What is 9.4: Trapezium rule and bounds?

The trapezium rule approximates an integral by replacing the curve over each strip with a straight chord and adding the resulting trapezium areas.

How many marks are 9.4 questions worth?

9.4 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 9.4 appeared in?

We have not yet indexed a verified past-paper appearance for 9.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

9.4

Understand and use numerical integration of functions, including the use of the trapezium rule and estimating the approximate area under a curve and limits that it must lie between.

Draft — not yet indexed

Trapezium rule and bounds

Worked answers and methods for 9.4 on Edexcel A-level Maths 9MA0.

Explanation

The trapezium rule approximates an integral by replacing the curve over each strip with a straight chord and adding the resulting trapezium areas.
For ordinates $y_0,\ldots,y_n$ at equal spacing $h$ , use $\frac{h}{2}[y_0+y_n+2(y_1+\cdots+y_{n-1})]$ and include each interior ordinate twice.
A convex curve with $f''(x)>0$ lies below its chords, so the trapezium rule overestimates its integral; more, narrower strips usually improve the estimate.
Numerical area is not automatically the signed integral: a common error is to leave contributions below the $x$ -axis negative when the question asks for total area.

The trapezium rule joins consecutive ordinates with straight segments; convexity indicates whether this polygonal boundary lies above or below the curve.

Worked example

Use four equal strips and the trapezium rule to estimate $\int_0^2(x^2+1)\,dx$ . State whether the estimate is an overestimate or an underestimate, with a reason.

1.Here $h=0.5$ and the ordinates at $x=0,0.5,1,1.5,2$ are $1,1.25,2,3.25,5$ .
2.Thus the estimate is $\frac{0.5}{2}[1+5+2(1.25+2+3.25)]=0.25(19)=4.75$ .
3.Since $f''(x)=2>0$ , each chord lies above the curve, so the result is an overestimate.

Answer: $4.75$ ; It is an overestimate because $f''(x)=2>0$ , so the curve is convex.

Common mistakes

Don't apply weight two to the endpoints and weight one to the interior ordinates, reversing the trapezium-rule coefficients.
Don't use the number of ordinates as the number of strips, giving the wrong strip width and endpoint weights.

Exam tip

For the trapezium rule, calculate strip width from the interval and use half weight only on the two endpoints.

Worked practice

Tier 1 · Easy

The values of a function are

y=1,3,9

x=0,1,2

respectively. Use two strips and the trapezium rule to estimate

\int_0^2 y\,dx

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$8$	2
Notes
The strip width is $h=1$ . The trapezium rule gives $\frac12[1+9+2(3)]=\frac12(16)=8$ .

(2 marks)

Tier 2 · Standard

The values of a function at

x=0,1,2,3,4

are respectively

2,2.5,4,6.5,10

. Use the trapezium rule to estimate

\displaystyle\int_0^4 f(x)\,\mathrm dx

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$19$	3
Notes
The strip width is $h=1$ . The trapezium rule gives $\dfrac12[2+10+2(2.5+4+6.5)]=\dfrac12[12+26]=19$ .

(3 marks)

Tier 3 · Hard

For

f(x)=\frac1x

1\leq x\leq2

, use four equal strips to find a trapezium-rule estimate. By using left- and right-endpoint rectangles, also give lower and upper bounds for

\int_1^2\frac1x\,dx

. Give decimals to

4

decimal places.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Trapezium estimate $=0.6970$ . $0.6345<\int_1^2\frac1x\,dx<0.7595$ .	6
Notes
The width is $h=0.25$ and the ordinates are $1,\frac45,\frac23,\frac47,\frac12$ . The trapezium estimate is $\frac18[1+\frac12+2(\frac45+\frac23+\frac47)]=\frac{1171}{1680}=0.6970$ . Since $1/x$ is decreasing, right endpoints give the lower sum $\frac14(\frac45+\frac23+\frac47+\frac12)=\frac{533}{840}=0.6345$ , while left endpoints give the upper sum $\frac14(1+\frac45+\frac23+\frac47)=\frac{319}{420}=0.7595$ .

(6 marks)

Tier 1 · Easy

x=0,1,2

, the corresponding ordinates of a curve are

2,k,6

. The trapezium rule with two strips gives an estimate of

10

for the integral from

0

2

. Find

k

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$k=6$	3
Notes
The strip width is $1$ , so the trapezium estimate is $\frac12[2+6+2k]=4+k$ . Equating this to $10$ gives $k=6$ .

(3 marks)

Tier 2 · Standard

The table gives values of

y=\sqrt{x+1}

: for

x=0,0.5,1,1.5,2

, the corresponding values of

y

are

1,1.225,1.414,1.581,1.732

. Obtain a trapezium-rule estimate for

\int_0^2\sqrt{x+1}\,\mathrm dx

. Report the result to

3

decimal places and state, with a reason, whether it is an overestimate or an underestimate.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$2.793$ ; it is an underestimate because $y=\sqrt{x+1}$ is concave on the interval, so its chords lie below the curve.	4
Notes
Here $h=0.5$ . The estimate is $\frac{0.5}{2}[1+1.732+2(1.225+1.414+1.581)]=2.793$ . Also $y''=-1/[4(x+1)^{3/2}]<0$ for $0\leq x\leq2$ , so the graph is concave and each trapezium chord lies below the curve. The estimate is therefore an underestimate.

