Integration: Additional Mathematics Tingkatan 5 questions

Section A: objective questions

10 marks

Choose the correct answer. Each question carries 1 mark.

Given d/dx (x⁵ + 3x) = 5x⁴ + 3, find ∫ (5x⁴ + 3) dx.

x⁵ + 3x + c
20x³ + c
5x⁴ + 3x + c
x⁵ + 3x

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Integration reverses differentiation, so the integral is x⁵ + 3x. An indefinite integral also needs + c, which the last option leaves out.

For the teacher: Remembering, low difficulty, standard 3.1.1

Textbook p. 83

Find ∫ (3x² − 4x + 5) dx.

6x − 4 + c
x³ − 2x² + 5x + c
x³ − 4x² + 5x + c
3x³ − 4x² + 5x + c

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Integrate term by term: x³ − 2x² + 5x + c. The first option is the derivative; the third forgets to divide −4x by 2.

For the teacher: Applying, low difficulty, standard 3.2.2

Textbook p. 87

Find ∫ x(2x − 3) dx.

x² (x² − 3x) + c
2x³3 − 3x² + c
x²2 (2x²2 − 3x) + c
2x³3 − 3x²2 + c

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Expand first: 2x² − 3x. Then ∫ (2x² − 3x) dx = 2x³3 − 3x²2 + c. A product cannot be integrated factor by factor.

For the teacher: Applying, low difficulty, standard 3.2.2

Textbook p. 87

∫ (2x − 5)⁴ dx = ______

(2x − 5)⁵10 + c
(2x − 5)⁵5 + c
4(2x − 5)³ + c
(2x − 5)⁵2 + c

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Use ∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹ ÷ [a(n + 1)]. Here a = 2 and n + 1 = 5, so divide by 10. Dividing by only 5 forgets the a = 2.

For the teacher: Applying, low difficulty, standard 3.2.3

Textbook p. 88

A curve has gradient function dy/dx = 6x − 4 and passes through the point (1, 3). Find the value of y when x = 2.

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y = 3x² − 4x + c. At (1, 3): 3 = 3 − 4 + c, so c = 4. Then y = 3(4) − 4(2) + 4 = 8. The value 4 comes from leaving out c.

For the teacher: Analysing, medium difficulty, standard 3.2.4

Textbook p. 90

∫₁³ (2x + 1) dx = ______

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[x² + x] from 1 to 3 = (9 + 3) − (1 + 1) = 10. Taking only the upper limit gives 12.

For the teacher: Applying, low difficulty, standard 3.3.1

Textbook p. 92

Given ∫₁⁵ f(x) dx = 7 and ∫₁³ f(x) dx = 3, find ∫₃⁵ f(x) dx.

4
10
−4
21

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∫₁³ + ∫₃⁵ = ∫₁⁵, so ∫₃⁵ = 7 − 3 = 4. Adding the two values gives 10.

For the teacher: Understanding, low difficulty, standard 3.3.1

Textbook p. 94

Given ∫₀² f(x) dx = 6, find the value of k if ∫₀² [k f(x) − 2] dx = 14.

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k(6) − [2x] from 0 to 2 = 14, so 6k − 4 = 14 and k = 3. Forgetting that ∫ 2 dx over 0 to 2 is 4, not 2, gives 6k − 2 = 14, so k = 83.

For the teacher: Analysing, high difficulty, standard 3.3.1

Textbook p. 94

Diagram 1 shows the curve y = x² + 1. Find the area of the shaded region, in units².

Curve y = x squared plus 1; the region under it between x = 0 and x = 3 and above the x-axis is shaded — Diagram 1

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Area = ∫₀³ (x² + 1) dx = x³3 + x from 0 to 3 = 9 + 3 = 12. The value 9 forgets the +1 and 30 is y at x = 3 times 3.

For the teacher: Applying, medium difficulty, standard 3.3.3

Textbook p. 97

The region bounded by the curve y = √x, the x-axis and the line x = 4 is revolved through 360° about the x-axis. Find the volume generated, in units³.

4π
16π
2π
8π

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V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = πx²2 from 0 to 4 = 8π. Forgetting the ÷ 2 gives 16π.

For the teacher: Applying, medium difficulty, standard 3.3.5

Textbook p. 107

Section B: structured questions

35 marks

Answer all questions. Show your working.

Find each integral with respect to x.

[2 marks] (a)∫ (3x + 1)⁵ dx

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Use ∫ (ax + b)ⁿ dx formula [1]1
(3x + 1)⁶18 + c [1]1

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[2 marks] (b)∫ 8(2x − 1)² dx

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Write as 8(2x − 1)⁻² [1]1
−4(2x − 1)⁻¹ + c, or −42x − 1 + c [1]1

8 ÷ (2 × (−1)) = −4.

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For the teacher: Applying, medium difficulty, standard 3.2.3

Textbook p. 88

The gradient function of a curve is dy/dx = 3x² − 6x. The curve passes through the point (2, 1).

