Edexcel International A Level Maths · P2 · Pure Mathematics 2
Definite integration, areas and the trapezium rule
28 free, high-quality past-paper-style questions on Definite integration, areas and the trapezium rule for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 238 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-8-B-01P2 · ASEasy5 marks
Use algebraic integration to find the exact value of each of the following. Show your working.
(a)
∫12 (6x2 − 4x + 3) dx
(3)
(b)
∫14 3/√x dx
(2)
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Answer
Mark
Notes
Part (a)
raises the power of at least one term by one
M1
[2x3 − 2x2 + 3x]12, all correct
A1
(16 − 8 + 6) − (2 − 2 + 3) = 11
A1
Part (b)
writes 3x−½ and integrates to 6x½
M1
[6√x]14 = 12 − 6 = 6
A1
Common error: Ignoring the lower limit. Here F(1) = 3 in part (a), not 0: always substitute both limits and subtract, upper minus lower.
Figure 1 shows part of the curve C with equation y = 2 + x − x2. C crosses the y-axis at the point A and the positive x-axis at the point B. The shaded region R is bounded by C, the positive x-axis and the positive y-axis.
(a)
Find the coordinates of A and of B.
(2)
(b)
Use algebraic integration to find the exact area of R.
(3)
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Answer
Mark
Notes
Part (a)
solves 2 + x − x2 = 0: (2 − x)(1 + x) = 0
M1
A(0, 2) and B(2, 0)
A1
Part (b)
raises the power of at least one term by one
M1
[2x + ½x2 − ⅓x3]02, all correct
A1
4 + 2 − 8/3 = 10/3
A1
Common error: Using x = −1 as the lower limit because it is a root of 2 + x − x2 = 0. Figure 1 shows that R stops at the y-axis, so the limits are 0 and 2, not the two roots.
Figure 1 shows the cross-section of a straight river, 12 m wide, between the banks P and Q. The depth of the water was measured at points 2 m apart across the surface, and the depths in metres are shown on the figure. The depth is zero at both banks.
(a)
Use the trapezium rule with all the depths shown to estimate the area of the cross-section of the water.
(4)
(b)
The water flows through this cross-section at an average speed of 0.75 m s−1. Estimate the volume of water that passes through the cross-section each second.
(1)
(c)
State how the depth measurements could be changed to give a more reliable estimate in part (a).
(1)
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Answer
Mark
Notes
Part (a)
7 ordinates, so 6 strips of width h = 2
B1
correct structure: ½h{(first + last) + 2(sum of the others)}
measure the depth at more points, closer together (more, narrower strips), because the river bed between measurements is curved, not straight
B1
Exam tip: Count strips, not ordinates. Seven depths (including the two zeros at the banks) make six strips, so h = 12 ÷ 6 = 2. The zero end values still belong in the formula.
Figure 1 shows part of the curve C with equation y = 1 + 8/x2, x > 0. The points P and Q lie on C and have x-coordinates 1 and 4 respectively. The shaded region R is bounded by C, the x-axis and the lines x = 1 and x = 4. The straight line PQ is also shown.
(a)
Find the y-coordinates of P and Q.
(1)
(b)
Use algebraic integration to find the area of R.
(3)
(c)
Find the area of the trapezium bounded by the line PQ, the x-axis and the lines x = 1 and x = 4. Explain, with reference to Figure 1, why this area is greater than the area of R.
(2)
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Answer
Mark
Notes
Part (a)
P(1, 9) and Q(4, 1.5), both correct
B1
Part (b)
writes 8x−2 and integrates it to −8x−1
M1
[x − 8/x]14 = (4 − 2) − (1 − 8)
M1
area = 9
A1
Part (c)
½ × 3 × (9 + 1.5) = 15.75 ft from their (a)
B1
C bends upwards between P and Q, so the straight line PQ lies above the curve and the trapezium includes extra area that is not part of R
B1
Exam tip: The trapezium in part (c) is the trapezium rule with a single strip. The gap between the chord and the curve in Figure 1 is exactly the error of that estimate, which is why one strip is so poor here.
