Numerical Integration (AI HL)
Some functions can't be integrated by any algebraic technique you know, and some "functions" are really just a table of measurements with no formula at all. The trapezoidal rule gets you a usable estimate of the area under the curve in either case, by replacing the true curve with a series of straight-line strips you can add up by hand. It's part of the broader Numerical Methods topic.
19 questions on this sub-topic.
The trapezoidal rule
Covered under IB syllabus reference SL5.8. The rule is given in the formula booklet, so you don't need to memorise it - you need to build the ordinates correctly and substitute them in the right places.
General rule (n strips)
\(\displaystyle\int_a^b y\,dx \approx \frac{h}{2}\Big[(y_0+y_n) + 2(y_1+y_2+\cdots+y_{n-1})\Big]\)
where \(h = \dfrac{b-a}{n}\) is the strip width and \(y_0, y_1, \ldots, y_n\) are the function values at each of the \(n+1\) equally spaced x-values.
Over- or under-estimate?
Concave up \(\Rightarrow\) over-estimate.
Concave down \(\Rightarrow\) under-estimate.
The straight-line tops of the trapezia sit above a curve that bulges downward (concave up), and below a curve that bulges upward (concave down) - so the estimate leans the same way each time.
Need the wider numerical-methods picture, including your GDC's built-in solvers? See Numerical Methods.
Worked examples
Estimate \(\displaystyle\int_0^2 x^2\,dx\) using the trapezoidal rule with 2 intervals.
Worked solution
\(h = 1\); \(y = x^2\) gives \(0, 1, 4.\) M1
\(\int \approx \tfrac{h}{2}[y_0 + 2y_1 + y_2] = \tfrac12[0 + 2 + 4].\) M1
\(= 3.\) A1
exact \(\tfrac83 \approx 2.67\); estimate is an over-estimate (concave up). A1
Estimate \(\displaystyle\int_0^4 x^2\,dx\) using the trapezoidal rule with 4 intervals.
Worked solution
\(h = 1\); values of \(x^2\): \(0,1,4,9,16\); \(\int \approx \tfrac12[0 + 16 + 2(1+4+9)]\) M1
A1
\(= 22.\) A1
Use your GDC to solve \(2^x = 3x\) for the larger positive root, to 3 s.f.
Worked solution
Graph \(y = 2^x\) and \(y = 3x.\) M1
Intersections near \(x \approx 0.458\) A1
and \(x \approx 3.31.\) A1
Common mistakes
- Doubling the wrong ordinates. Only the interior y-values \(y_1\) through \(y_{n-1}\) get doubled - the two end values \(y_0\) and \(y_n\) are each counted once. Doubling everything (or nothing) is the single most common slip.
- Using the wrong number of strips. "4 intervals" means 5 ordinates \((y_0,\ldots,y_4)\), not 4. Miscounting here throws off both \(h\) and the sum.
- Not commenting on over/under-estimate when asked. If a question asks whether the estimate is too high or too low, you need the shape of the curve (concave up or down) over that interval, not just the numerical answer.
Ready to practise properly?
20 numerical-integration questions, marked instantly like the real exam.
Quick answers
What is the trapezoidal rule used for?
It estimates the area under a curve (a definite integral) by splitting the region into equal-width strips and treating each strip as a trapezium: \(\int_a^b y\,dx \approx \tfrac{h}{2}[(y_0+y_n)+2(y_1+\cdots+y_{n-1})]\).
Does the trapezoidal rule over-estimate or under-estimate the true area?
It over-estimates when the curve is concave up and under-estimates when the curve is concave down, because the straight-line tops of the trapezia sit on the outside or inside of the curve accordingly.