​dx.

The structure is intuitive: each thin band of the surface is approximately a strip of circumference $2\pi y$2πy (the circle traced by the point at height $y$y) and width equal to the arc-length element $\sqrt{1 + (\mathrm{d}y/\mathrm{d}x)^2}\,\mathrm{d}x$1+(dy/dx)2​dx. Multiplying these and integrating gives the area. For rotation about the $y$y-axis the $2\pi y$2πy factor becomes $2\pi x$2πx, and you express the rest in terms of the appropriate variable.

A worked example. Find the surface area when $y = 2\sqrt{x}$y=2x​ for $0 \le x \le 3$0≤x≤3 is rotated about the $x$x-axis. Here $\tfrac{\mathrm{d}y}{\mathrm{d}x} = x^{-1/2}$dxdy​=x−1/2, so $\left(\tfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \tfrac{1}{x}$(dxdy​)2=x1​ and $\sqrt{1 + \tfrac{1}{x}} = \sqrt{\tfrac{x+1}{x}}$1+x1​​=xx+1​​. Then $S = 2\pi\displaystyle\int_0^{3} 2\sqrt{x}\cdot\sqrt{\frac{x+1}{x}}\,\mathrm{d}x = 4\pi\displaystyle\int_0^{3}\sqrt{x+1}\,\mathrm{d}x = 4\pi\left[\tfrac{2}{3}(x+1)^{3/2}\right]_0^{3} = 4\pi\cdot\tfrac{2}{3}(8 - 1) = \frac{56\pi}{3}$S=2π∫03​2x​⋅xx+1​​dx=4π∫03​x+1​dx=4π[32​(x+1)3/2]03​=4π⋅32​(8−1)=356π​. Notice how the $\sqrt{x}$x​ factors cancel cleanly — a hallmark of a well-posed question.

A common pitfall is forgetting the $2\pi y$2πy circumference factor and computing only the arc length, or dropping it to $\pi y$πy. Another is failing to simplify the integrand, which usually signals an arithmetic slip earlier. For worked surfaces about each axis, see the Surface Area of Revolution lesson.

First-Order Differential Equations

A differential equation relates a function to its derivatives, and it is the natural language for any process described by a rate of change. AQA's first-order content covers two main solution techniques. The first is separation of variables, used when the equation can be written so that all the $y$y-terms are on one side and all the $x$x-terms on the other:

$\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)g(y) \quad\Longrightarrow\quad \int \frac{1}{g(y)}\,\mathrm{d}y = \int f(x)\,\mathrm{d}x.$dxdy​=f(x)g(y)⟹∫g(y)1​dy=∫f(x)dx.

The second is the integrating factor method, for linear first-order equations of the form $\tfrac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)$dxdy​+P(x)y=Q(x). Multiplying through by the integrating factor $I = e^{\int P\,\mathrm{d}x}$I=e∫Pdx turns the left-hand side into the exact derivative $\tfrac{\mathrm{d}}{\mathrm{d}x}(Iy)$dxd​(Iy), after which you integrate both sides. A general solution contains an arbitrary constant; a boundary (or initial) condition pins that constant down to give a particular solution.

A worked example. Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} = xy$dxdy​=xy given that $y = 2$y=2 when $x = 0$x=0. Separate: $\displaystyle\int\frac{1}{y}\,\mathrm{d}y = \int x\,\mathrm{d}x$∫y1​dy=∫xdx, so $\ln|y| = \tfrac{1}{2}x^2 + c$ln∣y∣=21​x2+c. Exponentiate: $y = A e^{x^2/2}$y=Aex2/2 where $A = e^c$A=ec. Apply the condition $y(0) = 2$y(0)=2: $2 = A e^0 = A$2=Ae0=A, so $y = 2e^{x^2/2}$y=2ex2/2.

A common pitfall is forgetting the constant of integration, or applying the boundary condition too late (after some algebra has already lost it). Another is choosing separation when the equation is genuinely linear and needs an integrating factor, or vice versa. For both methods and modelling contexts, see the First Order Differential Equations lesson.

