WMA12/01 Jan 2024, Question 10: Definite Integrals and Areas
Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 9 marks
The question, in short
Show by calculus that the curve y = ½x² + 1458/√(x³) − 74 has its only stationary point at x = 9, then find the exact area of the shaded region R bounded by the curve, the horizontal line through P, and x = 4.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 4 marks
Differentiate y = ½x² + 1458/√(x³) − 74 and solve for the stationary point, showing the x coordinate of P equals 9.
- 1y = ½x² + 1458x^(−3/2) − 74
- 2dy/dx = x − 1458 × (3/2)x^(−5/2) = x − 2187x^(−5/2)
- 3At P, dy/dx = 0: x = 2187x^(−5/2), so x^(7/2) = 2187
- 42187 = 3⁷, so x^(7/2) = 3⁷ and x = 3² = 9
AnswerShown: dy/dx = x − 2187x^(−5/2) = 0 ⇒ x^(7/2) = 2187 ⇒ x = 2187^(2/7) = 9
Part (b) · 5 marks
Using algebraic integration, find the exact area of R between x = 4 and P, bounded above by the curve and below by the horizontal line through P.
- 1At x = 9: y = 40.5 + 1458/27 − 74 = 40.5 + 54 − 74 = 20.5
- 2∫ (½x² + 1458x^(−3/2) − 74) dx = x³/6 − 2916x^(−1/2) − 74x
- 3From 4 to 9: (121.5 − 972 − 666) − (32/3 − 1458 − 296)
- 4= −1516.5 + 1743.33… = 226 5/6 (area under the curve)
- 5Subtract the rectangle under l: 5 × 20.5 = 102.5
- 6Area R = 226 5/6 − 102 1/2 = 124 1/3 = 373/3
Answer373/3 (= 124⅓)
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