I'm having some trouble with the properties of definite integrals, I understand that if $$m\leq f(x)\leq M \;\mathrm{on}\; [a,b] \;\mathrm{then,}\;$$ $$m(b-a)\leq \int_{a}^{b}f(x)\,dx\leq M(b-a)$$ but I don't know how to apply that to these questions that our professor set for us as a past quiz, no need for a detailed answer just wondering how you would approach a question like this (algebraically, geometrically, etc.) Any useful source on this topic?
$$\mathrm{Suppose\;two\;Riemann\;integrable\;functions}\; f,g:[a,b]\rightarrow \mathbb{R} \;\mathrm{satisfy}\; m\leq f(x)\leq M \;\mathrm{and}\; m\leq g(x)\leq M \;\mathrm{for\; all\;} x \in [a,b], \;\mathrm{which\; of\; the\; following\; are\; true?}$$
$$\mathrm{if\;} m\cdot M>0 \mathrm{\;then\;} \int_{a}^{b} \frac{f(x)}{g(x)}\,dx \geq \left |\frac{m}{M}\right |(b-a)$$
$$\mathrm{if\;} m >0 \mathrm{\;then\;} \int_{a}^{b} \frac{f(x)}{g(x)}\,dx \leq \frac{M}{m} (b-a)$$
$$\mathrm{if\;} M <0 \mathrm{\;then\;} \int_{a}^{b} \frac{f(x)}{g(x)}\,dx \leq \frac{M}{m}$$
$$\mathrm{if\;} m\cdot M<0 \mathrm{\;then\;} \int_{a}^{b} \frac{f(x)}{g(x)}\,dx \mathrm{\;does\;not\;exist}$$
I know that if the integrand is positive over an interval, the integral is also positive, and the property involving infimums and supremums of functions
you should try to find a proof or a counterexample.
I will do the first problem here: If $mM>0$ then either $m, M < 0$ or $m, M>0$. In either case $g$ is bounded away from zero and $\frac fg$ is well-defined. In this case $\frac fg$ is also integrable.
If $m, M < 0$ then $f,g < 0$ on the intervall or if $m, M > 0$ then $f, g > 0 $ on the intervall. Thus $\frac fg>0$ and thus $\frac fg = \frac{|f|}{|g|}$.
Now since $|f|\geq |m|$ and $|g|\leq |M|$ we have $\frac{f}{g}\geq \left|\frac mM\right|$. We can integrate that inequality on both sides over the interval $[a,b]$ to get the desired result.
The last line: When $mM < 0$ then this means that either $m<0$ and $M>0$ or $m>0$ and $M<0$. Since $m\leq M$ only the first case is possible. And in that case this means that $g$ might become zeros. Thus $\frac fg$ is not always well-defined. But there are cases in which $-5=m<\frac 12\leq f,g\leq M$. Both $f,g$ are positive and their quotient is well-defined and bounded. Thus the integral exists.
Furthermore even if $g$ might be zero at $t\in[a,b]$. $f$ might be such that the discontinuity is bounded (e.g. $f=2g$). In this case, even though $\frac fg$ is not well defined on the interval, we could get a continuous extension here or use the fact that $\frac fg$ can be undefined at finitely many points on the interval $[a,b]$.
This is just elementary arguing and should be straight-forward.