OCR C2 2007 June — Question 4 6 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2007
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeOver/underestimate justification with graph
DifficultyModerate -0.3 This is a straightforward trapezium rule application with standard strip width and function evaluation, followed by a conceptual question about concavity. The calculation is routine C2 content requiring only substitution into the trapezium rule formula. The reasoning part about over/under-estimate requires understanding that the curve is concave down (second derivative negative), making it slightly more demanding than pure calculation but still standard for this topic.
Spec1.09f Trapezium rule: numerical integration

4 \includegraphics[max width=\textwidth, alt={}, center]{e429080f-8634-46bc-b451-7b13b871e518-2_543_857_1155_644} The diagram shows the curve \(\mathrm { y } = \sqrt { 4 \mathrm { X } + 1 }\).
  1. Use the trapezium rule, with strips of width 0.5 , to find an approximate value for the area of the region bounded by the curve \(y = \sqrt { 4 x + 1 }\), the \(x\)-axis, and the lines \(x = 1\) and \(x = 3\). Give your answer correct to 3 significant figures.
  2. State with a reason whether this approximation is an under-estimate or an over-estimate.

(i) area \(= \frac{1}{2} \times \frac{1}{x} \times [\sqrt{5} + 2(\sqrt{7} + \sqrt{9} + \sqrt{11}) + \sqrt{13}]\)
\(\approx 0.25 \times 23.766...\)
AnswerMarks Guidance
\(\approx 5.94\)M1, M1, A1, A1 4 Attempt \(y\)-values for at least 4 of the \(x = 1, 1.5, 2, 2.5, 3\) only. Attempt to use correct trapezium rule. Obtain \(\frac{1}{4} \times [\sqrt{5} + 2(\sqrt{7} + \sqrt{9} + \sqrt{11}) + \sqrt{13}]\), or decimal equiv. Obtain 5.94 or better (answer only is 0/4).
(ii) This is an underestimate...... ...as the tops of the trapezia are below the curve*B1, B1dep*B 2 State underestimate. Correct statement or sketch.
(i) area $= \frac{1}{2} \times \frac{1}{x} \times [\sqrt{5} + 2(\sqrt{7} + \sqrt{9} + \sqrt{11}) + \sqrt{13}]$

$\approx 0.25 \times 23.766...$
$\approx 5.94$ | M1, M1, A1, A1 4 | Attempt $y$-values for at least 4 of the $x = 1, 1.5, 2, 2.5, 3$ only. Attempt to use correct trapezium rule. Obtain $\frac{1}{4} \times [\sqrt{5} + 2(\sqrt{7} + \sqrt{9} + \sqrt{11}) + \sqrt{13}]$, or decimal equiv. Obtain 5.94 or better (answer only is 0/4).

(ii) This is an underestimate...... ...as the tops of the trapezia are below the curve | *B1, B1dep*B 2 | State underestimate. Correct statement or sketch.
4\\
\includegraphics[max width=\textwidth, alt={}, center]{e429080f-8634-46bc-b451-7b13b871e518-2_543_857_1155_644}

The diagram shows the curve $\mathrm { y } = \sqrt { 4 \mathrm { X } + 1 }$.\\
(i) Use the trapezium rule, with strips of width 0.5 , to find an approximate value for the area of the region bounded by the curve $y = \sqrt { 4 x + 1 }$, the $x$-axis, and the lines $x = 1$ and $x = 3$. Give your answer correct to 3 significant figures.\\
(ii) State with a reason whether this approximation is an under-estimate or an over-estimate.

\hfill \mbox{\textit{OCR C2 2007 Q4 [6]}}