Figure 1 shows the line with equation y = 3x + 2. The region under the line, above the x-axis and between the lines x = 0 and x = 4 is split by the line x = 8/3 into the two shaded regions R1 and R2.
(a)
Find the total area of R1 and R2 by using the formula for the area of a trapezium.
(2)
(b)
Verify your answer to part (a) by evaluating ∫04 (3x + 2) dx.
(2)
(c)
Show that R1 and R2 have equal areas.
(2)
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Answer
Mark
Notes
Part (a)
parallel sides are the heights 2 and 14 at x = 0 and x = 4
M1
½ × 4 × (2 + 14) = 32
A1
Part (b)
[(3/2)x2 + 2x]04
M1
24 + 8 − 0 = 32
A1
Part (c)
area of R1 = [(3/2)x2 + 2x]08/3 = (3/2)(64/9) + 16/3
M1
= 32/3 + 16/3 = 16, which is half of 32, so R2 also has area 16 cso
A1
Exam tip: For a straight-line graph the definite integral and the trapezium formula give the same area. The split is not at the midpoint x = 2 because the right-hand part of the region is taller.
Find the x-coordinates of the points where C meets the x-axis.
(2)
(b)
Evaluate ∫03 (x2 − 4x + 3) dx.
(2)
(c)
The region R1 is bounded by C and the coordinate axes, and the region R2 is bounded by C and the x-axis. Find the total area of R1 and R2, and explain why your answer to part (b) is not this total.
(3)
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Answer
Mark
Notes
Part (a)
(x − 1)(x − 3) = 0
M1
x = 1 and x = 3
A1
Part (b)
[x3/3 − 2x2 + 3x]03
M1
(9 − 18 + 9) − 0 = 0
A1
Part (c)
∫01 = 1/3 − 2 + 3 = 4/3, the area of R1
M1
∫13 = 0 − 4/3 = −4/3, so R2 has area 4/3 and the total is 8/3
A1
R2 lies below the x-axis, so its integral is negative and cancels the positive integral for R1 in part (b)
B1
Exam tip: A definite integral of 0 is a warning sign, not an area. Split the integral at every point where the curve crosses the axis.
The curve C has equation y = x2 − 6x + 10 and the line L has equation y = x + 4. C and L meet at the points A and B, where A has the smaller x-coordinate.
Figure: the curve C, the line L and the finite region between them
(a)
Find the coordinates of A and B.
(3)
(b)
Find the exact area of the finite region bounded by C and L.
Exam tip: Check which graph is on top with a test value between the meeting points: at x = 2 the line gives 6 and the curve gives 2, so the line is on top and the integrand is line − curve.
The integral I = ∫02 3x dx is to be estimated numerically, because 3x cannot be integrated with the methods of this unit. Values of 3x that are not exact are rounded to 4 decimal places.
x
0
0.5
1
1.5
2
3x
1
3
9
Figure: y = 3ˣ for 0 ≤ x ≤ 2 with the ordinates used in the trapezium rule
(a)
Complete the table.
(1)
(b)
Use the trapezium rule with all the values in the completed table to find an estimate for ∫02 3x dx, giving your answer to 3 significant figures.
(3)
(c)
State, with a reason, whether your answer to part (b) is an overestimate or an underestimate.
(2)
(d)
Using your answer to part (b), find an estimate for ∫02 (3x + 1 + 4x) dx.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
1.7321 at x = 0.5 and 5.1962 at x = 1.5, both correct
B1
Part (b)
h = 0.5
B1
½ × 0.5 × {(1 + 9) + 2(1.7321 + 3 + 5.1962)}
M1
7.46
A1
Part (c)
the curve y = 3x bends upwards (convex) on 0 ≤ x ≤ 2, so each chord lies above the curve
B1
so it is an overestimate (dependent on the reason)
A1
Part (d)
3x + 1 = 3 × 3x, so the estimate is 3 × 7.46… + [2x2]02 = 22.39… + 8
M1
30.4 ft from their (b)
A1
Exam tip: A ‘using your answer’ part never needs a new table. Constant multiples and added polynomial terms can be dealt with separately: integrate the polynomial exactly and scale the estimate.
