# Monotonic functions?

• May 11th 2009, 10:28 PM
fardeen_gen
Monotonic functions?
Let $\displaystyle a + b =4$, $\displaystyle a < 2$ and $\displaystyle g(x)$ be monotonically increasing function of x. Prove that $\displaystyle f(x) = \int_{0}^{a} g(x)\ dx + \int_{0}^{b} g(x)\ dx$ increases with increase in $\displaystyle (b - a)$.
• May 12th 2009, 03:25 AM
Plato
Quote:

Originally Posted by fardeen_gen
Let $\displaystyle a + b =4$, $\displaystyle a < 2$ and $\displaystyle g(x)$ be monotonically increasing function of x. Prove that $\displaystyle \color{red}f(x) = \int_{0}^{a} g(x)\ dx + \int_{0}^{b} g(x)\ dx$ increases with increase in $\displaystyle (b - a)$.

You need to repair this post.
As written $\displaystyle f(x)$ is constant.

Also what does "increases with increase in" mean?
• May 12th 2009, 04:51 AM
fardeen_gen
I have no idea what the question means! Thats why I posted it.
• May 12th 2009, 09:05 AM
pankaj
You are right Plato.

The question says that prove that $\displaystyle \int_{0}^{a}g(x)dx+\int_{0}^{b}g(x)dx$ increases as $\displaystyle (b-a)$ increases.

Let $\displaystyle b-a=t$.
Therefore,$\displaystyle a=\frac{4-t}{2}$
$\displaystyle b=\frac{4+t}{2}$

Let $\displaystyle F(t)=\int_{0}^{a}g(x)dx+\int_{0}^{b}g(x)dx=\int_{0 }^{\frac{4-t}{2}} g(x)dx+\int_{0}^{\frac{4+t}{2}} g(x)dx$

$\displaystyle F'(t)=g(\frac{4-t}{2})(-\frac{1}{2})+g(\frac{4+t}{2})(\frac{1}{2})$
(Using Leibnitz rule)

$\displaystyle F'(t)=\frac{1}{2}(g(b)-g(a))$

Given that $\displaystyle a<2$,therefore $\displaystyle b>2,i.e.b>a.$

Now,since $\displaystyle g(x)$ is monotonically increasing,therefore $\displaystyle g(b)>g(a)$

Hence,$\displaystyle F'(t)>0$

Therefore,$\displaystyle F(t)$ increases as $\displaystyle t$ increases

i.e.$\displaystyle \int_{0}^{a}g(x)dx+\int_{0}^{b}g(x)dx$ increases as $\displaystyle (b-a)$ increases.