Monotonic functions?

**fardeen_gen** · May 11th 2009, 10:28 PM

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

**Plato** · May 12th 2009, 03:25 AM

Originally Posted by fardeen_gen

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

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

Also what does "increases with increase in" mean?

**fardeen_gen** · May 12th 2009, 04:51 AM

I have no idea what the question means! Thats why I posted it.

**pankaj** · May 12th 2009, 09:05 AM

You are right Plato.

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

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

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

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

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

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

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

Hence, $F'(t)>0$

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

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

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