[3 marks] (a)Find the equation of the curve.

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y = x³ − 3x² + c [1]1
Substitute (2, 1): 1 = 8 − 12 + c [1]1
c = 5, so y = x³ − 3x² + 5 [1]1

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[1 mark] (b)Find the value of y when x = −1.

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−1 − 3 + 5 = 1 [1]1

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For the teacher: Applying, low difficulty, standard 3.2.4

Textbook p. 90

Given ∫₀⁴ f(x) dx = 10 and ∫₀⁴ g(x) dx = 3, find

[1 mark] (a)∫₄⁰ f(x) dx

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−10 [1]1

Swapping the limits changes the sign.

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[1 mark] (b)∫₀⁴ [2f(x) − g(x)] dx

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2(10) − 3 = 17 [1]1

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[2 marks] (c)∫₀⁴ [g(x) + x] dx

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∫₀⁴ x dx = x²2 from 0 to 4 = 8 [1]1
3 + 8 = 11 [1]1

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For the teacher: Understanding, low difficulty, standard 3.3.1

Textbook p. 94

Diagram 2 shows part of the curve y = x² − 4x. Region P is below the x-axis and region Q is above it.

Curve y = x squared minus 4x crossing the x-axis at 0 and 4; region P is below the axis between 0 and 4 and region Q is above the axis between 4 and 5 — Diagram 2

[2 marks] (a)Find the area of region P.

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∫₀⁴ (x² − 4x) dx = x³3 − 2x² from 0 to 4 = −323 [1]1
Area = 323 units² (area is positive) [1]1

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[1 mark] (b)Find the area of region Q.

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x³3 − 2x² from 4 to 5 = −253 + 323 = 73 units² [1]1

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[2 marks] (c)Hence find the total area of P and Q. Explain why ∫₀⁵ (x² − 4x) dx does not give this area.

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323 + 73 = 13 units² [1]1
The integral over 0 to 5 is −253: the part below the x-axis is negative and cancels part of Q [1]1

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For the teacher: Analysing, high difficulty, standard 3.3.3

Textbook p. 98

Diagram 3 shows the curve y = x² and the straight line y = 2x, which intersect at the origin and at point P.

Line y = 2x above the curve y = x squared from the origin to the point P; the region between them is shaded — Diagram 3

[2 marks] (a)Find the coordinates of P.

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x² = 2x, x(x − 2) = 0, x = 2 [1]1
P = (2, 4) [1]1

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[3 marks] (b)Find the area of the shaded region.

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Area = ∫₀² (2x − x²) dx [1]1
[x² − x³3] from 0 to 2 [1]1
4 − 83 = 43 units² [1]1

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For the teacher: Applying, medium difficulty, standard 3.3.3

Textbook p. 103

The region bounded by the curve y = x², the x-axis and the line x = 2 is revolved through 360°.

[3 marks] (a)About the x-axis, find the volume generated in terms of π.

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V = π ∫₀² y² dx = π ∫₀² x⁴ dx [1]1
πx⁵5 from 0 to 2 [1]1
32π5 units³ [1]1

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[2 marks] (b)A different region is bounded by the curve y = x², the y-axis and the line y = 4. Find the volume generated when it is revolved about the y-axis, in terms of π.

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V = π ∫₀⁴ x² dy = π ∫₀⁴ y dy [1]1
πy²2 from 0 to 4 = 8π units³ [1]1

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For the teacher: Applying, medium difficulty, standard 3.3.5

Textbook p. 107

Water flows into an empty tank at the rate dV/dt = 6t + 4, where V is the volume in litres and t is the time in minutes.

[2 marks] (a)Find V in terms of t.

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V = 3t² + 4t + c [1]1
V = 0 when t = 0, so c = 0 and V = 3t² + 4t [1]1

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[2 marks] (b)Find the time taken for the tank to hold 64 litres.

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3t² + 4t − 64 = 0, (3t + 16)(t − 4) = 0 [1]1
t = 4 minutes (reject t = −163) [1]1

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[1 mark] (c)The tank holds only 40 litres. Will it overflow within 3 minutes? Give a reason.

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No. V(3) = 27 + 12 = 39 litres, which is less than 40 [1]1

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For the teacher: Evaluating, medium difficulty, standard 3.4.1

Textbook p. 112

A curve passes through (0, 5) and has a stationary point where x = 2.

[3 marks] (a)Write down one possible gradient function dy/dx for the curve and use it to find the equation of the curve.

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Any dy/dx = k(x − 2), k ≠ 0, with its matching curve is accepted.

A gradient function that is 0 at x = 2, for example dy/dx = 2x − 4 [1]1
y = x² − 4x + c [1]1
Through (0, 5): c = 5, so y = x² − 4x + 5 [1]1

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For the teacher: Creating, medium difficulty, standard 3.2.4

Textbook p. 90

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Section B: structured questions

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18