AI SL · 13.5
0 of 41 skills done
The trapezium rule with data, and its error
13.5.8
Screenshot and video to come
Screenshot and video to come
Choose a function and the limits a and b. Change the number of strips n and compare the estimate with the calculator value. Or add the strips one at a time.
| x | y | Times used |
|---|
Both ends once, the middle values twice. The working shows y to 3 decimal places; the estimate uses more.
Change n to see the estimate move.
The graph shows part of \(y=f(x)\). The table gives some values of \(f(x)\). x: 1.5, 2, 2.5, 3, 3.5 y: 1.7, 3.3, 4.2, 3.5, 1.9 Estimate the area between the graph and the x-axis, from x = 1.5 to x = 3.5.
Have a go first. Then open the working one line at a time.
h = 0.5
The x-values are 0.5 apart, so each strip is 0.5 wide. There are 4 strips.
\(\tfrac12(0.5)[(1.700+1.900)+2(3.300+4.200+3.500)]\) = 6.400
Both ends once, the middle values twice. The y-values are given, so there is no table to build.
≈ 6.40
Dashed lines mark the five y-values. Four strips.
The width of a pond is measured every 4 m along its length. The widths are 0, 6.2, 9.8, 11.4, 7.7 and 0 m. Estimate the area of the pond.
Have a go first. Then open the working one line at a time.
h = 4
The widths are measured 4 m apart. Five strips.
\(\tfrac12(4)[(0+0)+2(6.200+9.800+11.40+7.700)]\) = 140.4
Both ends are 0, because the pond has no width there. Only the middle widths are counted, twice.
≈ 140 m²
Area, so square metres.
Use the trapezium rule with 4 strips to estimate \(\int_{2}^{6} \sqrt{3x-5}\,dx\).
Have a go first. Then open the working one line at a time.
\(h=\dfrac{6-2}{4}=1\)
The interval is cut into 4 equal strips.
x: 2, 3, 4, 5, 6; y: 1.000, 2.000, 2.646, 3.162, 3.606
4 strips need 5 y-values. Keep 4 significant figures in the table.
\(\tfrac12(1)[(1.000+3.606)+2(2.000+2.646+3.162)]\) = 10.11
Both ends once, the middle values twice, then ½h.
≈ 10.1
Give 3 significant figures. Check the first value: for x = 2, 3 × 2 − 5 = 1, so y = 1.
A field is drawn on a map. Points on its top edge are (10, 88), (20, 101), (30, 94), (40, 108), (50, 99), (60, 90). Points on its bottom edge are (10, 42), (20, 36), (30, 44), (40, 39), (50, 45), (60, 40). Distances are in metres. Estimate the area of the field.
Have a go first. Then open the working one line at a time.
h = 10
The x-values are 10 m apart. Five strips.
\(\tfrac12(10)[(88.00+90.00)+2(101.0+94.00+108.0+99.00)]\) = 4910
Apply the rule to the top y-values.
\(\tfrac12(10)[(42.00+40.00)+2(36.00+44.00+39.00+45.00)]\) = 2050
The same h, with the bottom y-values.
4910 − 2050 = 2860 ≈ 2860 m²
The field is the area under the top edge minus the area under the bottom edge.
The trapezium rule with 3 strips is used to estimate \(\int_1^4 \frac{6}{x}\,dx\). (a) Find the estimate. (b) Say whether it is an overestimate or an underestimate, and give a reason.
Have a go first. Then open the working one line at a time.
h = 1: \(\tfrac12(1)[(6.000+1.500)+2(3.000+2.000)]\) = 8.750
Part (a): the estimate with 3 strips.
The curve is concave up.
It is steep on the left and flattens to the right.
Each chord lies above the curve.
A straight line between two points of a concave-up curve is above the curve.
The estimate 8.75 is an overestimate.
Chords above the curve mean too much area.
The trapezium rule with 4 strips is used to estimate \(\int_0^5 (5x-x^2)\,dx\). Find the percentage error of the estimate.
Have a go first. Then open the working one line at a time.
h = 1.25; x: 0, 1.25, 2.5, 3.75, 5; y: 0, 4.688, 6.250, 4.688, 0; \(\tfrac12(1.25)[(0+0)+2(4.688+6.250+4.688)]\) = 19.53
The trapezium rule with 4 strips.
Calculator: \(\int_0^5(5x-x^2)\,dx = 20.83\)
Use the definite integral on the calculator.
\(\dfrac{|19.53-20.83|}{20.83}\times100 = 6.25\%\)
The estimate is below the exact value, because the curve is concave down.