OxfordAQA International AS and A-level Mathematics: Integration, areas and the trapezium rule (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section P1.4 (Integration) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, which sits in International AS, Unit P1 (Pure Maths). Learn the methods from the study guide and revision notes. Course pages: hub and checklist. Give exact answers unless a question asks for decimals.

Questions

1. Find ∫ (10x^(3/2) − 6x^(−1/4) + 7) dx. [3]

2. Find ∫ (x + 3)²/√x dx. [4]

3. A curve has gradient dy/dx = 12x^(1/3) and passes through the point (8, 50). Find the equation of the curve. [4]

4. Show that ∫₁⁴ (3√x + 2/x²) dx = 31/2. [4]

5. The curve C has equation y = 9√x − x√x for x ≥ 0. The region R is bounded by C and the x-axis. Find the exact area of R. [5]

6. Find the area of the finite region between the curve y = x² − 7x + 6 and the x-axis. [4]

7. The curve y = 2x² and the line y = x + 6 cross at the points P and Q. (a) Work out the x-coordinates of P and Q. [2] (b) Calculate the area enclosed by the curve and the line. [4]

8. Find the area of the finite region between the curves y = x^(1/3) and y = x². [6]

9. The function y = x^(3/2) is defined for x ≥ 0. (a) Estimate ∫₁³ x^(3/2) dx by the trapezium rule with four equal strips, giving the ordinates to 3 decimal places and your estimate to 4 significant figures. [4] (b) Explain, with reference to the shape of the curve, whether your estimate is an over-estimate or an under-estimate. [2] (c) Find the exact value of the integral, and its value to 4 significant figures. [3] (d) State how the estimate could be improved. [1]

10. The curve y = 4√x and the line y = x intersect at the origin and at the point A. (a) Find the coordinates of A. [3] (b) Calculate, in exact form, the area enclosed by the curve and the line. [5] (c) Use the trapezium rule with four strips to estimate the area under y = 4√x from x = 0 to x = 16. Compare it with the exact area under the curve and say why the estimate is not equal to it. [4]

11. Given that k > 1 and ∫₁ᵏ 3√x dx = 14, find the value of k. [5]

Answers

1. 10x^(3/2) → 10 ÷ (5/2) = 4, giving 4x^(5/2) [1]. −6x^(−1/4) → −6 ÷ (3/4) = −8, giving −8x^(3/4) [1]. Constant term 7x, plus c [1]. 4x^(5/2) − 8x^(3/4) + 7x + c [3] Examiner insight: an indefinite integral needs its constant of integration, and dividing by a fractional power is a place where answers often go wrong, so show the division.

2. Expand and divide: (x² + 6x + 9)/x^(1/2) = x^(3/2) + 6x^(1/2) + 9x^(−1/2) [1]. Integrate: (2/5)x^(5/2) [1], 6 × (2/3)x^(3/2) = 4x^(3/2) [1], 9 × 2x^(1/2) = 18x^(1/2), plus c [1]. (2/5)x^(5/2) + 4x^(3/2) + 18x^(1/2) + c [4] Examiner insight: dividing the top and bottom of the integrand separately is not integration; rewriting as a sum of powers is the step that makes the rest of the working possible.

3. y = 12x^(4/3) ÷ (4/3) [1] = 9x^(4/3) + c [1]. At x = 8: 8^(4/3) = 16, so 144 + c = 50 [1], giving c = −94 [1]. y = 9x^(4/3) − 94 [4] Examiner insight: when a point is given, you must use it to find c; leaving + c in the final equation means the curve has not been identified.

4. 3x^(1/2) integrates to 2x^(3/2) [1]; 2x^(−2) integrates to −2x^(−1) [1]. F(4) = 16 − 1/2 = 31/2 [1]. F(1) = 2 − 2 = 0, so the integral is 31/2 − 0 = 31/2 [1]. 31/2 [4] Examiner insight: on a “show that” question, write F(4) and F(1) separately so every substitution can be seen before the subtraction.

5. y = √x (9 − x), so C meets the x-axis at x = 0 and x = 9 [1]; y is positive in between (for example y = 10 at x = 4), so the area is the integral [1]. ∫ (9x^(1/2) − x^(3/2)) dx = 6x^(3/2) − (2/5)x^(5/2) [1] [1]. At x = 9: 162 − 486/5 = 324/5 [1]. 324/5 [5] Examiner insight: the limits are not given, so finding them is part of the method; look for a common factor to get them quickly.