Write (x√x + 6)/√x in the form xp + 6xq, stating the values of p and q.
(2)
(b)
Hence show that ∫28 (x√x + 6)/√x dx = 30 + 12√2.
(4)
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Answer
Mark
Notes
Part (a)
p = 1
B1
q = −½
B1
Part (b)
integrates x + 6x−½, raising a power by one
M1
[½x2 + 12x½]28
A1
substitutes the limits: (32 + 12√8) − (2 + 12√2)
M1
uses √8 = 2√2: 30 + 24√2 − 12√2 = 30 + 12√2 cso
A1
Exam tip: In a ‘show that’ with surds, write the unsimplified substitution and the surd simplification (√8 = 2√2) as separate lines. Every step must be visible.
The curve C has equation y = 3√x, x ≥ 0, and the line L has equation y = 18 − 3x. The finite region R is bounded by C, L and the y-axis.
Figure: the curve C, the line L and the region R
(a)
Show that the point P(4, 6) lies on both C and L.
(2)
(b)
Use algebraic integration to find the area of R.
(5)
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Answer
Mark
Notes
Part (a)
C: 3√4 = 6
B1
L: 18 − 3 × 4 = 6
B1
Part (b)
area = ∫04 ((18 − 3x) − 3x½) dx, line minus curve
M1
integrates to 18x − (3/2)x2 − 2x3/2
A1
uses the limits 0 and 4
M1
72 − 24 − 16
A1
area = 32
A1
Common error: Integrating only the curve. R is bounded above by the line and below by the curve, so the integrand is line − curve; the area under the curve alone, 16, is the wrong region.
The function f is defined by f(x) = x + 4/x2, x > 0. The region S lies under the graph of f, above the x-axis, between the lines x = 1 and x = 3. Values of f at equally spaced points, rounded to 4 decimal places where necessary, are listed below.
x
1
1.5
2
2.5
3
f(x)
5
3
3.14
3.4444
Figure: the region S with the ordinates used in the trapezium rule
(a)
Complete the table.
(1)
(b)
Use the trapezium rule with all the values in the table to find an estimate for ∫13 (x + 4/x2) dx, giving your answer to 3 significant figures.
(3)
(c)
Use algebraic integration to find the exact value of ∫13 (x + 4/x2) dx.
(3)
(d)
Hence find the percentage error in using your answer to part (b) as the area of S.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
at x = 1.5: 1.5 + 4/2.25 = 3.2778
B1
Part (b)
h = 0.5
B1
0.25 × {(5 + 3.4444) + 2(3.2778 + 3 + 3.14)}
M1
6.82
A1
Part (c)
integrates 4x−2 to −4x−1
M1
[½x2 − 4/x]13 = (9/2 − 4/3) − (1/2 − 4)
M1
20/3
A1
Part (d)
(6.82 − 20/3) ÷ 20/3 × 100
M1
2.3% (accept 2.30%)
A1
Exam tip: Percentage error divides by the exact value, not by the estimate. Find the exact value first if the question gives you a way to do it.
Figure 1 shows the curves C1 and C2 with equations C1: y = x2 − 4x + 6 and C2: y = 6 + 2x − x2. The curves meet at the point A on the y-axis and at the point B. The shaded region R is enclosed by C1 and C2.
Exam tip: Between two curves, subtract first and integrate once: upper − lower = 6x − 2x2, and the constant terms cancel. Figure 1 shows that C2 is the upper curve between A and B.
Figure 1 shows the curve C with equation y = x3 − x2 − 6x. C crosses the x-axis at the point P, at the origin O and at the point Q. The shaded region R1 is bounded by C and the x-axis between P and O; the shaded region R2 is bounded by C and the x-axis between O and Q.
