A2 June 2023 Q1

EdexcelCurrent spec5 marksNumerical Methods

(a) Use Simpson’s rule with 4 intervals to find an estimate for\[\int_0^2 \mathrm{e}^{\sin^2 x}\,\mathrm{d}x\]Give your answer to 3 significant figures. (4)

Given that \(\displaystyle \int_0^2 \mathrm{e}^{\sin^2 x}\,\mathrm{d}x = 3.855\) to 4 significant figures,

(b) comment on the accuracy of your answer to part (a). (1)

Mark scheme (a)

Scheme

Marks

Step length \(= 0.5\)

1.1b

\(x\)	0	0.5	1	1.5	2
\(y\)	\(y_0\)	\(y_1\)	\(y_2\)	\(y_3\)	\(y_4\)
	\(\mathrm{e}^0\)	\(\mathrm{e}^{\sin^2 \frac{1}{2}}\)	\(\mathrm{e}^{\sin^2 1}\)	\(\mathrm{e}^{\sin^2 \frac{3}{2}}\)	\(\mathrm{e}^{\sin^2 2}\)
	1	1.258…	2.030…	2.704…	2.286…

1.1b

\(\displaystyle \int_0^2 \mathrm{e}^{\sin^2 x}\,\mathrm{d}x \approx \frac{0.5}{3}\left\{y_0 + 4y_1 + 2y_2 + 4y_3 + y_4\right\} = \frac{0.5}{3} \times \{23.198\ldots\}\ (= 3.8664\ldots)\)

1.1b

\(= 3.87\)

A1cao

1.1b

(4)

Notes

B1: Correct step length of 0.5 which may be implied e.g. by their 0, 0.5, etc.

M1: Attempts to find \(y\) values for all their \(x\) values – may be in terms of e or numerical values. Must be trying to find at least 3 values (e.g. if step length 1 is used). If substitution is not seen at least 2 of the values other than 1 must be correct to at least 2 s.f. rounded or truncated.

M1: Correct application of Simpson’s rule: \(\dfrac{h}{3}\)(ends + 2evens + 4odds) (must have an odd number of ordinates). Must be \(y\) values not \(x\) values.

A1cao: 3.87. Must be to 3 s.f.

Note: must see evidence of Simpson’s Rule so answer with no working scores no marks.

Mark scheme (b)

Scheme	Marks	AO
eg It is accurate to 2 significant figures	B1	3.2b
	(1)
	(5 marks)

Notes

B1: Must have awrt 3.87 for part (a), makes a sensible comment regarding the accuracy. Must relate to the accuracy, not just the value.

It is correct to 2 significant figures
It is correct to 1 decimal place
It is not correct to 3 significant figures
It is awrt 99.6% accurate (using 3.87) or 99.7% accurate (using 3.866…)
Percentage error is awrt 0.3%/0.4% so very accurate
Very accurate as only out by 0.015 (requires reference to accuracy and evidence)

Do not accept e.g. it is out by 0.015 with no reference to what this means about the accuracy. Do not accept e.g. “very close to” as evidence, some quantification (absolute error, number of s.f. it agrees to) must be given.