AS June 2018 Paper 1 Q11
11 Four finite regions \(A\), \(B\), \(C\) and \(D\) are enclosed by the curve with equation
\[y = x^3 - 7x^2 + 11x + 6\]and the lines \(y = k\), \(x = 1\) and \(x = 4\), as shown in the diagram below.
The areas of \(B\) and \(C\) are equal.
Find the value of \(k\). [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| States integral(s) of the cubic with limits that include 1 and 4 | M1 | 1.1a |
| Integrates the function and substitutes correct limits. Condone one incorrect term. Note: \(\displaystyle\int_1^4 (x^3 - 7x^2 + 11x + 6)\,\mathrm{d}x = \frac{69}{4} \Rightarrow \text{M1M1}\) | M1 | 1.1a |
| Obtains the correct value of \(k\). Do not apply ISW. NMS can score 3/3. | A1 | 1.1b |
| (3 marks) |
Typical solution
\[\text{mean} = k = \frac{1}{4 - 1}\int_1^4 (x^3 - 7x^2 + 11x + 6)\,\mathrm{d}x\]\[\therefore k = \frac{1}{3}\left[\frac{x^4}{4} - \frac{7x^3}{3} + \frac{11x^2}{2} + 6x\right]_1^4\]\[= \frac{1}{3}\left(\frac{4^4}{4} - \frac{7 \times 4^3}{3} + \frac{11 \times 4^2}{2} + 6 \times 4\right) - \frac{1}{3}\left(\frac{1^4}{4} - \frac{7 \times 1^3}{3} + \frac{11 \times 1^2}{2} + 6 \times 1\right)\]\[= \frac{1}{3} \times \frac{80}{3} - \frac{1}{3} \times \frac{113}{12}\]\[= 5.75\]Alternative
\[\int_1^4 (x^3 - 7x^2 + 11x + 6 - k)\,\mathrm{d}x = 0\]\[\left[\frac{x^4}{4} - \frac{7x^3}{3} + \frac{11x^2}{2} + 6x - kx\right]_1^4 = 0\]\[\left(\frac{4^4}{4} - \frac{7 \times 4^3}{3} + \frac{11 \times 4^2}{2} + 6 \times 4 - k \times 4\right) - \left(\frac{1}{4} - \frac{7}{3} + \frac{11}{2} + 6 - k\right) = 0\]\[64 - \frac{448}{3} + 88 + 24 - 4k - \frac{113}{12} + k = 0\]\[\frac{69}{4} = 3k\]\[k = 5.75\]Alternative
Area B = Area C
\[\therefore \int_1^p (x^3 - 7x^2 + 11x + 6 - k)\,\mathrm{d}x = \int_p^4 (k - x^3 + 7x^2 - 11x - 6)\,\mathrm{d}x\]\[\left[\frac{x^4}{4} - \frac{7x^3}{3} + \frac{11x^2}{2} + 6x - kx\right]_1^p = \left[kx - \frac{x^4}{4} + \frac{7x^3}{3} - \frac{11x^2}{2} - 6x\right]_p^4\]\[\begin{aligned}&\left(\frac{p^4}{4} - \frac{7p^3}{3} + \frac{11p^2}{2} + 6p - kp\right) - \left(\frac{1}{4} - \frac{7}{3} + \frac{11}{2} + 6 - k\right) \\ &\quad = \left(4k - 64 + \frac{7 \times 64}{3} - \frac{11 \times 16}{2} - 24\right) - \left(kp - \frac{p^4}{4} + \frac{7p^3}{3} - \frac{11p^2}{2} - 6p\right)\end{aligned}\]\[-\frac{1}{4} + \frac{7}{3} - \frac{11}{2} - 6 + k = 4k - 64 + \frac{448}{3} - 88 - 24\]\[\frac{69}{4} = 3k\]\[k = 5.75\]