6. x² − 7x + 6 = (x − 1)(x − 6), so the limits are 1 and 6 [1]. F(x) = x³/3 − 7x²/2 + 6x [1]. F(6) − F(1) = −18 − 17/6 = −125/6 [1]. The region is below the axis, so the area is the magnitude [1]. 125/6 [4] Examiner insight: a negative integral shows the region is below the axis; the final answer for “area” needs to be positive.

7. (a) 2x² = x + 6 gives 2x² − x − 6 = 0 [1], so (2x + 3)(x − 2) = 0 and x = −3/2 or x = 2 [1]. [2] (b) At x = 0 the line gives 6 and the curve gives 0, so the line is above [1]. Area = ∫ (x + 6 − 2x²) dx [1] = x²/2 + 6x − (2/3)x³ [1]. F(2) = 26/3 and F(−3/2) = −45/8, so area = 26/3 + 45/8 = 343/24 [1]. [4] Examiner insight: decide which graph is on top before integrating, and substitute the negative limit with brackets to avoid a sign slip.

8. x^(1/3) = x² gives x = 0 or x^(5/3) = 1, so x = 1 [2]. For 0 < x < 1, x^(1/3) is above x² (at x = 1/8: 1/2 against 1/64) [1]. ∫₀¹ (x^(1/3) − x²) dx [1] = [(3/4)x^(4/3) − x³/3]₀¹ [1] = 3/4 − 1/3 = 5/12 [1]. 5/12 [6] Examiner insight: solve for both intersections: x = 0 is easy to lose when you divide through by a power of x, so factorise or check by substitution instead.

9. (a) h = 0.5. Ordinates: 1, 1.837, 2.828, 3.953, 5.196 [2]. Estimate = (0.5/2)(1 + 5.196 + 2(1.837 + 2.828 + 3.953)) [1] = 0.25 × 23.43 = 5.858 [1]. [4] (b) The curve bends upwards [1], so each chord lies above the curve and the rule over-estimates [1]. [2] (c) ∫ x^(3/2) dx = (2/5)x^(5/2) [1]; value (2/5)(3^(5/2) − 1) = (2/5)(9√3 − 1) [1] = 5.835 [1]. This is below 5.858, which agrees with part (b). [3] (d) Use more strips (a smaller strip width) [1]. [1] Examiner insight: in part (b), the shape of the curve is the reason that earns credit; a result such as “5.858 is bigger” is only a comparison.

10. (a) 4√x = x. Let u = √x: u² − 4u = 0 [1], so u = 0 or u = 4 [1]. x = 16 and y = 16: A = (16, 16) [1]. [3] (b) The curve is above the line (at x = 4: 8 against 4) [1]. Area = ∫₀¹⁶ (4x^(1/2) − x) dx [1] = [(8/3)x^(3/2) − x²/2]₀¹⁶ [1] [1] = 512/3 − 128 = 128/3 [1]. [5] (c) h = 4. Ordinates: 0, 8, 11.314, 13.856, 16 [1]. Estimate = (4/2)(0 + 16 + 2(8 + 11.314 + 13.856)) [1] = 164.7 [1]. The exact area under the curve is (8/3)(64) = 512/3 = 170.7. The curve bends downwards, so the chords lie below it and the estimate is smaller [1]. [4] Examiner insight: keep part (b) exact, and use unrounded ordinates in part (c); rounding early can shift the fourth significant figure.

11. ∫ 3x^(1/2) dx = 2x^(3/2) [1]. Upper limit minus lower limit: 2k^(3/2) − 2 [1]. Set equal to 14 [1]: 2k^(3/2) = 16, so k^(3/2) = 8 [1]. Raise to the power 2/3: k = 4 [1]. [5] Examiner insight: both limits must be substituted; forgetting F(1) gives k^(3/2) = 7 and a wrong answer.

Where marks are usually lost

• Dividing by the old power rather than the new power.
• Keeping brackets or roots unexpanded before integrating.
• Leaving out + c, or failing to use the given point to find it.
• Reversing the limits so the sign of a result is wrong.
• Giving a negative value for an area below the axis.
• Using (lower − upper) between two graphs.
• Missing the limit x = 0 when two curves meet at the origin.
• Using n instead of n + 1 ordinates for the trapezium rule.
• Not linking over- or under-estimate to the shape of the curve.

Next steps

• Revisit the revision notes or the study guide.
• Return to the course hub and the checklist.
• Take all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section P1.4 Integration (International AS exams from May/June 2018, International A-level from May/June 2019), published by OxfordAQA.