(a)
Factorise x3 − x2 − 6x completely, and hence find the coordinates of P and Q.
(2)
(b)
Find the area of R1.
(3)
(c)
Find the total area of the two shaded regions.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
x(x − 3)(x + 2)
M1
P(−2, 0) and Q(3, 0)
A1
Part (b)
integrates to x4/4 − x3/3 − 3x2
M1
0 − (4 + 8/3 − 12)
M1
area of R1 = 16/3
A1
Part (c)
∫03 (…) dx = 81/4 − 9 − 27 = −63/4
M1
area of R2 = 63/4, as the region is below the x-axis
A1
total = 16/3 + 63/4 = 253/12
A1
Common error: Integrating from −2 to 3 in one step. That gives 16/3 − 63/4 = −125/12, because the part of the integral for R2 is negative. Split the integral at O and add the sizes of the two areas.
Figure 1 shows the curve with equation y = √(16 − x2), 0 ≤ x ≤ 4, which is a quarter of the circle x2 + y2 = 16. The region R is bounded by the curve and the coordinate axes. Four trapezium strips of equal width are also drawn. Values of y are given to 4 decimal places where necessary.
x
0
1
2
3
4
y
4
0
(a)
Complete the table.
(2)
(b)
Use the trapezium rule with the four strips shown in Figure 1 to estimate the area of R, giving your answer to 4 significant figures.
(3)
(c)
Find the percentage error in your estimate, giving your answer to 2 significant figures.
(2)
(d)
Explain, with reference to Figure 1, why the trapezium rule gives an underestimate here.
the curve bends downwards (it is concave), so the straight top of each strip lies below the arc
B1
each strip misses a thin sliver of R, so the estimate is too small dependent on the reason
A1
Exam tip: When the exact area is known from geometry, as for a quarter circle (¼πr²), the percentage error of a trapezium estimate can be found without any integration.
Figure 1 shows part of the curve C with equation y = a√x + bx, x ≥ 0, where a and b are constants. C passes through the points P and Q, whose coordinates are shown. The shaded region R is bounded by C, the x-axis and the lines x = 1 and x = 4.
(a)
Use the coordinates of P and Q to find the value of a and the value of b.
(3)
(b)
Use algebraic integration to find the exact area of R.
(4)
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Answer
Mark
Notes
Part (a)
substitutes both points: a + b = 5 and 2a + 4b = 6
M1
solves the simultaneous equations
M1
a = 7, b = −2
A1
Part (b)
raises the power of at least one term by one
M1
[(14/3)x3/2 − x2]14, all correct ft from their a, b
A1
(14/3 × 8 − 16) − (14/3 − 1)
M1
area = 53/3
A1
Common error: Substituting Q as a × 4 + 4b = 6. The term is a√x, and √4 = 2, so the equation from Q is 2a + 4b = 6.
Figure 1 shows the cross-section of a tunnel. The floor lies along the x-axis and the roof is modelled by the curve with equation y = 5 − 5x2/16, where x and y are in metres and O is the middle of the floor. The rectangle shows the cross-section of a vehicle, 4 m wide and 3.5 m high, standing centrally on the floor.
(a)
Find the width of the tunnel at floor level.
(2)
(b)
Show that the vehicle fits inside the tunnel.
(2)
(c)
Use algebraic integration to find the area of the cross-section of the tunnel.
(3)
(d)
Find the area of the cross-section of the tunnel that is not occupied by the vehicle.
(1)
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Answer
Mark
Notes
Part (a)
5 − 5x2/16 = 0, so x2 = 16
M1
x = ±4, width = 8 m
A1
Part (b)
at x = ±2: y = 5 − 20/16 = 3.75
M1
3.75 > 3.5, and the roof is higher still for −2 < x < 2, so the top corners, and the whole top of the vehicle, are below the roof
A1
Part (c)
∫−44 (5 − 5x2/16) dx
M1
[5x − 5x3/48]−44
A1
(20 − 20/3) − (−20 + 20/3) = 80/3 m2 (26.7 m2)
A1
Part (d)
80/3 − 4 × 3.5 = 38/3 m2 (12.7 m2) ft from their (c)
B1
Exam tip: For a region symmetrical about the y-axis you may find ∫04 (…) dx and double it. Fewer negative numbers to substitute means fewer sign slips.
