5.8 Trapezoidal ruleIB Maths: Applications and Interpretation HL: Revision notes

The area under a curve (or under a graph made from data) can be estimated by splitting it into vertical strips of equal width $h$ and treating each strip as a trapezoid. A strip between $y_0$ and $y_1$ has area $\frac h2(y_0+y_1)$. Adding all the strips gives the trapezoidal rule: $A\approx\frac h2\left[y_0+y_n+2(y_1+y_2+\dots+y_{n-1})\right],$ where there are $n$ intervals and $n+1$ values. For a function on $a\le x\le b$, $h=\frac{b-a}{n}$. The interior values are used twice (once for each neighbouring strip), the first and last values once.

When the data is given in a table, the values must be equally spaced. Example: widths 12, 30, 44, 50, 36 and 8 metres measured every 20 m give five intervals with $h=20$. Area $\approx\frac{20}{2}[12+8+2(30+44+50+36)]=10\times340=3400$ m². Check units: width (m) times distance (m) gives m². Multiplying by an average depth of 2.5 m gives a volume of 8500 m³.

When you are given a function, choose $n$ and find $h=\frac{b-a}{n}$. Use your GDC table or $y$-values to find each ordinate. For $R(t)=t^2+1$ on $0\le t\le4$ with $n=4$, $h=1$ and the values are $1,2,5,10,17$. Then $A\approx\frac12[1+17+2(2+5+10)]=26$. In examinations you can use the GDC to generate the table, but you should show the substitution into the rule.

The exact area under $y=f(x)$ is found by integration with your GDC (SL 5.5): $\int_0^4(t^2+1)\,dt=25.3$. The trapezoidal estimate 26 is an over-estimate because the curve bends upwards and the chords lie above it. If the curve bends downwards (concave down), as for $y=\sqrt x$, the estimate is an under-estimate. Percentage error $=\frac{|\text{approximate}-\text{exact}|}{|\text{exact}|}\times100$. Using more, narrower strips usually reduces the error.

Data given to a stated accuracy lies in an interval: a reading of 4.2 to 1 decimal place lies between 4.15 and 4.25 (SL 1.6). To find the lower bound for a trapezoidal estimate, use the lower bound of every reading; for the upper bound use the upper bounds. For the rainfall readings 0.4, 4.2, 9.6, 12.4, 8.8, 3.6, 0.8 (mm per hour, every 0.5 hours) the estimate is 19.6 mm, with lower bound 19.45 mm and upper bound 19.75 mm. A value outside these bounds cannot be explained by rounding of the readings alone.

The area under a rate–time graph is the total amount: a rainfall rate in mm per hour multiplied by hours gives mm of rain; a flow rate in litres per minute multiplied by minutes gives litres. Always state the unit and what the area represents, and note that the trapezoidal rule assumes straight lines between readings, so the result is an approximation.