Edexcel International A Level Maths · P3 · Pure Mathematics 3
Integration of standard functions
28 free, high-quality past-paper-style questions on Integration of standard functions 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 · 229 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-7-B-01P3 · A2Easy5 marks
Throughout this question, x > 0.
(a)
Find ∫(6ex − 5/x + 3 cos x) dx.
(3)
(b)
Show that ∫1e (2 + 3/x) dx = 2e + 1.
(2)
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Answer
Mark
Notes
Part (a)
integrates at least two of the three terms correctly
M1
6ex − 5 ln x
A1
6ex − 5 ln x + 3 sin x + c, all correct including + c
Figure 1 shows part of the curve C with equation y = 8/(2x + 1), x ≥ 0. C meets the y-axis at A(0, 8) and the line x = 4 at the point B. The shaded region R is bounded by C, the coordinate axes and the line x = 4. The chord AB is shown dashed.
(a)
Find ∫ 8/(2x + 1) dx.
(2)
(b)
Show that the area of R is 8 ln 3.
(2)
(c)
Find the exact area of the region between C and the chord AB.
(3)
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Answer
Mark
Notes
Part (a)
k ln(2x + 1)
M1
4 ln(2x + 1) + c
A1
Part (b)
[4 ln(2x + 1)]04 = 4 ln 9 − 4 ln 1
M1
= 4 ln 9 = 8 ln 3
A1
Part (c)
B is (4, 8/9); area under AB is the trapezium ½ × 4 × (8 + 8/9)
M1
= 160/9
A1
area between chord and curve = 160/9 − 8 ln 3
A1
Common error: Writing 8 ln(2x + 1) and forgetting to divide by the coefficient of x. Differentiate the answer: d/dx[4 ln(2x + 1)] = 4 × 2/(2x + 1) = 8/(2x + 1), as required.
Figure 1 shows the curve with equation y = 1 + 2 sin 2x, 0 ≤ x ≤ π. The curve crosses the x-axis at the points P and Q. The shaded region R is bounded by the curve and the coordinate axes, between the y-axis and P.
Exam tip: Read the order of the roots from the figure. P is the first crossing, so take the smallest solution of sin 2x = −½ in range; 2x = 7π/6 comes before 2x = 11π/6.
Figure 1 shows the curve C with equation y = ae−kx, where a and k are positive constants. C passes through the points A(0, 12) and B(ln 4, 6). The region R is bounded by C, the coordinate axes and the line x = ln 16. The vertical line through B divides R into the two regions R1 and R2.
(a)
Find the value of a and the value of k.
(2)
(b)
Find ∫ 12e−x/2 dx.
(2)
(c)
Show that the areas of R1 and R2 are in the ratio 2 : 1.
Figure 1 shows part of the curve with equation y = 6x/(x2 + 3), x ≥ 0. The shaded region R is bounded by the curve, the x-axis and the lines x = 1 and x = 3.
(a)
Find ∫ 6x/(x2 + 3) dx.
(2)
(b)
Find the exact area of R, giving your answer in the form p ln q, where p and q are integers.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
recognises the numerator as a multiple of the derivative of x2 + 3: k ln(x2 + 3)
M1
3 ln(x2 + 3) + c
A1
Part (b)
[3 ln(x2 + 3)]13
M1
3 ln 12 − 3 ln 4
A1
3 ln 3
A1
Common error: Writing 6 ln(x2 + 3). The derivative of x2 + 3 is 2x, and 6x is three times that, so the logarithm is multiplied by 3, not 6.
Gas leaks from a damaged container. The rate of leakage is modelled by R = 12e−0.4t, where R is measured in kg per hour and t is the time in hours after the leak begins. The container held 35 kg of gas when the leak began.
(a)
Show that the mass of gas that leaks in the first 5 hours is 30(1 − e−2) kg, and find this mass to 3 significant figures.
(4)
(b)
Explain why, according to the model, the container never becomes empty.
