_0^5((x-[x/2])),dx, where [t] denotes greatest integer less than or equal to t, is equal t…

Question

$\displaystyle\int_0^5\cos\big(\pi(x-[x/2])\big)\,dx$, where $[t]$ denotes greatest integer less than or equal to $t$, is equal to:

Accepted Answer

We know,

$$\left[ {{x \over 2}} \right]$$ is discontinuous at 1, 2, 3, 4 ........

$$\therefore$$ [ x ] is discontinuous at 2, 4, 6, 8 .....

In between 0 to 5 it is discontinuous at 2 and 4.

Break the integration into 3 parts

(1) 0 to 2

(2) 2 to 4

(3) 4 to 5

$$\therefore$$ $$\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $$

$$ = \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } } $$

$$ = \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } } $$

$$ = \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } } $$

$$ = \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5$$

$= 0 - 0 + 0$

$= 0$

Solutions · 1