Let $f(x)\ge g(x)$ for every $x$ in $[a,b]$ and $f$ & $g$ are both bounded and Riemann integrable on [a,b]. At a point $c\in[a,b]$, let $f$ and $g$ be continuous and $f(c)>g(c)$ then prove that $\int_a^b f(x) \ dx>\int_a^b g(x) \ dx$ and hence show that $$\frac{-1}{2}<\int_a^b\frac{x^3\cos {5x}}{2+x^2}<\frac{1}{2}$$
My work
I have been able to prove the first part of the problem. My problem is to prove the inequality.
$$|f(x)|=|\frac{x^3\cos {5x}}{2+x^2}|\le |\frac{x^3}{2+x^2}|\le |\frac{x^3}{x^2}|=|x|$$ Taking this as $|x|=g(x)$
So now $$f(x)\le g(x)$$
Now $$\int_a^bg(x) \ dx=\int_a^b |x| \ dx=\frac{b^2-a^2}{2}$$
According to the result obtained in first part,
$$|\int_a^b\frac{x^3\cos {5x}}{2+x^2}|\le\frac{b^2-a^2}{2}$$
$$\implies\frac{a^2-b^2}{2}\int_a^b\frac{x^3\cos {5x}}{2+x^2}\le\frac{b^2-a^2}{2}$$
So I would be able to prove the result if $a=0$ & $b=1$
But how can I prove it for any interval $(a,b)$?
