Integration (AA SL)
Integration reverses differentiation - given a gradient function, it finds the original function back. This topic covers finding indefinite integrals of polynomials, using a boundary condition to pin down the constant of integration, evaluating definite integrals to find areas under and between curves, and using integration to recover displacement from velocity in kinematics.
What the syllabus says
This topic maps onto three points in the official IB Analysis & Approaches syllabus.
| 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 \(f(x)>0\). |
| SL5.10 | Indefinite integral of \(x^n\) (\(n\in\mathbb{Q}\)), \(\sin x\), \(\cos x\), \(e^x\) and \(\dfrac{1}{x}\). The composites of any of these with the linear function \(ax+b\). Integration by inspection or substitution. |
| SL5.11 | Definite integrals, including an analytical approach: \(\int_a^b g'(x)\,dx = g(b)-g(a)\). Areas of a region enclosed by a curve \(y=f(x)\) and the \(x\)-axis, where \(f(x)\) can be positive or negative, found without the use of technology. Areas between curves. |
Kinematics questions on this topic (displacement recovered from velocity by integration) draw on the related SL5.9 syllabus point.
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 indefinite integral?
An indefinite integral is the reverse of differentiation - it finds a family of functions whose derivative is the given function. Because any constant differentiates to zero, you always add "\(+C\)" to show that family.
e.g. \(\int 3x^2\,dx = x^3+C\), since \(\dfrac{d}{dx}(x^3+C)=3x^2\).
What is a definite integral?
A definite integral has an upper and lower limit and evaluates to a single number, usually representing an area. You find the indefinite integral, then subtract its value at the lower limit from its value at the upper limit.
e.g. \(\int_0^2 2x\,dx = \left[x^2\right]_0^2 = 4-0=4\).
What is the constant of integration?
The constant of integration, \(C\), represents the unknown vertical shift left behind after differentiating. If you're given one point the original function passes through, you can substitute it in to solve for \(C\) exactly.
e.g. If \(\dfrac{dy}{dx}=2x\) and \(y=5\) when \(x=1\): \(y=x^2+C\), so \(5=1+C\Rightarrow C=4\), giving \(y=x^2+4\).
What is the area under a curve?
The area between a curve and the \(x\)-axis, over an interval where the curve stays above the axis, is given directly by the definite integral of the function over that interval.
e.g. Area under \(y=x\) from \(x=0\) to \(x=4\) is \(\int_0^4 x\,dx = \left[\dfrac{x^2}{2}\right]_0^4 = 8\).
What is kinematics using integration?
Since velocity is the derivative of displacement, displacement can be recovered by integrating velocity with respect to time, using an initial condition to fix the constant.
e.g. If \(v(t)=2t\) and \(s(0)=0\), then \(s(t)=t^2\), so \(s(3)=9\) m.
Key formulas
A small set of results cover every integration 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 indefinite integral of \(x^n\) and \(\dfrac{1}{x}\), and the fundamental theorem linking definite integrals to antiderivatives, are all on the official formula booklet.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\displaystyle\int ax^n\,dx = \dfrac{ax^{n+1}}{n+1}+C,\ n\neq-1\) | Indefinite integral of a power | ✓ Yes |
| \(\displaystyle\int \dfrac{1}{x}\,dx = \ln|x|+C\) | Indefinite integral of \(1/x\) | ✓ Yes |
| \(\displaystyle\int_a^b g'(x)\,dx = g(b)-g(a)\) | Fundamental theorem (definite integral) | ✓ Yes |
| Area \(=\displaystyle\int_a^b f(x)\,dx\) | Area under a curve, \(f(x)\ge0\) | Not in booklet - application of the definite integral |
Indefinite vs definite integral
The two forms use the same integration rules, but they answer different questions.
| Feature | Indefinite integral | Definite integral |
|---|---|---|
| Notation | \(\int f(x)\,dx\) | \(\int_a^b f(x)\,dx\) |
| Result | A function, plus \(+C\) | A single number |
| Needs a constant? | Yes - \(+C\) | No - \(C\) cancels |
| Typical use | Recovering the original function | Area, displacement, accumulated change |
Finding the integral
These are the core mechanics you'll use before applying integration to any exam context.