(1)
(c)
Find the time after the leak begins at which 15 kg of gas has leaked, giving your answer in hours to 3 significant figures.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
integrates to ke−0.4t
M1
−30e−0.4t
A1
limits 0 and 5: −30e−2 + 30 = 30(1 − e−2)
M1
25.9 kg
A1
Part (b)
mass leaked by time T is 30(1 − e−0.4T), which is always less than 30, and 30 < 35
B1
Part (c)
30(1 − e−0.4T) = 15
M1
e−0.4T = ½, so T = (ln 2)/0.4
M1
1.73 hours
A1
Model sketch
Figure: the mass leaked in the first 5 hours is the shaded area
Exam tip: The integral of a rate between two times is the total change in that time. Here the period starts when the leak begins, so the lower limit is t = 0.
Figure 1 shows the curve C with equation y = 4 cos²x, 0 ≤ x ≤ π, and the line L with equation y = 1. L meets C at the points P and Q. The shaded region R is bounded by C, L and the y-axis.
(a)
Show that 4 cos²x ≡ 2 + 2 cos 2x.
(1)
(b)
Find the exact x-coordinates of P and Q.
(3)
(c)
Find the exact area of R.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
uses cos 2x ≡ 2 cos²x − 1, so 4 cos²x ≡ 2(1 + cos 2x) ≡ 2 + 2 cos 2x
B1
Part (b)
4 cos²x = 1, so cos x = ±½
M1
P: x = π/3
A1
Q: x = 2π/3
A1
Part (c)
area = ∫0π/3 (2 + 2 cos 2x − 1) dx
M1
x + sin 2x
A1
substitutes the limits 0 and their π/3
M1
π/3 + √3/2
A1
Common error: Forgetting the negative root. cos²x = ¼ gives cos x = ½ or cos x = −½, and the second one gives Q. Figure 1 shows two crossings, so you need both.
A cyclist starts from rest. The speed of the cyclist, v m s−1, t seconds after starting is modelled by v = V(1 − e−t/2), t ≥ 0, where V is a constant. Figure 1 shows the graph of v against t; the dashed line is the asymptote that the graph approaches as t increases.
(a)
Use Figure 1 to state the value of V.
(1)
(b)
Find the distance travelled by the cyclist in the first 6 seconds, giving your answer in the form p + qe−3 and then to 3 significant figures.
(4)
(c)
Find the time taken for the cyclist to reach a speed of 6 m s−1, giving your answer in seconds to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
as t → ∞, e−t/2 → 0, so v → V; from the asymptote V = 8
B1
Part (b)
distance = ∫06 8(1 − e−t/2) dt
M1
8t + 16e−t/2
A1
(48 + 16e−3) − (0 + 16)
M1
32 + 16e−3 = 32.8 m
A1
Part (c)
8(1 − e−t/2) = 6, so e−t/2 = ¼
M1
t = 2 ln 4 = 2.77 s
A1
Exam tip: Distance is the area under a speed–time graph. Here the speed is given by a formula, so the area is found by integrating with respect to time between the two times.
Figure 1 shows the curve C with equation y = x√(16 − x2), 0 ≤ x ≤ 4. C has a maximum point M(2√2, 8). The region R is bounded by C and the x-axis. The region S, shaded darker, is bounded by C and the straight line OM, where O is the origin.
(a)
By considering the derivative of (16 − x2)3/2, find ∫ x(16 − x2)½ dx.
(3)
(b)
Show that the area of R is 64/3.
(2)
(c)
Find the exact area of S, giving your answer in the form (p + q√2)/3, where p and q are integers.
(4)
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Answer
Mark
Notes
Part (a)
k(16 − x2)3/2
M1
d/dx(16 − x2)3/2 = −3x(16 − x2)½, so k = −⅓
M1
−⅓(16 − x2)3/2 + c
A1
Part (b)
[−⅓(16 − x2)3/2]04 = 0 − (−⅓ × 64)
M1
= 64/3
A1
Part (c)
area under C from 0 to 2√2: 64/3 − ⅓ × 83/2
M1
= 64/3 − 16√2/3
A1
subtracts the triangle under OM: ½ × 2√2 × 8 = 8√2
M1
(64 − 40√2)/3
A1
Exam tip: The x outside the bracket is (apart from a constant) the derivative of the expression inside. Guess k(16 − x2)3/2, differentiate, and choose k to match.
