IB · Mathematics: Analysis & Approaches SL & HL · Calculus

Integration techniques: IB Mathematics: Analysis & Approaches questions with answers

Free Integration techniques questions for IB Mathematics: Analysis & Approaches (SL & HL), HL only, part of Calculus. Each question has the answer and the marking points an examiner would look for.

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Integration techniques practice questions

Question 1Structured · 14 marksA* challenge

Let In = ∫01xnexdx, for n ∈ ℕ.

(a)

Find the exact value of I0.

[2]
(b)

Use integration by parts to show that In = e − nIn−1, for n ≥ 1.

[5]
(c)

Hence find the exact value of I2.

[4]
(d)

Explain why 0 < I2 < e − 1.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)e − 1
[ex]01 (M1)
e − 1 A1
2
(b)shown
u = xn, dv = exdx M1
du = nxn−1dx, v = ex A1
[xnex]01 − n∫01xn−1exdx A1
[xnex]01 = e A1
In = e − nIn−1 AG A1
5
(c)e − 2
repeated use of the reduction formula M1
I1 = 1 (A1)
I2 = e − 2 A1
correct working A1
4
(d)explanation
xnex > 0 on (0, 1), so the integral is positive R1
xn < 1 on (0, 1), so xnex < ex R1
hence I2 < I0 = e − 1 R1
3
Question 2Structured · 6 marks

Find the exact value of ∫1ex3ln x dx.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
3e4 + 116
attempt at integration by parts with u = ln x M1
x44ln x − ∫x34dx A1A1
x44ln x − x416 A1
substituting limits (M1)
3e4 + 116 A1
6
Question 3Structured · 6 marks

Use the substitution u = x + 1 to find the exact value of ∫08x√x + 1dx.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
403
x = u − 1, dx = du (A1)
∫19u − 1√udu A1
= ∫(u1/2 − u−1/2)du (M1)
[23u3/2 − 2u1/2] A1
substituting limits (M1)
403 A1
6
Question 4Structured · 14 marksA* challenge

Let In = ∫01xnexdx, for n ∈ ℕ.

(a)

Find the exact value of I0.

[2]
(b)

Use integration by parts to show that In = e − nIn−1, for n ≥ 1.

[5]
(c)

Hence find the exact value of I3.

[4]
(d)

Explain why 0 < I3 < e − 1.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)e − 1
[ex]01 (M1)
e − 1 A1
2
(b)shown
u = xn, dv = exdx M1
du = nxn−1dx, v = ex A1
[xnex]01 − n∫01xn−1exdx A1
[xnex]01 = e A1
In = e − nIn−1 AG A1
5
(c)−2e + 6
repeated use of the reduction formula M1
I1 = 1 (A1)
I2 = e − 2 (A1)
I3 = −2e + 6 A1
4
(d)explanation
xnex > 0 on (0, 1), so the integral is positive R1
xn < 1 on (0, 1), so xnex < ex R1
hence I3 < I0 = e − 1 R1
3
Question 5Structured · 5 marks

Find the exact value of ∫01xe2xdx.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
e2 + 14
attempt at integration by parts, with u = x M1
xe2x2 − ∫e2x2dx A1
xe2x2 − e2x4 A1
substituting limits (M1)
e2 + 14 A1
5
Question 6Structured · 6 marks

Use the substitution u = x + 1 to find the exact value of ∫015x√x + 1dx.

[6]
Show answers and mark scheme
PartAnswer and marking pointsMarks
36
x = u − 1, dx = du (A1)
∫116u − 1√udu A1
= ∫(u1/2 − u−1/2)du (M1)
[23u3/2 − 2u1/2] A1
substituting limits (M1)
36 A1
6

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in