Reverse the power rule
\[\int ax^n\,dx = \dfrac{ax^{n+1}}{n+1}+C\]
Add 1 to the exponent, then divide by the new exponent.
Booklet gives the general form directlyUsing a boundary condition
Integrate first to get the general answer with \(+C\), then substitute the given point into the equation and solve for \(C\).
Evaluating a definite integral
Integrate without \(+C\), write the antiderivative in square brackets with the limits, then substitute the upper limit minus the lower limit.
Applications
Once you can integrate, the same technique answers several different exam contexts.
Area under a curve
Where \(f(x)\ge0\) on \([a,b]\), the area equals \(\int_a^b f(x)\,dx\) directly - no adjustment needed.
Area below the x-axis
Where \(f(x)<0\), the definite integral comes out negative - take the absolute value of that piece before adding it to the total area.
Kinematics
Displacement is the integral of velocity, \(s(t)=\int v(t)\,dt\); total distance travelled is \(\int|v(t)|\,dt\).
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Find \(\displaystyle\int (6x^2-4x+5)\,dx\).
Worked solution
\(\displaystyle\int(6x^2-4x+5)\,dx\) M1
\(=2x^3-2x^2+5x+C.\) A1
A particle has velocity \(v(t) = 3t^2 - 4t\) m/s. At \(t=0\) its displacement is \(s = 2\) m.
(a) Find \(s(t)\).
(b) Find the displacement at \(t = 3\) s.
Worked solution
(a) Displacement is the integral of velocity. \(s(t)=\int(3t^2-4t)\,dt=t^3-2t^2+C.\) M1
\(t^3-2t^2+C.\) A1
At \(t=0,\ s=2\Rightarrow C=2\), so \(s(t)=t^3-2t^2+2.\) M1
(b) \(s(3)=27-18+2\) M1
\(=11\) m. 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 "\(+C\)" on an indefinite integral. Even when the question doesn't obviously need it, the accuracy mark is often tied to including the constant - drop it and you lose the mark.
- Treating a negative definite integral as a negative area. Where the curve dips below the \(x\)-axis the integral comes out negative - area itself is always positive, so take the absolute value.
- Substituting the limits the wrong way round. \(\int_a^b\) means "value at \(b\) minus value at \(a\)" - swapping the order flips the sign of the whole answer.
- Confusing displacement with total distance in kinematics. Displacement is \(\int v(t)\,dt\) and can be negative; total distance is \(\int|v(t)|\,dt\) and is never negative - they only agree when the particle never reverses direction.
Using your GDC
On Paper 2, your GDC can evaluate a definite integral numerically - useful for checking your working, or for integrals that are awkward to do by hand. Look for a numerical integral or "\(\int f(x)\,dx\)" function in the calculus/math menu of your model, enter the function and the limits, and it returns the value directly. Remember that a negative result still means the curve dips below the \(x\)-axis over that interval - take the absolute value if the question is asking for area rather than the signed integral. Paper 1 has no calculator, so you'll still need the integration rules above for that half of the exam.
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 an indefinite and a definite integral?
An indefinite integral gives you a family of functions plus a constant, written with \(+C\), because differentiation loses information about vertical shifts. A definite integral has limits and evaluates to a single number - the constant cancels out.
Why do I need +C for an indefinite integral but not a definite one?
Because any constant differentiates to zero, so infinitely many functions share the same derivative. A definite integral subtracts the antiderivative at the lower limit from the value at the upper limit, so the \(+C\) terms cancel automatically.
How do I find the area between a curve and the x-axis when the curve dips below it?
Split the integral at each \(x\)-intercept, integrate each piece separately, and take the absolute value of any piece where the curve is below the \(x\)-axis before adding the areas together.
Can I use my GDC to evaluate an integral?
Yes, on Paper 2 your GDC can compute a definite integral numerically, which is useful for checking work or for integrals that are awkward to do by hand. Paper 1 expects you to integrate analytically. See the GDC guide for more.
Sub-topics
Integration broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AA SL syllabus unit, in case you want to keep going.