C2 June 2017 Q3

EdexcelOld spec6 marksIntegration

(a) \[y = 5^x + \log_2(x + 1), \qquad 0 \leqslant x \leqslant 2\]

Complete the table below, by giving the value of \(y\) when \(x = 1\)

\(x\)	0	0.5	1	1.5	2
\(y\)	1	2.821		12.502	26.585

(1)

(b) Use the trapezium rule, with all the values of \(y\) from the completed table, to find an approximate value for\[\int_0^2 \left(5^x + \log_2(x + 1)\right)\mathrm{d}x\]giving your answer to 2 decimal places. (4)

(c) Use your answer to part (b) to find an approximate value for\[\int_0^2 \left(5 + 5^x + \log_2(x + 1)\right)\mathrm{d}x\]giving your answer to 2 decimal places. (1)

Scheme	Marks
\(\{\text{At } x = 1,\}\ y = 6\) (allow 6.000 or even 6.00)	B1 cao
	(1)

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2} \times 0.5\) ;	B1 oe
\(\underline{\left\{1 + 26.585 + 2\left(2.821 + \text{their } 6 + 12.502\right)\right\}}\) For structure of \(\{\ldots\ldots\ldots\ldots\}\) ;	M1A1ft
\(\tfrac{1}{2} \times 0.5\,\underline{\left\{1 + 26.585 + 2\left(2.821 + 6 + 12.502\right)\right\}}\ \left\{= \tfrac{1}{4}(70.231) = 17.557..\right\} =\) awrt 17.56	A1
	(4)

Notes

B1: for using \(\tfrac{1}{2} \times 0.5\) or \(\tfrac{1}{4}\) or equivalent.

M1: requires the correct \(\{\ldots\ldots\}\) bracket structure. It needs the first bracket 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 2nd bracket this may be regarded as a slip and the M mark can be allowed (An extra repeated term forfeits the M mark however). M0 if values used in brackets are \(x\) values instead of \(y\) values

A1ft: for the correct bracket \(\{\ldots\ldots\}\) following through candidate’s \(y\) value found in part (a).

A1: for answer which rounds to 17.56

NB: Separate trapezia may be used: B1 for 0.25, M1 for \(1/2\,h(a + b)\) used 3 or 4 times (and A1ft if it is all correct ) Then A1 as before.

Special case: Bracketing mistake \(0.25 \times (1 + 26.585) + 2\left(2.821 + \text{their } 6 + 12.502\right)\) scores B1 M1 A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). An answer of 49.542 usually indicates this error.

Mark scheme (c)

Scheme	Marks
\(10 + \text{“}17.56\text{”} = \text{“}27.56\text{”}\)	B1ft
	(1)
	[6]

Notes

B1ft: \(10 +\) their answer to part (b)
(May be obtained by using the trapezium rule again with all values for \(y\) increased by 5)