Second-Order Differential Equations

A second-order linear differential equation with constant coefficients has the form $a\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} + b\dfrac{\mathrm{d}y}{\mathrm{d}x} + cy = f(x)$adx2d2y​+bdxdy​+cy=f(x). The solution method is built on the auxiliary equation $am^2 + bm + c = 0$am2+bm+c=0, whose roots determine the complementary function (the general solution of the equation with $f(x) = 0$f(x)=0):

Roots of the auxiliary equation	Complementary function
Real and distinct $m_1, m_2$m1​,m2​	$y = A e^{m_1 x} + B e^{m_2 x}$y=Aem1​x+Bem2​x
Real and equal $m$m	$y = (A + Bx)e^{m x}$y=(A+Bx)emx
Complex $\alpha \pm \beta i$α±βi	$y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)$y=eαx(Acosβx+Bsinβx)

When $f(x)$f(x) is non-zero, you add a particular integral — a trial solution of the same form as $f(x)$f(x) (a polynomial, exponential or trig expression) whose coefficients you find by substitution. The full general solution is the complementary function plus the particular integral, and two boundary conditions fix the two arbitrary constants. The complex-roots case is exactly what produces oscillating solutions, which is why this topic underpins the mathematics of springs and circuits.

A worked example. Solve $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 5\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6y = 0$dx2d2y​−5dxdy​+6y=0. The auxiliary equation is $m^2 - 5m + 6 = 0$m2−5m+6=0, which factorises as $(m - 2)(m - 3) = 0$(m−2)(m−3)=0, so $m = 2$m=2 or $m = 3$m=3. The roots are real and distinct, so the general solution is $y = A e^{2x} + B e^{3x}$y=Ae2x+Be3x.

A common pitfall is choosing a trial particular integral that clashes with a term already in the complementary function (it must then be modified, usually by a factor of $x$x). Another is omitting one of the two arbitrary constants, or forgetting that the complex-roots case gives a trig-times-exponential form. For the particular-integral method and oscillatory cases, see the Second Order Differential Equations lesson.

Common Mark-Loss Patterns Across Further Calculus

Across the whole topic, a small set of habits accounts for a disproportionate share of lost marks. Most are not about content you do not know — they are about content you do know, applied carelessly under exam pressure.

• Treating $\infty$∞ as a number in an improper integral instead of taking a proper limit, and missing singularities inside the interval.
• Forgetting the $\frac{1}{b-a}$b−a1​ factor when finding a mean value.
• Integrating $y$y and then squaring in a volume of revolution — you must square $y$y first, and never drop the $\pi$π.
• Keeping $x$x-limits when rotating about the $y$y-axis, instead of converting to $y$y-limits and rearranging for $x^2$x2.
• Choosing $u$u and $\tfrac{\mathrm{d}v}{\mathrm{d}x}$dxdv​ the wrong way round in integration by parts, making the new integral harder.
• Making a derivation slip in a reduction formula, which corrupts every value, or using the wrong base case.
• Forgetting to square the derivative or the $+1$+1 inside the arc-length root.
• Dropping the $2\pi y$2πy circumference factor in a surface area, computing only the arc length.
• Losing the constant of integration, or applying a boundary condition too late, in a differential equation.
• Choosing a particular integral that clashes with the complementary function in a second-order equation.
• Not showing enough working. AQA mark schemes award method marks generously when the working is clear; a bare final answer can score fewer marks than a flawed answer with clean method.

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

How LearningBro's AQA Further Maths Further Calculus Course Helps

LearningBro's AQA A-Level Further Maths: Further Calculus course is built around the structure of this guide. Each lesson covers one focus area of the compulsory Pure content — improper integrals, the mean value of a function, volumes of revolution about each axis, repeated integration by parts, reduction formulae, arc length, surface area of revolution, and first- and second-order differential equations — with worked examples, mark-scheme-style solutions, and practice 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 technique on the integration foundation laid by the one before. As a revision tool, you can drop into any single lesson — drilling reduction formulae and volumes of revolution together before a mock, say — and work the practice independently. 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 further calculus well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the AQA A-Level Further Maths: Further Calculus course.

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