Figure 1 shows two copies of the curve y = x√x for 0 ≤ x ≤ 4. In panel (a) the area under the curve is approximated by 2 trapezium strips, and in panel (b) by 4 trapezium strips. Values of y are given to 4 decimal places where necessary.
x
0
1
2
3
4
y
0
1
2.8284
5.1962
8
(a)
Use the 2 strips in panel (a) to estimate ∫04x√x dx, giving your answer to 4 significant figures.
(2)
(b)
Use the 4 strips in panel (b) to find a second estimate, giving your answer to 4 significant figures.
(2)
(c)
Use algebraic integration to find the exact value of ∫04x√x dx.
(3)
(d)
Find the percentage error in each estimate, giving your answers to 2 significant figures, and comment on the effect of doubling the number of strips.
both are overestimates; doubling the number of strips cut the error to roughly a quarter
B1
Exam tip: In Figure 1 the straight tops of the strips lie above the curve in both panels: y = x3/2 bends upwards. More strips hug the curve more closely, so the overestimate shrinks.
The curve C has equation y = f(x), where f(x) = x3 − 3x2 + 4. The line L has equation y = 4.
Figure: the curve C, the line L and the regions R₁ and R₂
(a)
Show that (x + 1) is a factor of f(x), and hence factorise f(x) completely.
(3)
(b)
Explain why C touches the x-axis at the point (2, 0).
(1)
(c)
The finite region R1 is bounded by C and the x-axis. Find the area of R1.
(5)
(d)
L meets C at the point (0, 4) and at the point Q. The finite region R2 is bounded by C and L. Find the coordinates of Q, and show that R2 has the same area as R1.
(4)
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Answer
Mark
Notes
Part (a)
f(−1) = −1 − 3 + 4 = 0, so (x + 1) is a factor
B1
divides to obtain x2 − 4x + 4
M1
f(x) = (x + 1)(x − 2)2
A1
Part (b)
(x − 2) is a repeated factor, so x = 2 is a repeated root and C meets the axis there without crossing it
Exam tip: A repeated factor such as (x − 2)² means the curve touches the axis. When you need limits for an area, the touching point and the crossing points all come from the factorised form.
The curve C has equation y = √x(4 − x), 0 ≤ x ≤ 4. The region R is bounded by C and the x-axis. The table shows some values of y, given to 4 decimal places where appropriate.
x
0
1
2
3
4
y
0
3
0
Figure: the curve C with the four trapezium strips
(a)
Complete the table.
(2)
(b)
Use the trapezium rule with all the values in the table to estimate the area of R, giving your answer to 3 significant figures.
(3)
(c)
Use algebraic integration to find the exact area of R.
(4)
(d)
Find the percentage error in your estimate from part (b), giving your answer to 3 significant figures.
(2)
(e)
Explain, with reference to the shape of C, why the trapezium rule gives an underestimate here, and state one way of reducing the error.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
x = 2: 2√2 = 2.8284
B1
x = 3: √3 = 1.7321
B1
Part (b)
h = 1
B1
½ × 1 × {(0 + 0) + 2(3 + 2.8284 + 1.7321)}
M1
7.56
A1
Part (c)
expands to 4x½ − x3/2
M1
integrates to (8/3)x3/2 − (2/5)x5/2
A1
substitutes x = 4: 64/3 − 64/5
M1
128/15
A1
Part (d)
(128/15 − 7.56…) ÷ 128/15 × 100
M1
11.4%
A1
Part (e)
C bends downwards (it is concave) for 0 < x < 4, so each chord lies below the curve and part of R is missed
B1
use more strips / a smaller strip width
B1
Exam tip: Write a root times a bracket as separate powers before integrating: √x(4 − x) = 4x^(1/2) − x^(3/2). Multiplying the integrals of √x and (4 − x) does not work.
