Integration (AI HL)
Integration is the reverse of differentiation - given a rate of change, it recovers the original quantity. At AI HL this covers a wider set of standard integrals than at SL, and their main application: finding the area enclosed by a curve, the area between two curves, and the volume swept out when a region is rotated about an axis. The trapezoidal rule gives a numerical estimate when an integral can't easily be evaluated exactly.
What the syllabus says
This topic draws on the SL calculus content (examinable at HL) plus the AHL extension points on integration.
| Code | Syllabus content |
|---|---|
| SL5.5 | Introduction to integration as anti-differentiation of functions of the form \(f(x)=ax^n+bx^{n-1}+\dots\), \(n\in\mathbb{Z},\ n\neq-1\). Anti-differentiation with a boundary condition to determine the constant term. Definite integrals using technology. Area of a region enclosed by a curve \(y=f(x)\) and the \(x\)-axis, where \(f(x)>0\). |
| SL5.8 | Approximating areas using the trapezoidal rule, given a table of data or a function, with intervals of equal width. |
| AHL5.11 | Definite and indefinite integration of \(x^n\) where \(n\in\mathbb{Q}\), including \(n=-1\), \(\sin x\), \(\cos x\), \(\dfrac{1}{\cos^2x}\) and \(e^x\). Integration by inspection, or substitution of the form \(\int f(g(x))g'(x)\,dx\). |
| AHL5.12 | Area of the region enclosed by a curve and the \(x\)- or \(y\)-axis in a given interval, including negative integrals. Volumes of revolution about the \(x\)-axis or \(y\)-axis, \(V=\pi\displaystyle\int_a^b y^2\,dx\) or \(V=\pi\displaystyle\int_a^b x^2\,dy\). |
SL5.5 and SL5.8 are core content shared with AA, examinable at AI HL alongside the AHL5.11-5.12 extension.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is an antiderivative?
An antiderivative is a function whose derivative is the one you started with - integration undoes differentiation. Because differentiating a constant gives zero, every indefinite integral needs a "\(+c\)" to represent that unknown constant.
e.g. \(\displaystyle\int x^2\,dx = \dfrac{x^3}{3}+c\), since \(\dfrac{d}{dx}\left(\dfrac{x^3}{3}\right)=x^2\).
What is a definite integral?
A definite integral has an upper and lower limit and evaluates to a single number - the antiderivative at the upper limit minus its value at the lower limit. The constant \(c\) always cancels, so you never need it here.
e.g. \(\displaystyle\int_1^3 2x\,dx = \big[x^2\big]_1^3 = 9-1=8\).
What is the area under a curve?
The area between a curve \(y=f(x)\) and the \(x\)-axis on \([a,b]\) is found by evaluating \(\displaystyle\int_a^b f(x)\,dx\), provided the curve stays above the axis on that interval. If it dips below, the integral gives a negative (signed) value there.
e.g. Area under \(y=x^2\) from \(0\) to \(3\): \(\displaystyle\int_0^3 x^2\,dx = \Big[\dfrac{x^3}{3}\Big]_0^3 = 9\).
What is the area between two curves?
The area enclosed between two curves is found by integrating (top curve minus bottom curve) between their points of intersection. Finding those intersection points is usually the first step, not an afterthought.
e.g. Between \(y=x\) and \(y=x^2\) on \([0,1]\): \(\displaystyle\int_0^1 (x-x^2)\,dx = \Big[\dfrac{x^2}{2}-\dfrac{x^3}{3}\Big]_0^1 = \dfrac12-\dfrac13=\dfrac16\).
What is the trapezoidal rule?
The trapezoidal rule estimates a definite integral by approximating the region under the curve with a series of trapezia instead of finding an exact antiderivative. It's especially useful when a function can't be integrated algebraically, or when you only have a table of values.
e.g. Estimating \(\displaystyle\int_0^2 x^2\,dx\) with \(h=1\): \(\tfrac12[(0+4)+2(1)] = 3\) (true value \(\tfrac83\approx2.67\), so this over-estimates).
Key formulas
A short list of standard integrals and two area/volume formulas cover almost every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The standard integrals, the area and volume-of-revolution formulas, and the trapezoidal rule are all on the formula booklet.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c,\ n\neq-1\) | Power rule for integration | ✓ Yes |
| \(\displaystyle\int \dfrac1x\,dx = \ln|x|+c\) | Reciprocal function | ✓ Yes |
| \(\displaystyle\int e^x\,dx = e^x+c,\quad \int\sin x\,dx=-\cos x+c,\quad\int\cos x\,dx=\sin x+c\) | Standard integrals | ✓ Yes |
| \(\displaystyle\int \dfrac{1}{\cos^2x}\,dx = \tan x+c\) | Standard integral | ✓ Yes |
| \(V=\pi\displaystyle\int_a^b y^2\,dx\) or \(\pi\displaystyle\int_a^b x^2\,dy\) | Volume of revolution | ✓ Yes |
| \(\displaystyle\int_a^b y\,dx \approx \dfrac{h}{2}\big[(y_0+y_n)+2(y_1+\dots+y_{n-1})\big]\) | Trapezoidal rule | ✓ Yes |
Signed area vs total area
A definite integral and the "true" area you can measure with a ruler are not always the same thing.
| Feature | Definite integral (signed area) | Total area |
|---|---|---|
| Region above the axis | Counts as positive | Counts as positive |
| Region below the axis | Counts as negative | Counts as positive (take the magnitude) |
| Curve crosses the axis | Regions can cancel out | Split at each root and add magnitudes |
| Example | \(\int_{-1}^{1}(x^3-x)\,dx = 0\) | \(\int_{-1}^{0}\!|x^3-x|\,dx+\int_{0}^{1}\!|x^3-x|\,dx = \tfrac12\) |
Integration techniques
Recognising the shape of the integrand tells you which standard result or technique to reach for.
