Q: Let a + b = 4, where a < 2, and let g(x) be a differentiable function. If dg/dx > 0 for all x , Prove that $\displaystyle \int_{0}^{a} g(x) dx + \int_{0}^{b} g(x) dx $ increases as (b – a) increases.
Sol. Let b – a = t
given a + b = 4
so a = 2 -t/2 , b = 2 + t/2
Let $\displaystyle f(t) = \int_{0}^{a} g(x) dx + \int_{0}^{b} g(x) dx $
$\displaystyle f(t) = \int_{0}^{2-\frac{t}{2}} g(x) dx + \int_{0}^{2+\frac{t}{2}} g(x) dx $
$\displaystyle f'(t) = g(2-\frac{t}{2})(-\frac{1}{2}) + g(2+\frac{t}{2})(\frac{1}{2}) $
$\displaystyle f'(t) = \frac{1}{2}[g(b)- g(a)]$
Since $\frac{dg}{dx} > 0$ for all x, so g(x) is increasing since b > a
g(b) > g(a)
Hence, f'(t) > 0
f(t) increasing as t increases
i.e. f(t) increases as (b – a) increases