Figure 1 shows part of the curve C with equation y = (x − 4)/(x + 2), x > −2. C crosses the y-axis at (0, −2) and the x-axis at (4, 0). The region R1 lies below the x-axis and is bounded by C and the coordinate axes. The region R2 lies above the x-axis and is bounded by C, the x-axis and the line x = 10.
(a)
Show that y = 1 − 6/(x + 2).
(2)
(b)
Find the exact area of R1.
(4)
(c)
Find the exact value of ∫010y dx, and explain how it is related to the areas of R1 and R2.
(3)
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Answer
Mark
Notes
Part (a)
writes x − 4 as (x + 2) − 6, or uses division
M1
(x + 2)/(x + 2) − 6/(x + 2) = 1 − 6/(x + 2)
A1
Part (b)
integrates to x + k ln(x + 2)
M1
x − 6 ln(x + 2)
A1
(4 − 6 ln 6) − (0 − 6 ln 2) = 4 − 6 ln 3
M1
area is the positive value: 6 ln 3 − 4
A1
Part (c)
(10 − 6 ln 12) − (0 − 6 ln 2)
M1
10 − 6 ln 6
A1
the integral counts R1 as negative, so it equals area of R2 − area of R1
B1
Common error: Quoting 4 − 6 ln 3 as an area. Below the x-axis the integral is negative (about −2.59); the area is its size, 6 ln 3 − 4.
Figure 1 shows the curves with equations y = 2 sin x and y = tan x, 0 ≤ x < π/2. The curves meet at the origin O and at the point P. The shaded region R is bounded by the two curves between O and P.
(a)
By writing tan x as sin x/cos x, show that ∫ tan x dx = ln|sec x| + c.
(2)
(b)
Find the coordinates of P.
(3)
(c)
Find the exact area of R.
(4)
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Answer
Mark
Notes
Part (a)
sin x/cos x = −(−sin x)/cos x, the derivative of cos x over cos x, so the integral is −ln|cos x|
M1
−ln|cos x| = ln|1/cos x| = ln|sec x| + c
A1
Part (b)
2 sin x = sin x/cos x, so sin x(2 cos x − 1) = 0
M1
cos x = ½, so x = π/3
A1
P(π/3, √3)
A1
Part (c)
area = ∫0π/3 (2 sin x − tan x) dx
M1
−2 cos x − ln|sec x|
A1
substitutes: (−1 − ln 2) − (−2 − ln 1)
M1
1 − ln 2
A1
Common error: Dividing both sides of 2 sin x = tan x by sin x loses the solution sin x = 0, which is O. Factorise instead, so that both meeting points appear.
Figure 1 shows part of the curve C with equation y = tan²x, 0 ≤ x < π/2, and the line y = 3. The line meets C at A, where x = π/3. The point B on C has x-coordinate π/4. The region R is bounded by C, the line y = 3 and the y-axis. The region S is bounded by C, the x-axis and the vertical lines through B and A.
(a)
Show that ∫ tan²x dx = tan x − x + c.
(2)
(b)
Find the exact area of R.
(4)
(c)
Find the exact area of S.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
uses tan²x = sec²x − 1
M1
integrates sec²x to tan x: tan x − x + c
A1
Part (b)
area = ∫0π/3 (3 − tan²x) dx
M1
= ∫(4 − sec²x) dx = [4x − tan x]
A1
substitutes 0 and π/3
M1
4π/3 − √3
A1
Part (c)
[tan x − x]π/4π/3
M1
(√3 − π/3) − (1 − π/4)
M1
√3 − 1 − π/12
A1
Exam tip: R lies between the line (above) and the curve (below), so integrate (3 − tan²x). Rewriting tan²x as sec²x − 1 first turns the integrand into 4 − sec²x.
