| Date | May 2008 | Marks available | 7 | Reference code | 08M.1.hl.TZ1.6 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Find the area between the curves \(y = 2 + x - {x^2}{\text{ and }}y = 2 - 3x + {x^2}\) .
Markscheme
\(2 + x - {x^2} = 2 - 3x + {x^2}\) M1
\( \Rightarrow 2{x^2} - 4x = 0\)
\( \Rightarrow 2x(x - 2) = 0\)
\( \Rightarrow x = 0,{\text{ }}x = 2\) A1A1
Note: Accept graphical solution.
Award M1 for correct graph and A1A1 for correctly labelled roots.
\(\therefore {\text{A}} = \int_0^2 {\left( {(2 + x - {x^2}) - (2 - 3x + {x^2})} \right){\text{d}}x} \) (M1)
\( = \int_0^2 {(4x - 2{x^2}){\text{d}}x\,\,\,\,\,{\text{or equivalent}}} \) A1
\( = \left[ {2{x^2} - \frac{{2{x^3}}}{3}} \right]_0^2\) A1
\( = \frac{8}{3}\left( { = 2\frac{2}{3}} \right)\) A1
[7 marks]
Examiners report
This was the question that gained the most correct responses. A few candidates struggled to find the limits of the integration or found a negative area.
