Year: 2025
Paper: 3
Question Number: 1
Course: LFM Pure
Section: Integration
The majority of candidates focused solely on the pure questions, with questions 1, 2 and 8 the most popular. The statistics questions were more popular than the mechanics questions in this exam series. Candidates who did well on this paper generally: were careful to explain and justify the steps in their arguments, explaining what they had done rather than expecting the examiner to infer what had been done from disjointed groups of calculations; paid close attention to what was required by the questions; made fewer unnecessary mistakes with calculations; thought carefully about how to present rigorous arguments involving trig functions and their inverse functions, especially in relation to domain considerations; understood that questions set on the STEP papers require sufficient justification to earn full credit; knew the difference between 'positive' and 'non-negative'; attempted all parts of a question, picking up marks for later parts even when they had not necessarily attempted or completed previous parts. Candidates who did less well on this paper generally: did not pay attention to 'Hence' instructions: this means that you must use the previous part; presented explanations that were not precise enough (e.g. in Question 3 describing the transformations but not in the context of the graphs or in Question 8 not explaining use of trigonometric relationships sufficiently well); made additional assumptions, e.g. that a function was differentiable when this had not been given; tried to present if and only if arguments in a single argument when dealing with each direction separately would have been more appropriate and safer (note that this is not always the case; in general candidates need to consider what is the most appropriate presentation of an if and only if argument); tried to carry out too many steps in one go, resulting in them not justifying the key steps sufficiently; did not take sufficient care with graphs/curve sketching.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\textit{You need not consider the convergence of the improper integrals in this question.}
For $p, q > 0$, define
$$b(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx$$
\begin{questionparts}
\item Show that $b(p,q) = b(q,p)$.
\item Show that $b(p+1,q) = b(p,q) - b(p,q+1)$ and hence that $b(p+1,p) = \frac{1}{2}b(p,p)$.
\item Show that
$$b(p,q) = 2\int_0^{\pi/2} (\sin\theta)^{2p-1}(\cos\theta)^{2q-1} \, d\theta$$
Hence show that $b(p,p) = \frac{1}{2^{2p-1}}b(p,\frac{1}{2})$.
\item Show that
$$b(p,q) = \int_0^\infty \frac{t^{p-1}}{(1+t)^{p+q}} \, dt$$
\item Evaluate
$$\int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt$$
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1}\, \d x \\
u = 1-x, \d u = -\d x && &= \int_{u=1}^{u = 0} (1-u)^{p-1}u^{q-1} (-1) \, \d u \\
&&&= \int_0^1 (1-u)^{p-1}u^{q-1} \d u \\
&&&= \int_0^1 u^{q-1}(1-u)^{p-1} \d u \\
&&&= b(q,p)
\end{align*}
\item \begin{align*}
b(p+1,q) + b(p,q+1) &= \int_0^1 x^p(1-x)^{q-1} \d x + \int_0^1 x^{p-1}(1-x)^{q} \d x \\
&= \int_0^1 \left (x^p(1-x)^{q-1} + x^{p-1}(1-x)^{q}\right) \d x \\
&= \int_0^1 x^{p-1}(1-x)^{q-1} \left (x + (1-x) \right) \d x \\
&= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\
&= b(p,q)
\end{align*}
Therefore $b(p+1,q) = b(p,q) - b(p,q+1)$, in particular $2b(p+1,p) = b(p+1,p)+b(p,p+1) = b(p,p) \Rightarrow b(p+1,p) = \frac12 b(p,p)$ as required.
\item \begin{align*}
&& b(p,q) &= \int_0^1 x^{p-1} (1-x)^{q-1} \d x \\
x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta && &= \int_{u=0}^{u = \pi/2} \sin^{2p-2} \theta (1-\sin^2 \theta)^{q-1} \cdot 2 \sin \theta \cos \theta \d \theta \\
&&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-2} \cos \theta \d \theta \\
&&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-1} \theta \d \theta
\end{align*}
\begin{align*}
b(p,p) &= 2\int_0^{\pi/2} (\sin \theta)^{2p-1}(\cos \theta)^{2p-1} \d \theta \\
&= 2 \int_0^{\pi/2} \left (\frac12 \sin 2\theta \right)^{2p-1} \d \theta \\
&= \frac1{2^{2p-1}} 2 \int_0^{\pi/2} (\sin 2 \theta)^{2p-1} \d \theta \\
&= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi} (\sin x)^{2p-1} 2 \d x\\
&= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi/2} (\sin x)^{2p-1} \d x\\
&= \frac1{2^{2p-1}} 2 \int_{0}^{\pi/2} (\sin x)^{2p-1} (\cos x)^{0} \d x\\
&= \frac1{2^{2p-1}} b(p,\tfrac12)
\end{align*}
\item \begin{align*}
&&b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\
t = \frac{x}{1-x}, \d t = (1-x)^{-2} \d x &&&= \int_{t=0}^{t = \infty} \left ( \frac{t}{1+t} \right)^{p-1} \left ( 1-\frac{t}{1+t} \right)^{q+1} \d t\\
x = \frac{t}{1+t} && &=\int_0^\infty t^{p-1} (1+t)^{-(p-1)-(q+1)} \d t \\
&&&= \int_0^{\infty} \frac{t^{p-1}}{(1+t)^{p+q}} \d t
\end{align*}
\item
\begin{align*}
I &= \int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt \\
&= b( \tfrac52, \tfrac72) \\
&= b( \tfrac52, \tfrac52+1) \\
&= \tfrac12 b( \tfrac52, \tfrac52) \\
&= \frac12 \cdot \frac1{2^{4}} b(\tfrac52, \tfrac12) \\
&= \frac{1}{2^5} \cdot 2 \int_0^{\pi/2} (\sin \theta)^{4} \d \theta \\
&= \frac1{2^4} \int_0^{\pi/2}\left (\frac{1-\cos 2 \theta}{2} \right)^2 \d \theta \\
&= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \cos^{2} 2 \theta \right) \d \theta \\
&= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \frac{\cos 4 \theta + 1}{2} \right) \d \theta \\
&= \frac1{2^6} \left [\frac32 \theta - \sin 2 \theta + \frac18 \sin 4 \theta \right]_0^{\pi/2} \\
&= \frac1{2^6} \frac{3 \pi}{4} \\
&= \frac{3 \pi}{2^8}
\end{align*}
\end{questionparts}
This question was the most popular question in terms of the number of attempts, and it was generally well done. Some candidates spent significant time attempting methods involving integration by parts in the early parts of this question which did not work. In part (iii) the most common method in the 'Hence show that...' involved using the substitution u = 2θ at some point. After this the integral looks like what is required in the given answer but with the limits 0 to π rather than 0 to π/2. Candidates needed to point out that the symmetry in the integral enabled the limits to be changed back to 0 to π/2 with the appearance of a factor of 2. Part (iv) was generally well done by those who got that far. In part (v) some marks were given for piecing together the earlier results to get to an integral that is relatively easy to calculate. Some candidates did not provide sufficient justification to be awarded full credit. A good number of candidates got to a final correct value of the integral. Some candidates had success with alternative methods involving more difficult integration having not made so much use of the properties of b to simplify the calculation.