OCR C2 — Question 8 12 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule then exact integration comparison
DifficultyModerate -0.3 This is a straightforward multi-part C2 integration question requiring standard trapezium rule application, routine integration of a power function (converting √x to x^(1/2)), and percentage error calculation. All techniques are procedural with no problem-solving insight needed, making it slightly easier than average, though the multiple parts and arithmetic with surds prevent it from being trivial.
Spec1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration

8. The finite region \(R\) is bounded by the curve \(y = 1 + 3 \sqrt { x }\), the \(x\)-axis and the lines \(x = 2\) and \(x = 8\).
  1. Use the trapezium rule with three intervals, each of width 2 , to estimate to 3 significant figures the area of \(R\).
  2. Use integration to find the exact area of \(R\) in the form \(a + b \sqrt { 2 }\).
  3. Find the percentage error in the estimate made in part (a).

AnswerMarks Guidance
(i)\(x\) \(2\)
\(1 + 3\sqrt{x}\)\(5.243\) \(7\)
area \(= \frac{1}{2} \times 2 \times [5.243 + 9.485 + 2(7 + 8.348)]\)B1 M1
\(= 45.4\) (3sf)A1
(ii) \(= \int_2^8 (1 + 3\sqrt{x}) dx\)
\(= [x + 2x^{\frac{3}{2}}]_2^8\)M1 A1
\(= [8 + 2(2\sqrt{2})^3] - [2 + 2(2\sqrt{2})]\)M1
\(= (8 + 32\sqrt{2}) - (2 + 4\sqrt{2})\)M1
\(= 6 + 28\sqrt{2}\)A1
(iii) \(= \frac{(6 + 28\sqrt{2}) - 45.4}{6 + 28\sqrt{2}} \times 100\% = 0.43\%\)M1 A1 (12)
**(i)** | $x$ | $2$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|
| $1 + 3\sqrt{x}$ | $5.243$ | $7$ | $8.348$ | $9.485$ | | M1 A1 |

area $= \frac{1}{2} \times 2 \times [5.243 + 9.485 + 2(7 + 8.348)]$ | B1 M1 |
$= 45.4$ (3sf) | A1 |

**(ii)** $= \int_2^8 (1 + 3\sqrt{x}) dx$ | |
$= [x + 2x^{\frac{3}{2}}]_2^8$ | M1 A1 |
$= [8 + 2(2\sqrt{2})^3] - [2 + 2(2\sqrt{2})]$ | M1 |
$= (8 + 32\sqrt{2}) - (2 + 4\sqrt{2})$ | M1 |
$= 6 + 28\sqrt{2}$ | A1 |

**(iii)** $= \frac{(6 + 28\sqrt{2}) - 45.4}{6 + 28\sqrt{2}} \times 100\% = 0.43\%$ | M1 A1 | (12)

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8. The finite region $R$ is bounded by the curve $y = 1 + 3 \sqrt { x }$, the $x$-axis and the lines $x = 2$ and $x = 8$.\\
(i) Use the trapezium rule with three intervals, each of width 2 , to estimate to 3 significant figures the area of $R$.\\
(ii) Use integration to find the exact area of $R$ in the form $a + b \sqrt { 2 }$.\\
(iii) Find the percentage error in the estimate made in part (a).\\

\hfill \mbox{\textit{OCR C2  Q8 [12]}}