A-LevelMathematicsIntegration9709/12 · May/June 2014 · Q911 Marks
The diagram shows part of the curve y = 8 – √(4 – x) and the tangent to the curve at P (3, 7). [Figure 9.1]
- (i)Find expressions for dy/dx and ∫ y dx.[5]
- (ii)Find the equation of the tangent to the curve at P in the form y = mx + c.[2]
- (iii)Find, showing all necessary working, the area of the shaded region.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [B1 B1] dy/dx = – ¹/₂(4 – x)⁻¹/² × (–1) Without (–1). For (x⁻¹).
- [B1] ∫ y dx = 8x B1 for "8x".
- [B1] ∫ y dx = (4-x)^(3/2) / (3/2) B1 for all except *(-1).
- [B1] ∫ y dx = ... * (-1) B1 for *(-1).
- [M1 A1] Eqn y – 7 = ¹/₂(x – 3) → y = ¹/₂x + 5¹/₂ M1 unsimplified. A1 as y=mx+c
- [M1] Area under curve = ∫ from 0 to 3 (58/3) Use of limits – needs use of "0"
- [M1] Area under line = ¹/₂(5¹/₂ + 7)×3 Correct method
- [M1 A1] → 58/3 – 75/4 = 7/12 M1 Subtraction. A1 co
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