The curves y = 2 sin²x and y = sin x are drawn for 0 ≤ x ≤ π. For π/6 ≤ x ≤ 5π/6 the curve y = 2 sin²x lies on or above the curve y = sin x, and the finite region R lies between the two curves for π/6 ≤ x ≤ 5π/6.
(a)
Find the x-coordinates of all the points where the two curves meet for 0 ≤ x ≤ π.
The curve C has equation y = (x2 + 2x + 5)/(x + 1), x > −1. The line l has equation y = x + 3.
Figure: C, l and the line x = 3; the finite region bounded by C, l and the y-axis is shaded
(a)
Show that y = x + 1 + 4/(x + 1).
(2)
(b)
Find the exact area of the region bounded by C, the coordinate axes and the line x = 3, giving your answer in the form p + q ln 2, where p and q are rational.
(4)
(c)
Show that l meets C at exactly one point, where x = 1.
(2)
(d)
Find the exact area of the finite region bounded by C, l and the y-axis.
area = ∫01 (4/(x + 1) − 2) dx, using curve minus line
M1
[4 ln(x + 1) − 2x]
A1
substitutes 1 and 0
M1
4 ln 2 − 2
A1
Exam tip: Rewriting an improper fraction as a polynomial plus a proper fraction is what makes it integrable here: the proper fraction gives the logarithm.
The function f is defined by f(x) = e2x/(e2x + 3), x ∈ ℝ. The region R is bounded by the curve y = f(x), the coordinate axes and the line x = ln 3.
Figure: the region R
(a)
Find ∫f(x) dx.
(2)
(b)
Find ∫e2x(e2x + 3)3 dx.
(3)
(c)
Show that the area of R is ½ ln 3.
(3)
(d)
The area of the region bounded by the curve, the coordinate axes and the line x = a, where a > 0, is ln 2. Find the exact value of a.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
k ln(e2x + 3)
M1
½ ln(e2x + 3) + c
A1
Part (b)
k(e2x + 3)4
M1
d/dx(e2x + 3)4 = 8e2x(e2x + 3)3, or equivalent adjustment
M1
⅛(e2x + 3)4 + c
A1
Part (c)
½[ln(e2x + 3)] with limits 0 and ln 3
M1
uses e2 ln 3 = 9: ½ ln 12 − ½ ln 4
M1
= ½ ln 3
A1
Part (d)
½ ln((e2a + 3)/4) = ln 2
M1
(e2a + 3)/4 = 4, so e2a = 13
M1
a = ½ ln 13
A1
Exam tip: Both integrals use the same inside function, e2x + 3, whose derivative is 2e2x. Write that derivative down first; it tells you the constant to divide by.
The function f is defined for x > −½. Figure 1 shows the graph of y = f′(x), where f′(x) = 6/(2x + 1) − 2. The graph meets the y-axis at (0, 4) and the x-axis at P. The curve y = f(x) passes through the point (0, 3).
(a)
Find the x-coordinate of P.
(2)
(b)
Find f(x).
(4)
(c)
Using Figure 1, explain why the curve y = f(x) has a maximum point where x is the x-coordinate of P, and find the exact y-coordinate of this maximum point.
(3)
(d)
Show that the area of the shaded region in Figure 1 is 3 ln 3 − 2.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
6/(2x + 1) = 2, so 2x + 1 = 3
M1
x = 1
A1
Part (b)
integrates 6/(2x + 1) to k ln(2x + 1)
M1
3 ln(2x + 1) − 2x + c
A1
uses (0, 3): 3 ln 1 − 0 + c = 3
M1
f(x) = 3 ln(2x + 1) − 2x + 3
A1
Part (c)
f′(x) is positive to the left of P and negative to the right, so f changes from increasing to decreasing
B1
substitutes x = 1 into their f(x): 3 ln 3 − 2 + 3
M1
3 ln 3 + 1
A1
Part (d)
area = ∫01 f′(x) dx = [f(x)]01
M1
(3 ln 3 + 1) − 3 = 3 ln 3 − 2
A1
Exam tip: The area under a graph of f′(x) between two values of x is the change in f between them. Once you have f(x), you do not need to integrate again.
