C4 June 2013 (R) Q5

Notes

B1: 6.248 or awrt 6.248. Look for this on the table or in the candidate’s working.

Notes

B1: Outside brackets \(\dfrac{1}{2} \times 2\) or 1

M1: For structure of trapezium rule \(\left[\ \ldots\ldots\ldots\ldots\ \right]\). Allow one miscopy of their values.

A1: 49.37 or anything that rounds to 49.37

Note: It can be possible to award : (a) B0 (b) B1M1A1 (awrt 49.37)

Note: Working must be seen to demonstrate the use of the trapezium rule. Note: actual area is 50.828…

Bracketing mistake: Unless the final answer implies that the calculation has been done correctly,
Award B1M0A0 for \(1 + 3 + 2\left(7.107 + 7.218 + \text{their } 6.248\right) + 5.223\) (nb: answer of 50.369).

Alternative method for part (b): Adding individual trapezia

\(\text{Area} \approx 2 \times \left[\dfrac{3 + 7.107}{2} + \dfrac{7.107 + 7.218}{2} + \dfrac{7.218 + 6.248}{2} + \dfrac{6.248 + 5.223}{2}\right] = 49.369\)

B1: 2 and a divisor of 2 on all terms inside brackets.

M1: First and last ordinates once and two of the middle ordinates twice inside brackets ignoring the 2.

A1: anything that rounds to 49.37

Notes

M1: For \(4t\mathrm{e}^{-\frac{1}{3}t} \to \pm At\mathrm{e}^{-\frac{1}{3}t} \pm B\displaystyle\int \mathrm{e}^{-\frac{1}{3}t}\{\mathrm{d}t\},\ A \neq 0,\ B \neq 0\)

A1: For \(t\mathrm{e}^{-\frac{1}{3}t} \to \left(-3t\mathrm{e}^{-\frac{1}{3}t} - \displaystyle\int -3\mathrm{e}^{-\frac{1}{3}t}\right)\) (some candidates lose the 4 and this is fine for the first A1 mark).
or \(4t\mathrm{e}^{-\frac{1}{3}t} \to 4\left(-3t\mathrm{e}^{-\frac{1}{3}t} - \displaystyle\int -3\mathrm{e}^{-\frac{1}{3}t}\right)\) or \(-12t\mathrm{e}^{-\frac{1}{3}t} - \displaystyle\int -12\mathrm{e}^{-\frac{1}{3}t}\) or \(12\left(-t\mathrm{e}^{-\frac{1}{3}t} - \displaystyle\int -\mathrm{e}^{-\frac{1}{3}t}\right)\)
These results can be implied. They can be simplified or un-simplified.

B1: \(3 \to 3t\) or \(3 \to 3x\) (bod) .
Note: Award B0 for 3 integrating to \(12t\) (implied), which is a common error when taking out a factor of 4.
Be careful some candidates will factorise out 4 and have \(4\left(\ldots + \dfrac{3}{4}\right) \to 4\left(\ldots + \dfrac{3}{4}t\right)\)
which would then be fine for B1.
Note: Allow B1 for \(\displaystyle\int_0^8 3\,\mathrm{d}t = 24\)

A1: For correct integration of \(4t\mathrm{e}^{-\frac{1}{3}t}\) to give \(-12t\mathrm{e}^{-\frac{1}{3}t} - 36\mathrm{e}^{-\frac{1}{3}t}\) or \(4\left(-3t\mathrm{e}^{-\frac{1}{3}t} - 9\mathrm{e}^{-\frac{1}{3}t}\right)\) or equivalent.
This can be simplified or un-simplified.

dM1: Substitutes limits of 8 and 0 into an integrated function of the form of either \(\pm\lambda t\mathrm{e}^{-\frac{1}{3}t} \pm \mu\mathrm{e}^{-\frac{1}{3}t}\) or
\(\pm\lambda t\mathrm{e}^{-\frac{1}{3}t} \pm \mu\mathrm{e}^{-\frac{1}{3}t} + Bt\) and subtracts the correct way round.
Note: Evidence of a proper consideration of the limit of 0 (as detailed in the scheme) is needed for dM1.
So, just subtracting zero is M0.

A1: An exact answer of \(60 - 132\mathrm{e}^{-\frac{8}{3}}\). A decimal answer of 50.82818444... without a correct answer is A0.
Note: A decimal answer of 50.82818444... without a correct exact answer is A0.
Note: If a candidate gains M1A1B1A1 and then writes down 50.8 or awrt 50.8 with no method for substituting limits of 8 and 0, then award the final M1A0.

IMPORTANT: that is fine for candidates to work in terms of \(x\) rather than \(t\) in part (c).

Note: The "\(3t\)" is needed for B1 and the final A1 mark.

Notes

B1: 1.46 or awrt 1.46 or -1.46 or awrt -1.46.
Candidates may give correct decimal answers of 1.458184439... or 1.459184439...

Note: You can award this mark whether or not the candidate has answered part (c) correctly.