The following is from Bruckner's Real Analysis book (Ex.12:8.3):
Let $g(x)=x^2$ on $[−1, 1]$, and let $\mathcal{F} = {\{f \in \mathcal{C}[−1,1] : |f(x)| ≤ 5 \ \text{for all} \ x∈[−1,1]}\}$. Calculate $\sup_{f \in \mathcal{F}} \int_{-1}^1 f dg$.
I don't know any method of solving this, but I write some primitive thoughts. The integral becomes $\int_{-1}^1 2x f(x) dx$. $\sup$ cannot be for $f$ an even function of $x$. The integrand is bounded by $10|x|$. Because integration is on $[-1,1]$ thus the higher power polynomial the less value of the integral...
From exercise 12:8.2, we know that $$ \left | \int_a^b f dg \right|\leq \|f\|_\infty \|g\|$$
For the current case, $\|g\| = V(g, [-1, 1]) = 2$ and $\|f\|_\infty =5$. So we have $$ \left | \int_{-1}^1 f dg \right|\leq 10$$ that means $$ -10 \leq \int_{-1}^1 f dg \leq 10 \tag{1}$$
Now, you have already shown that $\int_{-1}^1 f dg = \int_{-1}^1 2x f(x) dx$. So, if we pick $f(x) = 5 \ \text{sgn}(x)$, we have
$$\int_{-1}^1 f dg = \int_{-1}^1 2x f(x) dx = \int_{-1}^1 10 |x| dx =10 \tag{2}$$
From $(1)$ and $(2)$, we have that $\sup_{f \in \mathcal{F}} \int_{-1}^1 f dg = 10$.
Remark: In the current case, to calculate $V(g, [-1, 1])$, note that $g$ is monotone on $[-1,0]$ and monotone on $[0,1]$, So
$$V(g, [-1, 0])= |g(0) - g(-1)| = 1 $$ and $$V(g, [0, 1])= |g(1) - g(0)| = 1 $$ Finally, $$V(g, [-1, 1]) =V(g, [-1, 0])+V(g, [0, 1])=2$$