# Tricky Calculus problem: Prove that the function increases as $b-a$ increases.

Problem: Let $a+b=4$, where $a<2$ and let $g(x)$ be a differentiable function. If $g'(x)>0$ $\forall x$, then show that $$I=\int^{a}_{0}g(x)dx+\int^{b}_{0}g(x)dx$$ increases as $(b-a)$ increases.

My Attempt: Let $$\phi(a)=I=\int^{a}_{0}g(x)dx+\int^{4-a}_{0}g(x)dx$$ $$\Rightarrow\phi'(a)=g(a)-g(4-a)$$ Now in order to show that $\phi(a)$ is increasing $\phi'(a)>0\Rightarrow g(a)>g(4-a)=g(b)$. But we know that $b>a$ and since $g(x)$ is an increasing function $g(b)>g(a)$, which is a contradiction. Where am I going wrong? Please explain.

• Your problem is that $(b-a)$ increasing is not the same as $a$ increasing. In fact, it's quite the opposite. As $a$ decreases, $(b-a)$ increases. Commented Oct 5, 2016 at 17:35
• Actually you must show that $\phi(a)$ is decreasing, because the problem ask you to show that $I$ increases when $b-a = 4 -2a$ increases, that is, when $a$ decreases. Thus, if $\phi'(a)<0$, then it's proved. Try make the same with $b$, It will be clearer. Commented Oct 5, 2016 at 17:36
• What you showed is correct, but you missed one point. Since $\phi$ is increasing as a function of $a$, it is decreasing as a function of $b-a=4-2a$. Specifically, let $u=b-a=4-2a$. Then $a(u)=(4-u)/2$, and so $\frac{d}{du} \phi(a(u))=\phi'(a(u)) a'(u) =-\frac 12 \phi'(a(u))\le 0$, as required. Commented Oct 5, 2016 at 17:39

Substitute $a=4-b$ in the first integral and find $d\phi/db$. It will come as $g(b)-g(4-b)$
$d\phi/db$=$g(b)-g(a)$ , where $g(b) > g(a)$
Now $b-a=b-4+b=2b-4$. As $b-a$ increases, this means $b$ increases and $a$ decreases. Moreover, $b>2$ which implies $2b-4>0$. Hence, $\phi '(b)>0$. Hence proved.
This means $\phi$ is increasing wrt $b$.
You have to make $b-a=:2u$ your working variable. Together with $a+b=4$ you have $$a=2-u,\quad b=2+u,$$ hence $u>0$. This gives $$I(u)=\int_0^{2-u} g(x)\>dx+\int_0^{2+u} g(x)\>dx\ ,$$ so that $$I'(u)=g(2+u)-g(2-u)=\int_{2-u}^{2+u}g'(x)\>dx>0\qquad(u>0)\ .$$