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WMA12/01 Jun 2023, Question 10: Definite Integrals and Areas

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 10 marks

The question, in short

Given y = (x − k)² ÷ √x for x > 0 with k positive, and line l through A(1, 9) and B(16, q) meeting the curve, work through the integral, find k, then the area between l and C.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 5 marks

Expand the integrand and integrate from 1 to 16, giving the result as ak² + bk + 2046/5 and state the integer values of a and b.

  1. 1(x − k)²/√x = x^(3/2) − 2kx^(½) + k²x^(−½)
  2. 2Integrate: (2/5)x^(5/2) − (4/3)kx^(3/2) + 2k²x^(½)
  3. 3At x = 16: (2/5)(1024) − (4/3)k(64) + 2k²(4) = 2048/5 − (256/3)k + 8k²
  4. 4At x = 1: 2/5 − (4/3)k + 2k²
  5. 5Subtract: 6k² − 84k + 2046/5, so a = 6, b = −84

AnswerShown: a = 6, b = −84, i.e. ∫₁¹⁶ (x − k)²/√x dx = 6k² − 84k + 2046/5

Part (b) · 2 marks

Use the fact that A(1, 9) lies on C to determine the value of k, showing it equals 4.

  1. 1A(1, 9) is on C: 9 = (1 − k)²/√1
  2. 2(1 − k)² = 9, so 1 − k = ±3
  3. 3k = −2 or k = 4
  4. 4k is positive, so k = 4

AnswerShown: 9 = (1 − k)²/1, so 1 − k = ±3, k = 4 or −2; k > 0 so k = 4

Part (c) · 3 marks

With k = 4, use earlier results to calculate the area of the shaded region R enclosed by the line and the curve.

  1. 1q = (16 − 4)²/√16 = 144/4 = 36, so B = (16, 36)
  2. 2Area under l from 1 to 16 (trapezium) = ½(9 + 36)(15) = 337.5
  3. 3Area under C from 1 to 16 = 6(4)² − 84(4) + 2046/5 = 96 − 336 + 409.2 = 169.2
  4. 4Area of R = 337.5 − 169.2 = 168.3 = 1683/10

AnswerArea of R = 1683/10 (= 168.3)

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