UNSW Maths Tutor Audition

For this problem, I began by writing down the question on the board while reading it aloud, ensuring that the students were engaged from the start. I then prompted them to recall the two versions of the Fundamental Theorem of Calculus:

1. If \(F(x) = \int_a^x f(t)\;dt\) then \(F'(x) = f(x)\).
2. \(\int_a^b f(t)\; dt = F(b) – F(a)\).

I highlighted that in the first version, the lower limit \(a\) is a constant, while in the second version, students could connect it to their high school experience of definite integrals. This contextualisation helps bridge their prior knowledge with the problem at hand.

I posed a question to the audience, asking why the first version of the Fundamental Theorem of Calculus couldn’t be applied directly to the problem. While pointing out the discrepancy between the lower limit in the problem, \(x\), and the constant lower limit, \(a\), in the theorem, an audience member correctly noted that the issue lies in the lower limit not being a constant.

Now by applying the second version of the theorem, we get:

\( y = F(x^{3}) – F(x)\), where \(F(x) = \int_a^x \sin(t^2)\;dt\).

Now we can apply the first version of the theorem, yielding:

\(\begin{align*}\frac{dy}{dx} &= F'(x^{3}) 3x^2 – F'(x)\\& = 3x^2\sin(x^6) – \sin(x^2).\end{align*}\)

During this explanation, I also answered questions from the audience about certain details ranging from why the chain rule had to be used to obtain the \(3x^2\) factor, and how to substitute \(x^3\) into \(\sin(x^2)\). Again, audience engagement is highly important.

To start, I asked the audience who had studied Mathematics Extension II or IB HL Mathematics during high school. This helped to establish rapport and a shared foundation. For those who had taken these courses, I pointed out that our approach to this question would likely differ from how it was taught in their schools, encouraging them to stay engaged. This strategy helped maintain the attention and interest of all students in the room.

The way the UNSW notes present this proof is with an emphasis on the Euler form:

\(\begin{align*}e^{in\theta}&=\cos(n\theta)+i\sin(n\theta)\\e^{-in\theta}&=\cos(-n\theta)+i\sin(-n\theta)\\&=\cos(n\theta)-i\sin(\theta)\end{align*}\)

I talked through how this then gives:

\(\cos(n\theta) = \frac{1}{2}\left(e^{in\theta}+e^{-in\theta}\right)\)

To find \(cos^4\theta\), we can substitute the above identity in:

\(\begin{align*}\cos^4\theta & = \left(\frac{1}{2}(e^{i\theta}+e^{-i\theta})\right)^4\\&=\frac{1}{16}(e^{4i\theta} + 4e^{2i\theta} + 6 + 4e^{-2i\theta} + e^{-4i\theta})\\&=\frac{1}{16}(e^{4i\theta}+e^{-4i\theta}+4(e^{2i\theta}+e^{-2i\theta}) +6)\\&=\frac{1}{16}(2\cos(4\theta)+8\cos(2\theta)+6)\\&=\frac{1}{8}\cos(4\theta)+\frac{1}{2}\cos(2\theta)+\frac{3}{8}\end{align*}\)

During this, I explained each step with detail – pointing students towards binomial theorem and advising them to revise it if it was a bit unfamiliar, answering questions about binomial theorem and its symmetrical nature, and also testing a particular value of \(\theta\) to see that the formula does indeed work.

I then went beyond the question and asked them to think about how we can apply this technique to solving the integral:

\[\int \cos^4\theta\;d\theta\]

and any higher powers as well.