Integration is differentiation run in reverse, and in the IB Mathematics Applications & Interpretation HL calculus strand it is the tool of accumulation: given a rate — a velocity, a marginal cost, a flow into a reservoir — integrating it recovers the total built up. AI treats it as a modelling skill rather than an exercise in algebra: a GDC will evaluate almost any definite integral for you, so the marks live in setting up the right integral, choosing between an exact area and a trapezoidal estimate, and reading the answer back into the context with the correct units.
This cheatsheet condenses the whole range — from the reverse power rule and the constant of integration in SL 5.5 to areas, volumes of revolution and kinematics in AHL 5.11–5.13 — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Integration as anti-differentiation SL 5.5
To integrate is to undo differentiation: you look for a function whose derivative is the one in front of you. Because the derivative of a constant is zero, an indefinite integral always carries an unknown constant of integration $+C$.
For example $\displaystyle\int\left(6x^{2}-4x+5\right)dx=2x^{3}-2x^{2}+5x+C$: each power goes up by one and is divided by that new power. Integration measures accumulation — the total built up from a rate of change — which is why every application on this sheet begins by integrating a rate.
§2 — The definite integral SL 5.5
Put limits on an integral and it stops being a function and becomes a number. Evaluate any antiderivative $F$ at the top limit and subtract its value at the bottom — the $+C$ cancels, so you never need it here.
So $\displaystyle\int_{1}^{3}6x^{2}\,dx=\left[2x^{3}\right]_{1}^{3}=2(27)-2(1)=52$. Work out the antiderivative first, wrap it in square brackets with the limits, then substitute the upper limit and subtract the lower-limit value.
§3 — The constant of integration: boundary conditions SL 5.5
An indefinite integral leaves $C$ unknown, so it really describes a whole family of parallel curves. One known point — a boundary condition — pins down exactly which curve you have.
If $\dfrac{dy}{dx}=4x-3$ and the curve passes through $(2,\,5)$, then $y=2x^{2}-3x+C$; substituting gives $5=2(4)-3(2)+C=2+C$, so $C=3$ and $y=2x^{2}-3x+3$. This is how a marginal rate plus a single observed reading rebuilds an entire model — in economics that constant is the fixed cost, $C(0)$ (see §9).
§4 — Area under a curve SL 5.5
The definite integral has a picture. When $f(x)\ge 0$ across $[a,b]$, $\displaystyle\int_{a}^{b}f(x)\,dx$ is exactly the area between the curve $y=f(x)$ and the $x$-axis — the meaning behind cross-sections, land areas and glass panels.
Because AI always allows a calculator, the fastest route on Paper 2 is to graph the model and let the GDC return the area — but you must still state the integral you are evaluating to secure the method mark.
§5 — The trapezoidal rule SL 5.5
When you only have a table of readings — not a formula — you cannot integrate, so you estimate the area by slicing it into thin trapezoidal strips. This is the single most common integration task in AI, and it needs no calculus at all.
The two end ordinates are counted once; every interior ordinate is doubled. With $h=2$ the front factor $\frac{h}{2}$ is simply $1$.
§6 — Signed area & area below the axis AHL 5.12
A definite integral returns a signed area: any part of the region below the $x$-axis, where $f(x)<0$, counts as negative. So the raw integral gives the net area, which can differ from the total physical area.
If the curve crosses the axis inside $[a,b]$, find the roots, integrate over each piece separately, and add the absolute values — otherwise the positive and negative parts partly cancel and you understate the area.
§7 — Reverse chain rule & further integrals AHL 5.11
AHL widens the toolkit beyond powers. Two standard results come straight from reversing the derivatives in Differentiation, and one technique undoes the chain rule.
| Integral | Result |
|---|---|
| $\displaystyle\int e^{x}\,dx$ | $e^{x}+C$ — unchanged |
| $\displaystyle\int \frac{1}{x}\,dx$ | $\ln|x|+C$ — the $n=-1$ case the power rule cannot reach |
| $\displaystyle\int \sin x\,dx$ | $-\cos x+C$ |
| $\displaystyle\int \cos x\,dx$ | $\sin x+C$ |
When an integrand is a function raised to a power, multiplied by a multiple of the derivative of its inside, the reverse chain rule (substitution $u=g(x)$) undoes it:
For instance $\dfrac{d}{dx}(x^{2}+1)=2x$, exactly the factor out front, so $\displaystyle\int 2x\,(x^{2}+1)^{3}\,dx=\frac{(x^{2}+1)^{4}}{4}+C$.
§8 — Volumes of revolution AHL 5.12
Spin the region under $y=f(x)$ right around the $x$-axis and it sweeps out a solid of revolution. Picture slicing it into thin discs, each a circle of radius $y$ and area $\pi y^{2}$; stacking them gives the volume.
