A Level · Mathematics 9709 · Pure Mathematics 3

Integration: A Level Mathematics questions with answers

Free Integration questions for A Level Mathematics (9709), part of Pure Mathematics 3. Each question has the answer and the marking points an examiner would look for.

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Integration practice questions

Question 1Structured · 10 marksA* challenge
(a)

Show that ∫1e x2 ln x dx = 2e3 + 19.

[5]
(b)

Express −x − 5(x + 1)(x + 2) in partial fractions.

[2]
(c)

Hence find ∫01 −x − 5(x + 1)(x + 2) dx, giving your answer in the form a ln 2 + b ln 3.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)integration by parts with u = ln x
u = ln x, dv = xn chosen M1
x33 ln x − ∫ x23 dx A1
integrate the second term M1
limits substituted correctly M1
given answer obtained with no errors A1
5
(b)−4x + 1 + 3x + 2
correct method M1
A = −4, B = 3 A1
2
(c)−7ln 2 + 3ln 3
integrate to ln terms M1
substitute limits M1
−7ln 2 + 3ln 3 A1
3
Question 2Structured · 5 marksA* challenge
(a)

Use the trapezium rule with 3 intervals to estimate ∫01 √(1 + x3) dx. Give your answer correct to 3 decimal places.

[3]
(b)

Explain, with reference to the shape of the curve, whether your answer to (a) is an over-estimate or an under-estimate.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)1.121
h = 0.333 B1
correct trapezium formula with y-values M1
1.121 A1
3
(b)the curve is convex (concave upwards) on this interval, so the chords lie above the curve
curve concave up / convex B1
trapezia lie above the curve B1
2
Question 3Structured · 4 marks

Find the exact value of ∫0π8 sin2(4x) dx.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
π16
sin2 = ½(1 − cos 2θ) M1
½x − 116 sin 8x A1
limits M1
π16 A1
4
Question 4Structured · 3 marks

Find the exact value of ∫ from 0 to 2 of e3x dx.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
13(e6 − 1)
13e3x M1
Correct limits M1
13(e6 − 1) A1
3
Question 5Structured · 11 marksA* challenge

For integers n ≥ 0, let In = ∫01 x2n + 1ex^{2} dx.

(a)

Show that In = e2 − nIn − 1 for n ≥ 1.

[4]
(b)

Find the exact value of I3.

[3]
(c)

Show that 14453 < e < 3011, and hence state the value of e correct to 2 decimal places.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)parts with u = x2n, dv = xex^{2} dx
writes x2n+1ex^{2} = x2n · xex^{2} M1
∫xex^{2} dx = ½ex^{2} A1
[½x2nex^{2}] − ∫ nx2n−1ex^{2} dx M1
= e2 − nIn−1 A1
4
(b)3 − e
I0 = e − 12 B1
I1 = ½, I2 = e2 − 1 M1
I3 = 3 − e A1
3
(c)I₅ = 60 − 22e > 0, I₆ = 265e/2 − 360 > 0; e = 2.72
the integrand is positive on (0, 1], so every In > 0 B1
continues the recurrence: I5 = 60 − 22e, I6 = 265e2 − 360 M1
I5 > 0 gives e < 3011; I6 > 0 gives e > 14453 A1
2.7170 < e < 2.7273, so e = 2.72 A1
4
Question 6Structured · 12 marksA* challenge

For n ≥ 0, let In = ∫0π/4 tann x dx.

(a)

Show that, for n ≥ 2, In + In − 2 = 1n − 1.

[4]
(b)

Prove that, for n ≥ 2, 12(n + 1) < In < 12(n − 1).

[4]
(c)

Hence show that 982315 < π < 334105.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)tann − 2 x(tan2 x + 1) = tann − 2 x sec2 x
In + In − 2 = ∫ tann − 2 x sec2 x dx M1
substitution u = tan x (or recognising the derivative) M1
1n − 1tann − 1 x between 0 and π4 A1
= 1n − 1 A1
4
(b)In + 2 < In < In − 2
0 < tan x < 1 on 0 < x < π4, so In + 2 < In B1
In + In − 2 = 1n − 1 with In < In − 2 gives In < 12(n − 1) M1
In + 2 + In = 1n + 1 with In + 2 < In gives In > 12(n + 1) M1
both inequalities fully justified A1
4
(c)I8 = π4 − 76105; 118 < I8 < 114
I0 = π4 and repeated use of (a) M1
I8 = π4 − 76105 A1
n = 8 in (b): 118 < π4 − 76105 < 114 M1
rearranged to the given result A1
4

More Pure Mathematics 3 questions

AlgebraLogarithmic & exponential functionsTrigonometryDifferentiationNumerical solution of equationsVectorsDifferential equationsComplex numbers

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in