Figure 1 shows the front of an arched gateway. The arch is modelled by the curve with equation y = 4 sin(πx/6), 0 ≤ x ≤ 6, where x metres is the horizontal distance from O and y metres is the height above the ground. A horizontal bar PQ is fixed across the arch at a height of 2 m, with P and Q on the arch. The region between the bar and the arch is a glass fanlight.
(a)
State the width of the gateway at ground level and its greatest height.
(2)
(b)
Find the exact area of the whole opening under the arch, that is, the region between the arch and the ground, including the fanlight.
(3)
(c)
Find the x-coordinates of P and Q.
(2)
(d)
Show that the area of the fanlight is (24√3/π − 8) m2, and find this area to 3 significant figures.
Exam tip: Integrating sin(πx/6) divides by π/6, which is the same as multiplying by 6/π. With the factor 4 this gives −(24/π) cos(πx/6); differentiate it to check.
Figure 1 shows part of the curve C with equation y = (ln x)/x, x > 0. C crosses the x-axis at the point A and has a maximum point at M. The shaded region R is bounded by C, the x-axis and the line x = e2.
(a)
Show that M has coordinates (e, 1/e).
(4)
(b)
By considering the derivative of (ln x)2, find ∫ (ln x)/x dx.
(2)
(c)
Show that the area of R is 2.
(2)
(d)
The region bounded by C, the x-axis and the lines x = e2 and x = k, where k > e2, has the same area as R. Find the exact value of k.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
uses the quotient rule
M1
dy/dx = (1 − ln x)/x2
A1
sets 1 − ln x = 0, so x = e
M1
y = (ln e)/e = 1/e
A1
Part (b)
k(ln x)2, since d/dx(ln x)2 = 2(ln x)/x
M1
½(ln x)2 + c
A1
Part (c)
limits 1 (at A) and e2: ½(ln e2)2 − ½(ln 1)2
M1
= ½ × 4 − 0 = 2
A1
Part (d)
½(ln k)2 − 2 = 2
M1
(ln k)2 = 8, ln k = 2√2 (taking the positive root as k > 1)
M1
k = e2√2
A1
Common error: Writing ∫ (ln x)/x dx = ln(ln x). That would be the integral of 1/(x ln x); here ln x is multiplied by its own derivative 1/x, so the answer is ½(ln x)2.
Figure 1 shows the curves C1 and C2 with equations y = 4x/(x2 + 1)2 and y = x/(x2 + 1) respectively, for x ≥ 0. The curves meet at the origin and at the point P. The shaded region R is bounded by C1 and C2 between the origin and P.
(a)
Find the exact x-coordinates of the points where C1 and C2 meet.
(3)
(b)
Find ∫ x/(x2 + 1) dx.
(2)
(c)
Find ∫ 4x/(x2 + 1)2 dx.
(2)
(d)
Show that the area of R is 3/2 − ln 2.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
4x/(x2 + 1)2 = x/(x2 + 1), so x[4 − (x2 + 1)] = 0
M1
x = 0
A1
x2 = 3, so x = √3 at P (x ≥ 0)
A1
Part (b)
k ln(x2 + 1)
M1
½ ln(x2 + 1) + c
A1
Part (c)
k(x2 + 1)−1
M1
−2/(x2 + 1) + c
A1
Part (d)
area = ∫0√3 (C1 − C2) dx, with C1 above
M1
[−2/(x2 + 1) − ½ ln(x2 + 1)]
M1
(−½ − ½ ln 4) − (−2 − 0)
M1
= 3/2 − ln 2
A1
Exam tip: Both integrands contain the derivative 2x of x2 + 1. A first power underneath gives a logarithm; a square underneath gives a power, (x2 + 1)−1.
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