Question

If $I=\underset{0}{\overset{x}{\int }}\left[\cos t\right]dt$, where $x\in \left[\right(4n+1)\frac{\pi }{2},(4n+3\left)\frac{\pi }{2}\right],n\in N$ and $[\cdot ]$ represents greatest integer function, then the value of $I$ is

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

Answer 1

The correct option is C $\frac{\pi }{2}(2n+1)-x$

$I=\underset{0}{\overset{x}{\int }}\left[\cos t\right]dt=\underset{0}{\overset{2n\pi }{\int }}\left[\cos t\right]dt+\underset{2n\pi }{\overset{x}{\int }}\left[\cos t\right]dt=n\underset{0}{\overset{2\pi }{\int }}\left[\cos t\right]dt+\underset{2n\pi }{\overset{2n\pi +\pi /2}{\int }}\left[\cos t\right]dt+\underset{2n\pi +\pi /2}{\overset{x}{\int }}\left[\cos t\right]dt=2n\left[\underset{0}{\overset{\pi /2}{\int }}\left(0\right)dt+\underset{\pi /2}{\overset{\pi }{\int }}(-1)dt\right]+\underset{2n\pi }{\overset{2n\pi +\pi /2}{\int }}\left(0\right)dt+\underset{2n\pi +\pi /2}{\overset{x}{\int }}(-1)dt=-n\pi -[x-(2n\pi +\frac{\pi }{2})]=-n\pi +2n\pi +\frac{\pi }{2}-x\therefore I=(2n+1)\frac{\pi }{2}-x$