1. ## Integration

Given $f(x)$ is an odd function and defined everywhere,periodic with period $2$ and integrable on every interval.

Let $g(x)=\int_{0}^xf(t)dt$.

Then prove that $g(2n)=0$ for every integer $n$.

2. Originally Posted by pankaj
Given $f(x)$ is an odd function and defined everywhere,periodic with period $2$ and integrable on every interval.

Let $g(x)=\int_{0}^xf(t)dt$.

Then prove that $g(2n)=0$ for every integer $n$.
Given that $f$ is defined everywhere and integrable on every interval we can consider

$I = \int _{-1}^{1} f(t) dt$

$f$ being odd implies (easy to prove if you're not allowed to state) $I = 0$

Now you can see that a simple substitution should give you that $I = g(2)$.

Using the periodicity of f you should can show that any integral of f over an interval of length 2 has the same value. Therefore

$g(2) = \int_0^2 f(t) dt = \int_{2}^4 f(t) dt = \int_4^6 f(t) = \ldots = \int_{2n-2}^{2n} f(t) dt = 0$

Adding together these integrals gives g(2n) = 0

Hope this helps.