2019 Paper 1 Q8

Year: 2019
Paper: 1
Question Number: 8

Course: LFM Pure
Section: Integration

Difficulty: 1500.0 Banger: 1500.0

Problem

The function \(f\) is defined, for \(x > 1\), by $$f(x) = \int_1^x \sqrt{\frac{t-1}{t+1}} dt.$$ Do not attempt to evaluate this integral.
  1. Show that, for \(x > 2\), $$\int_2^x \sqrt{\frac{u-2}{u+2}} du = 2f\left(\frac{1}{2}x\right).$$
  2. Evaluate in terms of \(f\), for \(x > 0\), $$\int_0^x \sqrt{\frac{u}{u+4}} du.$$
  3. Evaluate in terms of \(f\), for \(x > 5\), $$\int_5^x \sqrt{\frac{u-5}{u+1}} du.$$
  4. Evaluate in terms of \(f\) $$\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du.$$

Solution

  1. Let \(2t = u\), \begin{align*} \int_2^x \sqrt{\frac{u-2}{u+2}} du &= \int_{t=1}^{t=x/2} \sqrt{\frac{2t-2}{2t+2}}2 \d t \\ &= 2\int_{t=1}^{x/2} \sqrt{\frac{t-1}{t+1}} \d t \\ &= 2f\l\frac{x}{2}\r \end{align*}
  2. Let \(v = u-2\), \begin{align*} \int_0^x \sqrt{\frac{u}{u+4}} du &= \int_{v = 2}^{x+2} \sqrt{\frac{v-2}{v+2}} \d v \\ &= 2 f \l \frac{x+2}{2} \r \end{align*}
  3. Let \(v = u-2, \d v = \d u\) \begin{align*} \int_5^x \frac{u-5}{u+1} du &= \int_3^{x-2} \frac{v-3}{v+3} \d v \\ &= \int_1^{\frac{x-2}{3}} \frac{3t - 3}{3t+3} 3 \d t \\ &= 3 f \l \frac{x-2}{3} \r \end{align*}
  4. Let \(v = u^2, \d v = 2u \d u\)\begin{align*}\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du &= \int_1^2 \sqrt{\frac{u^2}{u^2+4}} u \d u \\ &= \int_1^4 \sqrt{\frac{v}{v+4}} \frac12 \d v \\ &= f \l \frac{4+2}{2} \r - f \l \frac{3}{2} \r \\ &= f(3) - f(\frac32) \end{align*}
Examiner's report
— 2019 STEP 1, Question 8
Mean: ~6.5 / 20 (inferred) ~28% attempted (inferred) Inferred ~28% from '~850 candidates' out of ~3000 entry. Inferred ~6.5/20: 350/850 (41%) scored ≤2; bimodal with full attempts earning high marks. Estimating 41%×1.5 + 59%×10 ≈ 6.5. Bounded 3.0-10.0.

This was a fascinating question to mark, particularly as many candidates couldn't seem to figure out what to do with integrals that they weren't required to integrate. Another confusing feature of the question for many was the rather different role played by the variable x … a lot of candidates had clearly not encountered the notion that x doesn't have to be the (dummy) variable of the integrand; even amongst those who did make good progress with the question, checking what happened to the upper limit was a major stumbling-block. Almost 850 candidates attempted this question, 350 of whom failed to score more than 2/20 on it. Those candidates who realised that this was a test of substitution integration generally coped very well with parts (i) and (ii), where the change-of-variable was more obvious; indeed, (ii) fell very readily to either one of two obvious tactics. Part (iii) required either a very keen appreciation of what was going on in the background or a careful build-up to the problem that started generally and then compared the result with the desired outcome. Quite a few candidates seemed to resort to guesswork and the accompanying working could often be vague, at best. Part (iv), requiring a quadratic substitution, proved a step too far for many candidates, who – in spirit if not on paper – gave up at this point. Many appreciated that a quadratic substitution was 'on the cards' but couldn't make the conceptual leap of turning the u² du involved into √u · u du so that the integrand could now be made to look more like the one in part (ii)'s integral. Overall, full attempts were usually accompanied by high marks.

In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document; all of which are available from the STEP and Cambridge Examinations Board websites. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions for several months beforehand. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.

Source: Cambridge STEP 2019 Examiner's Report · 2019-p1.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
The function $f$ is defined, for $x > 1$, by
$$f(x) = \int_1^x \sqrt{\frac{t-1}{t+1}} dt.$$
Do not attempt to evaluate this integral.
\begin{questionparts}
\item Show that, for $x > 2$,
$$\int_2^x \sqrt{\frac{u-2}{u+2}} du = 2f\left(\frac{1}{2}x\right).$$
\item Evaluate in terms of $f$, for $x > 0$,
$$\int_0^x \sqrt{\frac{u}{u+4}} du.$$
\item Evaluate in terms of $f$, for $x > 5$,
$$\int_5^x \sqrt{\frac{u-5}{u+1}} du.$$
\item Evaluate in terms of $f$
$$\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du.$$
\end{questionparts}
Solution source
\begin{questionparts}
\item Let $2t = u$, \begin{align*}
\int_2^x \sqrt{\frac{u-2}{u+2}} du &= \int_{t=1}^{t=x/2} \sqrt{\frac{2t-2}{2t+2}}2 \d t \\
&= 2\int_{t=1}^{x/2} \sqrt{\frac{t-1}{t+1}} \d t \\
&= 2f\l\frac{x}{2}\r
\end{align*} 
\item Let $v = u-2$, \begin{align*}
\int_0^x \sqrt{\frac{u}{u+4}} du &= \int_{v = 2}^{x+2} \sqrt{\frac{v-2}{v+2}} \d v \\
&= 2 f \l \frac{x+2}{2} \r
\end{align*}
\item Let $v = u-2, \d v = \d u$ \begin{align*}
\int_5^x \frac{u-5}{u+1} du &= \int_3^{x-2} \frac{v-3}{v+3} \d v \\
&= \int_1^{\frac{x-2}{3}} \frac{3t - 3}{3t+3} 3 \d t \\
&= 3 f \l \frac{x-2}{3} \r
\end{align*}
\item Let $v = u^2, \d v = 2u \d u$\begin{align*}\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du &= \int_1^2 \sqrt{\frac{u^2}{u^2+4}} u \d u \\
&= \int_1^4 \sqrt{\frac{v}{v+4}} \frac12 \d v \\
&=  f \l \frac{4+2}{2} \r -  f \l \frac{3}{2} \r \\
&= f(3) -  f(\frac32)
\end{align*}
\end{questionparts}