C2 June 2013 Q4

EdexcelOld spec7 marksIntegration

4.\[y = \frac{5}{(x^2 + 1)}\]

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

\(x\)	0	0.5	1	1.5	2	2.5	3
\(y\)	5	4	2.5		1	0.690	0.5

(1)

Figure 1: curve decreasing from y = 5 at x = 0, with the shaded region R under it from x = 0 to x = 3 — Figure 1

Figure 1 shows the region \(R\) which is bounded by the curve with equation \(y = \dfrac{5}{(x^2 + 1)}\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\)

(b) Use the trapezium rule, with all the values of \(y\) from your table, to find an approximate value for the area of \(R\). (4)

(c) Use your answer to part (b) to find an approximate value for\[\int_0^3 \left(4 + \frac{5}{(x^2 + 1)}\right)\mathrm{d}x\]giving your answer to 2 decimal places. (2)

Scheme	Marks
\(\{\text{At } x = 1.5,\}\ y = 1.538\) (only)	B1 cao
	(1)

Mark scheme (b)

Scheme	Marks
\(\dfrac{1}{2} \times 0.5\) ;	B1 oe
\(\underline{\left\{5 + 0.5 + 2\left(4 + 2.5 + \text{their } 1.538 + 1 + 0.690\right)\right\}}\) For structure of \(\{\ldots\ldots\ldots\ldots\ldots\}\) ;	M1A1ft
\(\tfrac{1}{2} \times 0.5 \times \underline{\left\{(5 + 0.5) + 2\left(4 + 2.5 + \text{their } 1.538 + 1 + 0.690\right)\right\}}\ \left\{= \tfrac{1}{4}(24.956) = 6.239\right\} =\) awrt 6.24	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 6.24.

NB: Separate trapezia may be used : B1 for 0.25, M1 for 1/2 \(h(a + b)\) used 5 or 6 times (and A1ft if it is all correct ) Then A1 as before.

Special case: Bracketing mistake \(0.25 \times (5 + 0.5) + 2\left(4 + 2.5 + \text{their } 1.538 + 1 + 0.690\right)\) 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 20.831 usually indicates this error.

Mark scheme (c)

Scheme	Marks
Adds Area of Rectangle or first integral \(= 3 \times 4\) or \(\left[4x\right]_0^3\) to previous answer	M1
So required estimate \(= \{\text{"}6.239\text{"} + 12 = \text{"}18.239\text{"}\} =\) "awrt 18.24" (or 12 + previous answer).	A1ft
N.B. \(7 \times 4\) + previous answer is M0A0 (added 4 seven times because 7 numbers in table)
	(2)
	[7]

Notes

M1: Relates previous answer ( not integral of previous answer) to this question by integrating 4 between limits, and adding, or by using geometry to find rectangle and adding.

A1ft: for 12 + answer to (b)

Alternative method (c)

Those who do a trapezium rule for part (b)- using the table from (a) with 4 added to each cell of the table Get: M1 for "\(their\ \tfrac{1}{4}\)"\(\times\underline{\left\{9 + 4.5 + 2\left(8 + 6.5 + \text{their } 5.538 + 5 + 4.690\right)\right\}} =\) (structure must be correct – allow one copying error only)
And A1ft: for awrt 18.24 (or 12 + previous answer).