Integration (AI SL)
Integration reverses differentiation - given a gradient function, it recovers the original function, or the accumulated total (like area) that produced it. This topic covers anti-differentiation with the power rule, using a boundary condition to pin down the constant of integration, evaluating definite integrals to find area under a curve, and estimating area numerically with the trapezoidal rule when a formula isn't enough.
What the syllabus says
This topic maps onto two points in the official IB Applications & Interpretation syllabus, both within the Calculus unit.
| Code | Syllabus content |
|---|---|
| SL5.5 | Introduction to integration as anti-differentiation of functions of the form \(f(x)=ax^n+bx^{n-1}+\cdots\), where \(n\in\mathbb{Z}\), \(n\neq-1\). 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. |
SL5.5 is shared content with Analysis & Approaches; SL5.8 (the trapezoidal rule) is specific to Applications & Interpretation.
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 anti-differentiation?
Anti-differentiation is the reverse of differentiation - finding a function \(F(x)\) whose derivative is a given function \(f(x)\). Because differentiating any constant gives zero, the result is always a family of functions, written with a "+C".
e.g. The antiderivative of \(4x\) is \(2x^2+C\), since \(\dfrac{d}{dx}(2x^2)=4x\).
What is a definite integral?
A definite integral is an integral evaluated between two limits, \(\displaystyle\int_a^b f(x)\,dx\) - it gives a single number, often representing an area or an accumulated total, rather than a family of functions.
e.g. \(\displaystyle\int_1^3 3x^2\,dx=\big[x^3\big]_1^3=27-1=26.\)
What does the constant of integration represent?
The "+C" accounts for the fact that many functions share the same derivative - shifting a curve up or down doesn't change its gradient. A boundary condition (a known point on the curve) lets you find the specific value of \(C\).
e.g. If \(\dfrac{dy}{dx}=4x-3\) and the curve passes through \((2,1)\): \(8-6+C=1\Rightarrow C=-1.\)
What is the trapezoidal rule?
The trapezoidal rule is a numerical method for estimating the area under a curve by splitting it into equal-width strips and approximating each strip as a trapezium rather than a true curved region.
e.g. With \(f(0)=2,\ f(1)=4,\ f(2)=5\) and width 1: \(\tfrac12[2+5+2(4)]=\tfrac12(15)=7.5.\)
What is the link between area and the definite integral?
When \(f(x)>0\) between \(x=a\) and \(x=b\), the definite integral \(\displaystyle\int_a^b f(x)\,dx\) gives exactly the area enclosed between the curve and the \(x\)-axis over that interval.
e.g. \(\displaystyle\int_0^4(4x-x^2)\,dx=\dfrac{32}{3}\approx10.7\) is the area between \(y=4x-x^2\) and the \(x\)-axis.
Key formulas
Two ideas 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 trapezoidal rule is given in the formula booklet in full; the integration power rule is treated as a technique rather than a listed formula.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\displaystyle\int ax^n\,dx = \dfrac{ax^{n+1}}{n+1}+C,\ n\neq-1\) | Power rule for integration | Not in booklet - prior knowledge |
| \(\dfrac{h}{2}\big[y_0+y_n+2(y_1+\cdots+y_{n-1})\big]\) | Trapezoidal rule | ✓ Yes |
Indefinite vs definite integral
Both use the same antiderivative - the difference is what you do with it once you have it.
| Feature | Indefinite integral | Definite integral |
|---|---|---|
| Notation | \(\displaystyle\int f(x)\,dx = F(x)+C\) | \(\displaystyle\int_a^b f(x)\,dx\) |
| Result | A family of functions | A single number |
| Needs | A boundary condition to find \(C\) | Limits \(a\) and \(b\) |
| Example | \(\displaystyle\int 4x\,dx = 2x^2+C\) | \(\displaystyle\int_1^3 3x^2\,dx=26\) |
Anti-differentiation
Reverse the power rule term by term, then use any given condition to pin down the constant.
Power rule for integration
\[\int ax^n\,dx=\dfrac{ax^{n+1}}{n+1}+C\]
Increase the exponent by one, then divide by the new exponent.
Constant of integration
Every indefinite integral needs a "+C" - it represents the unknown vertical shift of the original function, since a constant vanishes on differentiation.
Using a boundary condition
Substitute a known point \((x,y)\) into the general antiderivative and solve for \(C\) to find the specific function.
Definite integrals and area
A definite integral only equals an area directly when the curve stays above the \(x\)-axis on that interval.
