OCR MEI C2 — Question 6 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypePure definite integration
DifficultyEasy -1.2 This is a straightforward definite integration question requiring only basic power rule application (including negative powers) and substitution of limits. It's a standard C2 exercise with no problem-solving element—purely routine technique with two simple terms.
Spec1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits
6 Evaluate \(\int _ { 1 } ^ { 2 } \left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x\).
Question 6:
AnswerMarks Guidance
\(\int_1^2 \left(x^2 + \frac{1}{x^2}\right)dx = \left[\frac{x^3}{3} - \frac{1}{x}\right]_1^2\)M1, A1 A1 Correct integral of each term
\(= \left(\frac{8}{3} - \frac{1}{2}\right) - \left(\frac{1}{3} - 1\right)\)M1 Substitute
\(= \frac{13}{6} + \frac{2}{3} = \frac{17}{6}\)A1 Total: 5 
## Question 6:

$\int_1^2 \left(x^2 + \frac{1}{x^2}\right)dx = \left[\frac{x^3}{3} - \frac{1}{x}\right]_1^2$ | M1, A1 A1 | Correct integral of each term

$= \left(\frac{8}{3} - \frac{1}{2}\right) - \left(\frac{1}{3} - 1\right)$ | M1 | Substitute

$= \frac{13}{6} + \frac{2}{3} = \frac{17}{6}$ | A1 | **Total: 5**

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6 Evaluate $\int _ { 1 } ^ { 2 } \left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) \mathrm { d } x$.

\hfill \mbox{\textit{OCR MEI C2  Q6 [5]}}
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