(4 marks)

Tier 3 · Hard

A continuous curve passes through the points whose coordinates are given by

x=0,1,2,3

and

y=-2,-1,1,4

respectively. Use the trapezium rule to estimate the signed integral from

0

3

. Then, using the straight-line trapezium approximation between each pair of adjacent points, estimate the total area between the curve and the

x

-axis.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Estimated signed integral $=1$ . Estimated total area $=4.5$ square units.	5
Notes
With $h=1$ , the signed estimate is $\frac12[-2+4+2(-1+1)]=1$ . The straight segment from $(1,-1)$ to $(2,1)$ crosses the axis halfway, at $x=1.5$ . The estimated area below the axis is $\frac12(2+1)(1)+\frac12(0.5)(1)=1.75$ . The estimated area above it is $\frac12(0.5)(1)+\frac12(1+4)(1)=2.75$ . Hence the total area is $1.75+2.75=4.5$ square units.

(5 marks)

Tier 2 · Standard

The corresponding values of

x

and

y

are

x=0,0.5,1.5,3

and

y=2,4,3,7

. Using a straight-line trapezium between each pair of adjacent data points, estimate

\int_0^3 y\,\mathrm dx

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$12.5$	4
Notes
The unequal widths are $0.5$ , $1$ and $1.5$ , so the three trapezium areas are $\frac{0.5}{2}(2+4)=1.5$ , $\frac12(4+3)=3.5$ and $\frac{1.5}{2}(3+7)=7.5$ . Their sum is $1.5+3.5+7.5=12.5$ .

(4 marks)

Tier 3 · Hard

x=0,1,2,3,4

, the ordinates of

y=f(x)

are

8,7,9,12,14

and the ordinates of

y=g(x)

are

2,3,4,6,9

. Throughout the interval,

f(x)\geq g(x)

. Use the trapezium rule with four strips to estimate the area between the two curves. Confirm the result by separately estimating

\int_0^4 f(x)\,\mathrm dx

and

\int_0^4 g(x)\,\mathrm dx

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The estimated area between the curves is $20.5$ square units. The separate estimates are $39$ and $18.5$ , whose difference is $20.5$ .	5
Notes
The ordinates of $f-g$ are $6,4,5,6,5$ . With width $1$ , the trapezium estimate is $\frac12[6+5+2(4+5+6)]=20.5$ . Separately, the estimate for $\int_0^4f(x)\,\mathrm dx$ is $\frac12[8+14+2(7+9+12)]=39$ , while that for $\int_0^4g(x)\,\mathrm dx$ is $\frac12[2+9+2(3+4+6)]=18.5$ . Their difference is $39-18.5=20.5$ square units.

(5 marks)

Tier 3 · Hard

A solid of revolution has radius

r

at distance

x

along its axis. For successive unit distances

x=0

through

x=4

, the radii are

1,2,3,2,1

. Using

V=\pi\int_0^4 r^2\,\mathrm dx

, obtain a four-strip trapezium estimate of the volume. Give it exactly in terms of

\pi

and show the transformed ordinates used.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The transformed ordinates are $1,4,9,4,1$ , giving $V\approx18\pi$ .	5
Notes
The integrand is $r^2$ , not $r$ , so the ordinates for the trapezium rule are $1^2,2^2,3^2,2^2,1^2$ , namely $1,4,9,4,1$ . With strip width $h=1$ , $\int_0^4r^2\,\mathrm dx\approx(1/2)[1+2(4+9+4)+1]=18$ . Multiplying by $\pi$ gives the volume estimate $18\pi$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Five function values are equally spaced from

x=0

x=2

in steps of

0.5

. The ordinates at the first, third and fifth points are

2,5,8

. Using only these three ordinates, verify that the two-strip trapezium estimate is

10

. The four-strip estimate using all five ordinates is

9.5

, and the fourth ordinate is twice the second. Find the two missing ordinates.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	The two-strip estimate is $10$ . The missing ordinates are $3$ at $x=0.5$ and $6$ at $x=1.5$ .	5
Notes
For two strips of width $1$ , the estimate is $(1/2)[2+2(5)+8]=10$ . Let the missing ordinates at $x=0.5$ and $x=1.5$ be $u$ and $2u$ . For four strips of width $0.5$ , the estimate is $(0.5/2)[2+2(u+5+2u)+8]=5+1.5u$ . Since this equals $9.5$ , $1.5u=4.5$ and $u=3$ . Thus the two missing ordinates are $3$ and $6$ .

(5 marks)

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