For $y=\sqrt{kx}$ the square is simply $y^{2}=kx$, which turns the integral into a plain power. The two marks students lose here are forgetting to square $y$ and forgetting the factor $\pi$.
§9 — Kinematics & rate-to-total applications AHL 5.13
This is where AI integration pays off: whenever you know a rate, integrating it over an interval gives the total change. The headline case is motion — velocity is the derivative of displacement, so displacement is the integral of velocity.
The same "rate → total" logic runs through the other AI applications. In economics the marginal cost $C'(x)$ integrates to the total variable cost, with the constant of integration equal to the fixed set-up cost, $C(0)$. And when the rate arrives as a data table rather than a formula, first fit a model — a quadratic regression $R(t)=at^{2}+bt+c$ on the GDC — then integrate it.
§10 — Exam attack plan All sections
| Question cue | What to do | Watch for |
|---|---|---|
| "Find $\int f(x)\,dx$" (indefinite) | Reverse power rule term by term; add $+C$ | Divide by the new power; never drop $+C$ |
| "Evaluate $\int_a^b f\,dx$" | Antiderivative in brackets, upper $-$ lower (or GDC) | $+C$ cancels; substitute the upper limit first |
| "Curve through a point, given $\frac{dy}{dx}$" | Integrate, sub the point to find $C$, then evaluate | Solve for $C$ before the target $x$ |
| "Area between the curve and the $x$-axis" | $\int_a^b f\,dx$; if it dips below, use $\int|f|\,dx$ | Integrate, don't substitute; split at the roots |
| "Estimate the area from a table" | Trapezoidal rule $\frac{h}{2}\big(y_0+y_n+2\sum\text{interior}\big)$ | Interior ordinates doubled; concavity sets over/under |
| "Volume when rotated about the $x$-axis" | $V=\pi\int_a^b y^2\,dx$ | Square $y$ first; keep the $\pi$ outside |
| "Displacement / distance from velocity" | $s=\int v\,dt$; distance $=\int|v|\,dt$ | Signed vs total; split where $v=0$ |
| "Total from a rate or marginal" | Fit a model if given data, then integrate the rate | Fixed cost $=C(0)$; not the final reading |
Worked Example — Displacement, a Boundary Condition & the Trapezoidal Rule
Question (AI HL Paper 2 style — 9 marks)
A drone rises so that its vertical velocity is modelled by $v(t)=3t^{2}-6t+4$ (in m s$^{-1}$), for $0\le t\le 4$ seconds. At $t=0$ the drone is at a height of $s=5$ m.
(a) Find the drone's displacement during the first $4$ seconds.
(b) Find an expression for its height $s(t)$, and hence its height when $t=4$.
(c) The onboard sensor logs the velocity once per second, giving $v=4,\,1,\,4,\,13,\,28$ at $t=0,1,2,3,4$. Use the trapezoidal rule to estimate the displacement, and state, with a reason, whether it over- or under-estimates the true value.
Solution
- (a) Displacement is the integral of velocity: $s=\displaystyle\int_{0}^{4}\left(3t^{2}-6t+4\right)dt=\left[t^{3}-3t^{2}+4t\right]_{0}^{4}$. (M1)(A1) Evaluating, $=\left(64-48+16\right)-0=32$ m. (A1)
- (b) Integrating with a constant, $s(t)=t^{3}-3t^{2}+4t+C$. The boundary condition $s(0)=5$ gives $C=5$. (M1) So $s(t)=t^{3}-3t^{2}+4t+5$ (A1), and $s(4)=64-48+16+5=37$ m. (A1) (Check: the start height $5$ plus the displacement $32$ gives $37$.)
- (c) With $h=1$: $\;s\approx\dfrac{1}{2}\big(4+28+2(1+4+13)\big)=\dfrac{1}{2}(32+36)=34$ m. (M1)(A1) Since $v''(t)=6>0$ the velocity model is concave up, so the straight strip-tops lie above the curve and the rule over-estimates (true value $32$ m). (R1)
Examiner's note: in AI you would normally get part (a) straight from the GDC's $\int$ tool — but write the integral down first, as that earns the method mark. Keep the units (m) and give $3$ s.f. unless told otherwise. Note the wording: this $v$ stays positive, so displacement and distance agree here; if the model dipped below zero you would integrate $|v|$ for the total distance. The trapezoidal estimate ($34$) sitting just above the exact area ($32$) is exactly what a concave-up curve predicts.
Common Student Questions
Do I add $+C$ to a definite integral?
How do I know whether the trapezoidal rule over- or under-estimates?
What is the difference between displacement and distance travelled?
Can I just use my GDC for the area or the definite integral?
Why can't I skip the constant of integration?
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