under which situation inequality between two integral yeilds pointwise inequality?

Question

suppose I have

$$\int{f(x)} < \int{g(x)}$$

when I can conclude this : $f(x_{0})< g(x_{0})$ for some $x_{0}$

or is there such a $x_{0}$? how I can find it?

if there is not answer, how about if $f(x)=(ax+bx^{2})c\sin(x)$ and $f(x)=(kx+lx^{2})f\sin(x)$ where $a, b, c, f, k, l$ are constant.

Answer 1

I'll assume you mean the standard Riemann integral, over a compact interval $[\alpha, \beta] \subset \Bbb{R}$. In this case, the statement is always true. The contrapositive might be easier and more obvious to prove, namely:

If $f,g:[\alpha, \beta] \to \mathbb{R}$ are Riemann integrable functions on $[\alpha,\beta]$, and for every $x \in [\alpha, \beta]$, $f(x) \geq g(x)$, then \begin{equation} \int_{\alpha}^{\beta} f \geq \int_{\alpha}^{\beta} g \end{equation}

It suffices to prove the special case that if $h: [\alpha , \beta] \to \Bbb{R}$ is non-negative, then $\int_{\alpha}^{\beta} h \geq 0$, because once we prove this, by choosing $h = f-g$, the general case follows. However, this is obvious, because all the Riemann sums are non-negative, hence any limiting process also preserves the inequality.