C2 June 2014 Q1

EdexcelOld spec5 marksIntegration

Figure 1: curve through (0, 1) with region R shaded between x = 1 and x = 2 — Figure 1

Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt{(x^2 + 1)}\), \(x \geqslant 0\)

The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\)

The table below shows corresponding values for \(x\) and \(y\) for \(y = \sqrt{(x^2 + 1)}\).

\(x\)	1	1.25	1.5	1.75	2
\(y\)	1.414		1.803	2.016	2.236

(a) Complete the table above, giving the missing value of \(y\) to 3 decimal places. (1)

(b) Use the trapezium rule, with all the values of \(y\) in the completed table, to find an approximate value for the area of \(R\), giving your answer to 2 decimal places. (4)

Mark scheme (a)

Scheme

Marks

\(x\)	1	1.25	1.5	1.75	2
\(y\)	1.414	1.601	1.803	2.016	2.236

\(\{\text{At } x = 1.25,\}\ y = 1.601\) (only)
1.601 (May not be in the table and can score if seen as part of their working in (b))

B1 cao

(1)

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2} \times 0.25; \times \underline{\left\{1.414 + 2.236 + 2\left(\text{their } 1.601 + 1.803 + 2.016\right)\right\}}\)	B1; M1 A1ft
\(\left\{= \tfrac{1}{8}(14.49)\right\} = 1.81125\) 1.81 or awrt 1.81	A1
	(4)
	Total 5

Notes

B1; for using \(\dfrac{1}{2} \times 0.25\) or \(\dfrac{1}{8}\) or equivalent.

M1: Structure of \(\{\ldots\ldots\ldots\}\)

A1ft: for the correct expression as shown following through candidate’s \(y\) value found in part (a).

M1 requires the correct structure for the \(y\) values. It needs to contain first \(y\) value plus last \(y\) value and the second bracket to be multiplied by 2 and to be the summation of the remaining \(y\) values in the table with no additional values. If the only mistake is a copying error or is to omit one value from 2(…..) bracket this may be regarded as a slip and the M mark can be allowed (nb: an extra repeated term, however, forfeits the M mark). M0 if any values used are \(x\) values instead of \(y\) values.

A1ft: for the correct underlined expression as shown following through candidate’s \(y\) value found in part (a).

Bracketing mistakes: e.g.
\(\left(\dfrac{1}{2} \times \dfrac{1}{4}\right)(1.414 + 2.236) + 2\left(\text{their } 1.601 + 1.803 + 2.016\right) (= 11.29625)\)
\(\left(\dfrac{1}{2} \times \dfrac{1}{4}\right)1.414 + 2.236 + 2\left(\text{their } 1.601 + 1.803 + 2.016\right) (= 13.25275)\)
Both score B1 M1 A0 unless the final answer implies that the calculation has been done correctly (then full marks could be given).

Alternative

Separate trapezia may be used, and this can be marked equivalently.
\(\left[\dfrac{1}{8}(1.414 + 1.601) + \dfrac{1}{8}(1.601 + 1.803) + \dfrac{1}{8}(1.803 + 2.016) + \dfrac{1}{8}(2.016 + 2.236)\right]\)
B1 for \(\dfrac{1}{8}\) (aef), M1 for correct structure, 1st A1ft for correct expression, ft their 1.601

Correct answer only in (b) scores no marks
If required accuracy is not seen in (a), full marks can still be scored in (b) (e.g. uses 1.6)