1. ## Integration

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

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

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

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

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

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

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

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

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

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

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

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

Hope this helps.