Power rule (reversed)
\[\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c\]
Increase the exponent by 1, then divide by the new exponent. Valid for any rational \(n\neq-1\).
✓ In the formula bookletStandard integrals
\[\int e^x\,dx=e^x+c,\quad \int\dfrac1x\,dx=\ln|x|+c\]
Memorise these alongside \(\sin x\), \(\cos x\) and \(\dfrac{1}{\cos^2x}\) - they appear constantly in AI HL papers.
✓ In the formula bookletIntegration by substitution
\[\int f(g(x))\,g'(x)\,dx\]
If the integrand is a function of \(g(x)\) multiplied by \(g'(x)\), substitute \(u=g(x)\) to simplify it into a standard integral.
✓ Named in the formula bookletArea, volume and numerical estimates
Once you can integrate, these three results turn that skill into geometry.
Area under a curve
\[A=\int_a^b f(x)\,dx\]
Only gives the true area directly when \(f(x)>0\) throughout \([a,b]\) - otherwise split at the roots first.
✓ In the formula bookletArea between two curves
\[A=\int_a^b \big(f(x)-g(x)\big)\,dx\]
\(a\) and \(b\) are the \(x\)-coordinates where the curves meet; \(f(x)\) must be the upper curve on that interval.
Not in the formula booklet - derived from the area formulaVolume of revolution
\[V=\pi\int_a^b y^2\,dx\]
Rotating the region under \(y=f(x)\) a full turn about the \(x\)-axis sweeps out this volume; swap \(x\) and \(y\) to rotate about the \(y\)-axis instead.
✓ In the formula bookletTrapezoidal rule
\[\int_a^b y\,dx \approx \dfrac h2\big[(y_0+y_n)+2(y_1+\dots+y_{n-1})\big]\]
If the curve is concave-up, the trapezia sit above it and the rule over-estimates; concave-down under-estimates.
✓ In the formula bookletWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Estimate \(\displaystyle\int_1^5 \dfrac{1}{x}\,dx\) using the trapezoidal rule with 4 intervals.
(a) State your estimate.
(b) State, with a reason, whether it over- or under-estimates the true value.
Worked solution
(a) \(h = 1\); values of \(\tfrac1x\) at \(1,2,3,4,5\): \(1, 0.5, 0.333, 0.25, 0.2.\) M1
\(\int \approx \tfrac12[1.2 + 2(1.083)]\) M1
\(\approx 1.68.\) A1
(b) \(y = \tfrac1x\) is concave up, so trapezia lie above the curve M1 → over-estimate (true \(\ln 5\) A1 \(\approx 1.61\)). A1
The curves \(y = x^2\) and \(y = 2x\) intersect at two points.
(a) Find the \(x\)-coordinates of the intersection points.
(b) Find the area enclosed between the two curves.
Worked solution
(a) \(x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x\) M1
\(= 0, 2.\) A1
(b) \(\int_0^2(2x - x^2)\,dx\) M1
\(= [x^2 - \tfrac{x^3}{3}]_0^2\) A1
\(= 4 - \tfrac83 = \tfrac43\) square units. A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting the "\(+c\)" on an indefinite integral. It's an easy mark to lose - any general antiderivative needs the constant, even if the question doesn't seem to care about it.
- Treating a definite integral as area without checking the sign. If the curve dips below the \(x\)-axis anywhere in the interval, the raw integral gives a signed (net) value, not the true area - split at the roots and take magnitudes first.
- Subtracting the curves the wrong way round. Area between two curves needs (top curve \(-\) bottom curve). Getting the order backwards gives a negative answer that should have been positive.
- Misjudging over- or under-estimate for the trapezoidal rule. The rule over-estimates when the curve is concave-up (trapezia sit above the curve) and under-estimates when it's concave-down - not the other way round.
Using your GDC
Every calculator on the approved list has a numerical integration feature that evaluates a definite integral once you've written the correct expression and limits - useful for checking an algebraic answer, or for finding an integral that can't be done by hand. Look for it under the calculus or math menu (often alongside the numerical derivative tool), enter the function and the limits, and read off the value. It won't set the integral up for you, though - working out what to integrate, and between which limits, is still the part that earns the method marks.
See the full GDC guide for model-specific button sequences across every topic.
Ready to practise properly?
Integration questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between a definite and an indefinite integral?
An indefinite integral gives a general antiderivative plus a constant, like \(\int x^2\,dx = \tfrac{x^3}{3}+c\). A definite integral has limits and gives a number - you evaluate the antiderivative at the upper limit and subtract its value at the lower limit, and the \(+c\) cancels out.
Why doesn't a definite integral always equal the area under a curve?
A definite integral gives signed area - regions below the \(x\)-axis count as negative. If a curve crosses the axis within the interval, the integral gives the net area, not the total area, so you need to split the interval at each root and add the magnitudes to get the true total area.
How do I find the area between two curves?
Find where the curves intersect first - these give the limits of integration. Then integrate (top curve minus bottom curve) between those limits. Getting the order right matters: if you subtract the wrong way round you'll get a negative answer.
Can I use my GDC for integration on the exam?
Yes, on any calculator paper - your GDC can evaluate a definite integral numerically once you've written the correct expression and limits. You still need to set up the integral correctly by hand; the calculator just does the final evaluation.
Sub-topics
Integration broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AI HL syllabus unit, in case you want to keep going.