The curve C has equation y = x2 − 2x + 5. The point P(3, 8) lies on C, and the line T is the tangent to C at P. The finite region R is bounded by C, T and the y-axis.
Figure: the curve C, the tangent T at P and the region R
(a)
Verify that P lies on C, and show that T has equation y = 4x − 4.
(4)
(b)
Use algebraic integration to show that the area of R is 9.
(5)
(c)
The line x = k, where 0 < k < 3, divides R into two regions of equal area. Show that 2k3 − 18k2 + 54k − 27 = 0.
(2)
(d)
By first showing that the equation can be written as 2(k − 3)3 + 27 = 0, find the value of k, giving your answer to 3 significant figures.
Figure 1 shows the curves C1: y = 4√x and C2: y = x2/2, x ≥ 0, which meet at the origin O and at the point P. The line y = 8 passes through P, and together with the coordinate axes and the line x = 4 it forms a rectangle. The curves divide the rectangle into the three regions R1, R2 and R3 shown.
(a)
Find the coordinates of P.
(3)
(b)
Use algebraic integration to show that the area of R2 is 32/3.
(4)
(c)
Show that R1, R2 and R3 all have the same area.
(2)
(d)
The line x = k, where 0 < k < 4, divides R2 into two parts of equal area. Show that k3 − 16k3/2 + 32 = 0.
(2)
(e)
By using the substitution u = k3/2, find the value of k, giving your answer to 3 significant figures.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
4√x = x2/2, so 8√x = x2
M1
x3/2 = 8, so x = 4
A1
P(4, 8)
A1
Part (b)
area = ∫04 (4x½ − x2/2) dx
M1
[(8/3)x3/2 − x3/6]04
A1
(8/3)(8) − 64/6
M1
= 64/3 − 32/3 = 32/3 cso
A1
Part (c)
area of R3 = ∫04x2/2 dx = 64/6 = 32/3
B1
area of R1 = 4 × 8 − 32/3 − 32/3 = 32/3
B1
Part (d)
[(8/3)x3/2 − x3/6]0k = 16/3
M1
multiplies by 6: 16k3/2 − k3 = 32, so k3 − 16k3/2 + 32 = 0 cso
A1
Part (e)
u2 − 16u + 32 = 0
M1
u = 8 ± 4√2; k < 4 needs u < 8, so u = 8 − 4√2
A1
k = (8 − 4√2)2/3 = 1.76
A1
Exam tip: Some regions are easier as a rectangle minus pieces you can integrate. Here R1 needs no new integral: subtract R2 and R3 from the 4 by 8 rectangle.
Figure 1 shows the curve C with equation y = x3 − 5x2 + 6x and the line L with equation y = 2x. C and L meet at the origin O and at the points A and B. The shaded region R1 is bounded by C and L between O and A; the shaded region R2 is bounded by C and L between A and B.
(a)
Show that the x-coordinates of the points where C and L meet satisfy x(x − 1)(x − 4) = 0, and hence find the coordinates of A and B.
(4)
(b)
Find the area of R1.
(3)
(c)
Find the area of R2.
(3)
(d)
Evaluate ∫04 (x3 − 5x2 + 4x) dx and explain why it is not the total area of the shaded regions.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
x3 − 5x2 + 6x = 2x, so x3 − 5x2 + 4x = 0
M1
x(x2 − 5x + 4) = x(x − 1)(x − 4) = 0
A1
A(1, 2)
A1
B(4, 8)
A1
Part (b)
∫01 (x3 − 5x2 + 4x) dx, curve minus line
M1
[x4/4 − 5x3/3 + 2x2]01
A1
1/4 − 5/3 + 2 = 7/12
A1
Part (c)
∫14 (2x − (x3 − 5x2 + 6x)) dx, line minus curve
M1
= −[x4/4 − 5x3/3 + 2x2]14 = −(−32/3 − 7/12)
M1
area of R2 = 45/4
A1
Part (d)
−32/3
B1
it equals area R1 − area R2: between A and B the curve is below the line, so the integrand is negative there and that part cancels some of R1
B1
Exam tip: The area between two graphs is ∫(upper − lower) dx, even where part of the region dips below the x-axis, as R2 does in Figure 1. Only the order of the two graphs matters, not the axis.
