OCR MEI AS Paper 2 2020 November — Question 6 6 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea under polynomial curve
DifficultyModerate -0.8 This is a straightforward area calculation requiring expansion of the cubic, integration using standard power rule, and evaluation between roots. The question explicitly tells students what to show and the curve's roots are easily found, making this easier than average with minimal problem-solving required.
Spec1.08e Area between curve and x-axis: using definite integrals
6 Use integration to show that the area bounded by the \(x\)-axis and the curve with equation \(y = ( x - 1 ) ^ { 2 } ( x - 3 )\) is \(\frac { 4 } { 3 }\) square units.
Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(\int_a^b \left(x^3 - 5x^2 + 7x - 3\right)dx\)M1 (2.1) All three brackets expanded. Allow sign errors and one coefficient error. Must be 4 term cubic
A1 (1.1)All correct
\(a = 1\) and \(b = 3\)B1 (2.2a)
\(F[x] = \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 3x\)M1 (1.1) Integration of their expanded brackets; condone \(+c\). Allow sign errors and one coefficient error. Later M marks both dep on first M1
\(F[3] - F[1]\) evaluatedM1 (1.1) Subst seen, allow sign or num slip \(\left[\frac{81}{4} - 45 + \frac{63}{2} - 9\right] - \left[\frac{1}{4} - \frac{5}{3} + \frac{7}{2} - 3\right]\). \(\left[-\frac{9}{4}\right] - \left[-\frac{11}{12}\right]\)
\(= -\frac{4}{3}\) so required area is \(\left-\frac{4}{3}\right = \frac{4}{3}\) AG
[6] 
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_a^b \left(x^3 - 5x^2 + 7x - 3\right)dx$ | M1 (2.1) | All three brackets expanded. Allow sign errors and one coefficient error. Must be 4 term cubic |
| | A1 (1.1) | All correct |
| $a = 1$ and $b = 3$ | B1 (2.2a) | |
| $F[x] = \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 3x$ | M1 (1.1) | Integration of their expanded brackets; condone $+c$. Allow sign errors and one coefficient error. Later M marks both dep on first M1 |
| $F[3] - F[1]$ evaluated | M1 (1.1) | Subst seen, allow sign or num slip $\left[\frac{81}{4} - 45 + \frac{63}{2} - 9\right] - \left[\frac{1}{4} - \frac{5}{3} + \frac{7}{2} - 3\right]$. $\left[-\frac{9}{4}\right] - \left[-\frac{11}{12}\right]$ |
| $= -\frac{4}{3}$ so required area is $\left|-\frac{4}{3}\right| = \frac{4}{3}$ AG | A1 (2.4) | Sign change must be seen; no reliance on decimals |
| **[6]** | | |

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6 Use integration to show that the area bounded by the $x$-axis and the curve with equation $y = ( x - 1 ) ^ { 2 } ( x - 3 )$ is $\frac { 4 } { 3 }$ square units.

\hfill \mbox{\textit{OCR MEI AS Paper 2 2020 Q6 [6]}}
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