Create a Huge Problem

Question

I am wondering if any problems have been designed that test a wide range of mathematical skills. For example, I remember doing the integral $$\int \sqrt{\tan x}\;\mathrm{d}x$$ and being impressed at how many techniques (substitution, trig, partial fractions etc.) I had to use to solve it successfully.

I am looking for suggestions/contributions to help build up such a question. For example, one problem could have as its answer $\tan x$ which would then be used in the integral, and something about the answer to the integral could lead into the next part.

The relevant subjects would be anything covered in the first few years of an undergraduate degree in mathematics.

The main reason I ask is that I want to work on developing some more integrated ways to practice mathematics "holistically" which I feel is very lacking in the current educational model.

Answer 1

One thing is when the answer to the first parts leads into the next parts as you said, and I don't have experienced that, in particular.

One different thing is an exercise like the integral in your example, where several techniques must be applied in succession in order to solve. Thinking to this second model, comes to my mind the differential equation $$y''+3y'=t+2e^{-3t}$$ from an assignment back from my yr1 at BEng degree course.

Here the student must in succession:

1. solve the homogeneous equation;
2. solve $y''+3y'=t$ (with a polynomial in mind);
3. solve $y''+3y'=2e^{-3t}$, knowing how to face the fact that $-3$, in the exponent, is also a root of the characteristic equation;
4. apply superposition to express the complete solution.

Not trivial: if I had to solve your integral and my ODE in a test or exam, I would probably fail the year at Uni. Jokes (?) aside you can find other examples, more interesting than this one, in derivates, integrals and equations, where 3-4 different concepts/notions must be applied to really complete the task.

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Integrals at pages 5, 6 and 10 of this exercise paper need 2-3 notions (rather than a single one) to be applied in order to solve, even if I'd say they are not huge problems.

Answer 2

I don't know this is eligible answer or not but I think it's worth being shared. This solution I made when trying to evaluate $$ \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx $$ on Brilliant.org. These are the methods I use to evaluate that integral and post it there as a solution.

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Method 1:

Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\ &=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ &=\frac{2a}{\omega^2+a^2}. \end{align} $$ Next, the inverse Fourier transform of $F(\omega)$ is $$ \begin{align} f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega.\tag1 \end{align} $$ Now, rewrite $$ \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\mathbb{Re}\left(e^{2ix}\right)}{x^2+2^2}\,dx.\tag2 $$ Comparing $(2)$ to $(1)$ yield $t=2$ and $a=2$. Thus, $$ \begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx &=\frac{1}{2}\frac{\pi e^{-2\cdot|2|}}{2}\\ &=\frac{\pi}{4e^4}\\ \end{align} $$

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Method 2:

Note that: $$ \int_{y=0}^\infty e^{-(x^2+4)y}\,dy=\frac{1}{x^2+4}, $$ therefore $$ \int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx=\int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx $$ Rewrite $\cos2x=\Re\left(e^{-2ix}\right)$, then $$ \begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos2x}{x^2+4}\,dx\\ &=\frac{1}{2}\int_{-\infty}^{\infty}\int_{y=0}^\infty e^{-(x^2+4)y}\cos2x\,dy\,dx\\ &=\frac{1}{2}\int_{y=0}^\infty\int_{x=-\infty}^\infty e^{-(yx^2+2ix+4y)}\,dx\,dy\\ &= \int_{y=0}^\infty e^{-4y} \int_{x=-\infty}^\infty \frac{1}{2} e^{-(yx^2+2ix)}\,dx\,dy. \end{align} $$ In general $$ \begin{align} \int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=-\infty}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{x=-\infty}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ \end{align} $$ Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then $$ \begin{align} \int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=-\infty}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\ \end{align} $$ The last form integral is Gaussian integral that equals to $\sqrt{\dfrac{\pi}{a}}$. Hence $$ \int_{x=-\infty}^\infty e^{-(ax^2+bx)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right). $$ Thus $$ \frac{1}{2}\int_{x=-\infty}^\infty e^{-(yx^2+2ix)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(\frac{(2i)^2}{4y}\right)=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(-\frac{1}{y}\right). $$ Next $$ \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy. $$ In general $$ \begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\ &=2\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\ &=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ \end{align} $$ The trick to solve the last integral is by setting $$ I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv. $$ Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then $$ I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt. $$ Let $t=v\;\rightarrow\;dt=dv$, then $$ I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Adding the two $I_t$s yields $$ 2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then $$ I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}. $$ Thus $$ \begin{align} \int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ &=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}\\ \end{align} $$ and $$ \begin{align} \int_0^{\infty}\frac{\cos2x}{x^2+4}\,dx&=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy\\ &=\frac{\sqrt{\pi}}{2}\cdot\sqrt{\frac{\pi}{4}}e^{-2\sqrt{4\cdot1}}\\ &=\frac{\pi}{4e^4}. \end{align} $$ It took me hours to make a solution using method 2 because almost no one on that site knows Fourier transform (it is understandable since most of the user there are only school students, so they know nothing about this method) so I made a solution using standard methods. Other methods to solve it is using contour integral or residual theorem, but I'm not familiar with those methods.

Answer 3

In terms of an integral question, I tried to come up with a double integral for my students that would test many first year integral techniques. It requires both integration by parts and multiple subsitutions, as well as an understanding of double integral regions. Here it is -

Sketch the region below $y=\sqrt{\sin x}$ and above $\displaystyle y=\frac{2x}{\pi}$ in the first quadrant. Find and mark the lower intersection point, $a$, and the upper intersection point, $b$. The volume under the surface $f(x,y) = y$ over the region is given by $$ V=\int\limits^b_a \! \int\limits_{ 2x/\pi}^{\sqrt{\sin x}} y \ \mathrm{d}y \, \mathrm{d}x, $$ where $a$ and $b$ are the intersection points of the curves. Evaluate this double integral. Verify your result by switching the order of integration. Show all working.