Evaluate between limits
\[\int_a^b f(x)\,dx=\big[F(x)\big]_a^b=F(b)-F(a)\]
Find the antiderivative, then substitute the upper limit and subtract the lower limit.
Area under a curve
For \(f(x)>0\) on \([a,b]\), the definite integral gives the area between the curve and the \(x\)-axis directly.
Curve below the axis
If \(f(x)<0\) on the interval, the integral comes out negative - take the modulus (absolute value) to get the actual area.
The trapezoidal rule
A way to estimate area numerically, useful whenever you have data rather than (or in addition to) a formula.
Equal-width strips
Split the interval into strips of equal width \(h\), reading off the function value at each dividing point.
The formula
\[\dfrac{h}{2}\big[y_0+y_n+2(y_1+\cdots+y_{n-1})\big]\]
Add the two end values, double all the values in between, and multiply the total by half the strip width.
When to use it
Reach for the trapezoidal rule when you're given a table of values rather than an explicit function, or when the function isn't one you can integrate by hand.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
\(\displaystyle\int_1^3 (x^3-2x)\,dx.\)
(a) Evaluate it.
(b) Evaluate \(\displaystyle\int_1^3 3x^2\,dx.\)
(c) Evaluate \(\displaystyle\int_1^3 (3x^2-2)\,dx.\)
Worked solution
(a) \(=12.\) M1 A1
(b) \(\left[x^3\right]_1^3=27-1\) M1
\(=26.\) A1
(c) \(\left[x^3-2x\right]_1^3=(27-6)-(1-2)\) M1
\(=22.\) A1
Consider \(y=x^2\) and \(y=2x.\)
(a) Find the area enclosed between the two curves.
(b) Find the area enclosed between them for \(2\le x\le3\) (where \(x^2>2x\)).
(c) Hence, find the total area enclosed between the two curves for \(0\le x\le3.\)
Worked solution
(a) Intersect at \(x=0,2;\ \displaystyle\int_0^2 (2x-x^2)\,dx=\dfrac{4}{3}\) M1
\(\approx1.33.\) A1
(b) \(\displaystyle\int_2^3(x^2-2x)\,dx=\left[\tfrac{x^3}{3}-x^2\right]_2^3=0-\left(-\tfrac43\right)=\tfrac43\) M1
\(\approx1.33.\) A1
(c) \(\tfrac43+\tfrac43=\tfrac83\approx2.67.\) 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. Without a boundary condition, \(\displaystyle\int 4x\,dx\) is \(2x^2+C\), not just \(2x^2\) - the constant is part of the answer until you're given a point to solve for it.
- Treating a negative definite integral as the final area. If a curve dips below the \(x\)-axis, the integral comes out negative - take the modulus to report a sensible (positive) area.
- Increasing the exponent but forgetting to divide by the new value. \(\displaystyle\int x^3\,dx=\dfrac{x^4}{4}+C\), not \(x^4+C\) - both steps of the power rule for integration are needed.
- Using the trapezoidal rule when the exact integral is asked for. The trapezoidal rule gives an estimate from data points; if you're given an explicit function and asked to "evaluate" or "find" the integral, integrate it directly instead.
Using your GDC
Every AI SL calculator has a numerical integration feature, so most integration questions can be checked or solved directly on the GDC without finding the antiderivative by hand. Look for a menu item along the lines of "numerical integral" or "fnInt" - enter the function together with the lower and upper limits, and the calculator returns the value of the definite integral. This works even for functions that are awkward or impossible to integrate analytically, which is exactly why AI SL leans on it so heavily.
See the full GDC guide for model-specific button sequences.
Ready to practise properly?
Integration questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What is the difference between an indefinite and a definite integral?
An indefinite integral gives a family of antiderivative functions, written with a "+C". A definite integral has limits and evaluates to a single number - often an area.
How do I find the constant of integration?
You need a boundary condition - a known point the curve passes through. Substitute that point into the general antiderivative and solve for \(C\).
When should I use the trapezoidal rule instead of integrating directly?
Use it when you only have a table of data values rather than a formula for the function, or when the function is too awkward to integrate by hand - the trapezoidal rule estimates the area from equally spaced values.
Can my GDC evaluate a definite integral directly?
Yes - every AI SL calculator has a numerical integration feature that evaluates a definite integral between two limits without you needing to find the antiderivative by hand.
Sub-topics
Integration broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AI SL syllabus unit, in case you want to keep going.