June 2025 Paper 1 Q7
7 In this question you must show detailed reasoning.
Show that the area of the finite region enclosed by the curves \(y = x^2 - 7x + 2\) and \(y = 14 - 9x - x^2\) is \(\frac{125}{3}\). [7]
| Scheme | Marks | AO |
|---|---|---|
| DR \(x^2 - 7x + 2 = 14 - 9x - x^2\) | M1 | 3.1a |
| \(2x^2 + 2x - 12 = 0\) \((x+3)(x-2) = 0\) | B1 | 1.1 |
| \(x = -3,\ 2\) | A1 | 2.1 |
| Area \(= \int_{-3}^{2} \left((14 - 9x - x^2) - (x^2 - 7x + 2)\right)\mathrm{d}x\) \(= \int_{-3}^{2} (12 - 2x - 2x^2)\,\mathrm{d}x\) | M1* | 3.1a |
| \(= \left[12x - x^2 - \dfrac{2x^3}{3}\right]_{-3}^{2}\) | A1 | 1.1 |
| \(= \left(12\times 2 - 2^2 - \dfrac{2\times 2^3}{3}\right) - \left(12(-3) - (-3)^2 - \dfrac{2(-3)^3}{3}\right)\) | M1(dep) | 2.1 |
| \(= \dfrac{125}{3}\) | A1 | 2.1 |
| [7] |
Notes
DR: This question included the instruction: In this question you must show detailed reasoning.
M1: Attempt to eliminate \(y\)
B1: Correct three term quadratic or factors seen
A1: Both correct values
M1*: Attempts to integrate either function or the difference of two functions. Allow sign error and limits interchanged. Also allow for two separate integrals
A1: Correct indefinite integral. If done as two integrals, both need to be correct
M1(dep): Substitution of their limits into their expression seen and subtraction of results seen
Allow for limits reversed or subtraction back to front
Note: minimum evidence needed for full credit \(\left(24 - 4 - \frac{16}{3}\right) - \left((-36) - 9 + 18\right)\)
Must see substitution and subtraction in both integrals if done separately.
A1: AG From fully correct working
Allow if \(-\frac{125}{3}\) found if sign change justified eg “area is positive”
Note \(\int_{-3}^{2}(14 - 9x - x^2)\,\mathrm{d}x = \left[14x - \frac{9}{2}x^2 - \frac{1}{3}x^3\right]_{-3}^{2} = \frac{485}{6}\)
and \(\int_{-3}^{2}(x^2 - 7x + 2)\,\mathrm{d}x = \left[\frac{1}{3}x^3 - \frac{7}{2}x^2 + 2x\right]_{-3}^{2} = \frac{235}{6}\)