I will teach a analysis class to some olympiad students this month. The subject is the fundamental theorem of algebra. My approach will be to prove the following "modified" version of the bounded convergence theorem:
Let $\{f_n\}$ be a sequence of Riemann Integrable functions such that $f_n:[a,b]\rightarrow\mathbb{R}$ and $|f_n(x)|<M$ for all $n\geq1$ with $M>0$. Suppose that $f_n\to f$ pointwise where $f:[a,b]\rightarrow\mathbb{R}$ is Riemann Integrable. Then $$\lim\limits_{n\to\infty}\int_a^bf_n(x)\,\mathrm{d}x = \int_a^bf(x)\,\mathrm{d}x.$$
Then I will show that this implies in
Let $f(x,t)$ be integrable in $x$ for each $t$ and have a bounded partial derivative $f_t(x,t)$ which is integrable in $x$ for each $t$. Then $$\frac{\mathrm{d}}{\mathrm{d}t} \int_a^b f(x,t)\mathrm{d}x = \int_a^b f_t(x,t)\mathrm{d}x.$$
Last, but not least, I will show that this implies in the fundamental theorem of algebra.
The class knows a lot about elementary calculus but doesn't know anything about the Lebesgue integral. I want to know what is the best/easiest proof of this modified version of the bounded convergence theorem.
Thanks.
EDIT: In rough terms, the idea is as follows:
Let $P$ be a polynomial, define $g:[0,\infty)\times[0,2\pi]\to\mathbb{C}, \:\: g(r,\theta)=\displaystyle\frac{1}{P(re^{i\theta})}$ and $F:[0,\infty)\to\mathbb{C}, \quad F(r)=\int_0^{2\pi}g(r,\theta)\mathrm{d}\theta$. Suppose $P(z)\neq0$ for all $z\in\mathbb{C}$.
Differentiating $F$ under the integral sign (I want the bounded convergence theorem to justify this), we have that
$$riF'(r)=\int_0^{2\pi}\frac{-rie^{i\theta}P'(re^{i\theta})}{[P(re^{i\theta})]^2}\mathrm{d}\theta=\int_0^{2\pi}\frac{\partial g}{\partial \theta}(r,\theta)\mathrm{d}\theta=g(r,2\pi)-g(r,0)=0.$$
This implies $F(r)=\frac{2\pi}{P(0)}\neq0, \quad \forall r\geq0$.
On the other side, $|P(z)|\to\infty$ when $|z|\to\infty$ implies in $g(r,\theta)\to0$ when $r\to\infty$. Then $F(r)\to 0$ when $r\to\infty$, which is absurd.
I am having trouble justifying the use of the Leibniz Rule.