How to prove the second mean value theorem for definite integrals
Solution 1:
(ii) follows from (i) just by changing the orientation of the integration. More specifically, use (i) for $\tilde{g}(x)=g(a+b-x)$ and $\tilde{f}(x)=f(a+b-x)$. Note $\tilde{g}$ is decreasing when $g$ is increasing, applying (i), there is $e'\in[a,b]$ such that $\int_a^b\tilde{g}(x)\tilde{f}(x)dx=\tilde{g}(a)\int_a^{e'}\tilde{f}(x)dx$. So, $$\int_a^b g(a+b-x)f(a+b-x)dx=g(b)\int_a^{e'}f(a+b-x)dx.$$ After a change of variable $t=a+b-x$, we get $$\int_a^b g(t)f(t)dt=g(b)\int_{a+b-e'}^bf(t)dt.$$ Hence, $e=a+b-e'$ works.