Question

If $I=\underset{0}{\overset{x}{\int }}\left[\sin t\right]dt$, where $x\in (2n\pi ,(2n+1\left)\pi \right),n\in N$ and $[\cdot ]$ denotes the greatest integer function, then the value of $I$ is

• A. $-n\pi$
• B. $-(n+1)\pi$
• C. $-2n\pi$
• D. $-(2n+1)\pi$

Answer 1

The correct option is A $-n\pi$

Let $I=\underset{0}{\overset{x}{\int }}\left[\sin t\right]dt$
$\Rightarrow I=\underset{0}{\overset{2n\pi }{\int }}\left[\sin t\right]dt+\underset{2n\pi }{\overset{x}{\int }}\left[\sin t\right]dt$

We know that the period of $\left[\sin t\right]$ is $2\pi$, so using
$\underset{0}{\overset{nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx\Rightarrow I=n\underset{0}{\overset{2\pi }{\int }}\left[\sin t\right]dt+\underset{2n\pi }{\overset{x}{\int }}\left[\sin t\right]dt\Rightarrow I=n\left[\underset{0}{\overset{\pi }{\int }}\left(0\right)dt+\underset{\pi }{\overset{2\pi }{\int }}(-1)dt\right]+\underset{2n\pi }{\overset{x}{\int }}\left(0\right)dt\therefore I=-n\pi$