Figure 1 shows the curve with equation y = 4x and the line with equation y = 3x + 1. The curve and the line meet at the points A and B, where A lies on the y-axis and B has x-coordinate 1. The shaded region R is bounded by the curve and the line. Values of 4x that are not exact are given to 4 decimal places.
x
0
0.25
0.5
0.75
1
4x
1
2
4
(a)
Show that A and B both lie on the curve and on the line, and state their coordinates.
(2)
(b)
Complete the table.
(1)
(c)
Use the trapezium rule with all the values in the table to estimate ∫01 4x dx, giving your answer to 3 significant figures.
(3)
(d)
Hence estimate the area of R, giving your answer to 2 decimal places.
(3)
(e)
State, with a reason, whether your answer to part (d) is an overestimate or an underestimate of the area of R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
40 = 1 = 3 × 0 + 1, so A(0, 1)
B1
41 = 4 = 3 × 1 + 1, so B(1, 4)
B1
Part (b)
1.4142 at 0.25 and 2.8284 at 0.75, both correct
B1
Part (c)
h = 0.25
B1
½ × 0.25 × {(1 + 4) + 2(1.4142 + 2 + 2.8284)}
M1
2.19
A1
Part (d)
area of R = ∫01 (3x + 1) dx − ∫01 4x dx
M1
[(3/2)x2 + x]01 = 2.5
A1
2.5 − 2.1857 = 0.31 ft from their (c)
A1
Part (e)
y = 4x bends upwards (convex), so the trapezium rule overestimates ∫01 4x dx
B1
the area of R is found by subtracting this estimate from the exact 2.5
M1
subtracting too much gives an underestimate of the area of R
A1
Common error: Saying “the trapezium rule overestimates, so the area of R is overestimated”. Here the estimate is subtracted, so an overestimate of the integral becomes an underestimate of R.
Figure 1 shows the curve C with equation y = 4/x2, x > 0, and the line L with equation y = 5 − x. C and L meet at the point A, whose coordinates are shown, and at the point B. The shaded region R is bounded by C and L.
(a)
Show that the x-coordinates of A and B satisfy x3 − 5x2 + 4 = 0.
(2)
(b)
Using the x-coordinate of A, find the exact x-coordinate of B.
(3)
(c)
Use algebraic integration to show that the area of R is 8√2 − 13/2.
(6)
Show mark scheme
Answer
Mark
Notes
Part (a)
4/x2 = 5 − x
M1
multiplies by x2: 4 = 5x2 − x3, so x3 − 5x2 + 4 = 0 cso
A1
Part (b)
divides by (x − 1): (x − 1)(x2 − 4x − 4)
M1
x = (4 ± √32)/2 = 2 ± 2√2
M1
B is to the right of A with x > 0, so x = 2 + 2√2
A1
Part (c)
area = ∫12 + 2√2 (5 − x − 4x−2) dx, line minus curve
M1
[5x − x2/2 + 4/x]12 + 2√2
A1
uses (2 + 2√2)2 = 12 + 8√2
M1
rationalises 4/(2 + 2√2) = 2√2 − 2
M1
value at the upper limit = 10 + 10√2 − (6 + 4√2) + 2√2 − 2 = 2 + 8√2, and at the lower limit 17/2
A1
area = 2 + 8√2 − 17/2 = 8√2 − 13/2 cso
A1
Exam tip: A point you can read from the figure is a factor you already know: A tells you that (x − 1) divides the cubic, which leaves only a quadratic to solve.
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