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How do we estimate a definite integral when we cannot find an antiderivative?

Use the trapezoidal rule to approximate definite integrals and areas under curves.

How to approximate a definite integral with the trapezoidal rule, set up the calculation from a formula or a table of data, and judge whether the estimate is too high or too low for TCE Mathematics Methods Unit 4.

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What this dot point is asking

Sometimes you cannot integrate a function exactly, or you are only handed measurements at intervals rather than a formula. The trapezoidal rule gives a numerical estimate of the area under the curve in these situations.

The idea

Divide the interval from x=ax=a to x=bx=b into nn strips of equal width h=bβˆ’anh = \dfrac{b-a}{n}. Instead of finding the exact area of each strip, approximate it by a trapezium whose parallel sides are the two ordinates (function values) at the edges of the strip. Adding the areas of all the trapezia gives the estimate.

The pattern is easy to remember: the first and last ordinates count once, and every interior ordinate counts twice, all multiplied by half the strip width.

Worked example

Working from a table of data

Many TASC questions never give you a formula at all: instead you receive measured values, such as a vehicle's speed recorded every few seconds, or the cross-sectional width of a river at equal spacings. The trapezoidal rule is the natural tool here because it only needs the ordinates, not an antiderivative. Read the spacing of the measurements straight off the table to get hh, count the readings to get n+1n+1 ordinates (so nn strips), then apply the rule exactly as before. When the readings are a rate of change (a speed, a flow rate, a rate of production), the area you are estimating is the accumulated total: distance from a speed, volume from a flow rate, and so on. Always state the units of the answer, which are the units of the ordinate multiplied by the units of the horizontal spacing.

Is the estimate too high or too low?

Whether the trapezoidal rule over or underestimates depends on concavity. For a concave up curve the straight tops of the trapezia sit above the curve, so the rule overestimates the true area. For a concave down curve the trapezia tops sit below the curve, so the rule underestimates. Increasing the number of strips reduces the error in either case.

Comparing with the exact value

When an antiderivative does exist, you can measure the error of the rule directly. For ∫141x dx\displaystyle\int_{1}^{4}\frac{1}{x}\,dx the exact value is ln⁑4βˆ’ln⁑1=ln⁑4β‰ˆ1.3863\ln 4 - \ln 1 = \ln 4 \approx 1.3863. The three-strip trapezoidal estimate 3524β‰ˆ1.4583\dfrac{35}{24} \approx 1.4583 overestimates by about 0.0720.072, consistent with 1x\dfrac{1}{x} being concave up on [1,4][1, 4]. Doubling the number of strips to six roughly quarters the error, because the trapezoidal error falls in proportion to h2h^{2}: halving hh divides the error by about four. This relationship is worth quoting when a question asks how to improve an estimate.

Summary

Choose your strip width h=(bβˆ’a)/nh = (b-a)/n, list the n+1n+1 ordinates, then combine them as h2\dfrac{h}{2} times the first plus the last plus twice everything in between. The rule shines when no antiderivative exists or when only tabulated data is available. Use concavity to predict whether the estimate is high or low, and add more strips for a sharper result.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20244 marksCalculator-assumed. The speed of a car (in m/s) is recorded every 22 seconds: at t=0,2,4,6,8t = 0, 2, 4, 6, 8 the speeds are 0,7,12,15,160, 7, 12, 15, 16. (a) Use the trapezoidal rule to estimate the distance travelled in the first 88 seconds. (b) State, with a reason, whether the estimate is likely to be an over or under-estimate.
Show worked answer β†’

(a) Distance is ∫08v dt\displaystyle\int_{0}^{8} v\,dt. There are 55 ordinates, so n=4n = 4 strips and h=2h = 2.

∫08v dtβ‰ˆh2[y0+y4+2(y1+y2+y3)]=22[0+16+2(7+12+15)].\int_{0}^{8} v\,dt \approx \frac{h}{2}\big[y_{0} + y_{4} + 2(y_{1} + y_{2} + y_{3})\big] = \frac{2}{2}\big[0 + 16 + 2(7 + 12 + 15)\big].

The interior sum is 7+12+15=347 + 12 + 15 = 34, doubled to 6868. Total bracket =0+16+68=84= 0 + 16 + 68 = 84, times 11, so the distance is approximately 8484 m.

(b) The speed values rise but level off, so the curve is concave down. For a concave-down curve the trapezia tops sit below the curve, so the rule under-estimates the true distance. One mark for hh and the setup, two for the value, one for the justified over/under reason.

TCE 20233 marksCalculator-free. Use the trapezoidal rule with three strips to estimate ∫141x dx\displaystyle\int_{1}^{4} \dfrac{1}{x}\,dx, giving your answer as an exact fraction.
Show worked answer β†’

With a=1a = 1, b=4b = 4 and n=3n = 3, the strip width is h=4βˆ’13=1h = \dfrac{4 - 1}{3} = 1. The ordinates of f(x)=1xf(x) = \dfrac{1}{x} at x=1,2,3,4x = 1, 2, 3, 4 are 1,12,13,141, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}.

∫141x dxβ‰ˆ12[1+14+2(12+13)]=12[54+2β‹…56]=12[54+53].\int_{1}^{4}\frac{1}{x}\,dx \approx \frac{1}{2}\left[1 + \frac{1}{4} + 2\left(\frac{1}{2} + \frac{1}{3}\right)\right] = \frac{1}{2}\left[\frac{5}{4} + 2\cdot\frac{5}{6}\right] = \frac{1}{2}\left[\frac{5}{4} + \frac{5}{3}\right].

Common denominator: 54+53=15+2012=3512\dfrac{5}{4} + \dfrac{5}{3} = \dfrac{15 + 20}{12} = \dfrac{35}{12}, so the estimate is 12Γ—3512=3524\dfrac{1}{2}\times\dfrac{35}{12} = \dfrac{35}{24}. Marks for hh, the weighted bracket, and the exact fraction.

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