Question

Evaluate ${\int }_0^x\left[\cos t\right]dt$

where $nϵ(2n\pi ,(4n+1\left)\frac{\pi }{2}\right)$; $nϵN$ and $[.]$ denotes the greatest integer function

• A. $n\pi$
• B. $2n\pi$
• C. $-2n\pi$
• D. $-n\pi$

Answer 1

Answer: (D)

$-n\pi$

Let $I={\int }_0^x\left[\cos t\right]dt$
$={\int }_0^{2n\pi }\left[\cos t\right]dt+{\int }_{2n\pi }^x\left[\cos t\right]dt$
$=n{\int }_0^{2\pi }\left[\cos t\right]dt+{\int }_{2n\pi }^x\left[\cos t\right]dt$
$=n({\int }_0^{\frac{\pi }{2}}\left[\cos t\right]dt+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[\cos t\right]dt+{\int }_{\frac{3\pi }{2}}^{2\pi }\left[\cos t\right]dt)+{\int }_{2n\pi }^x\left[\cos t\right]dt$
$=n({\int }_0^{\frac{\pi }{2}}0dt+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}(-1)dt+{\int }_{\frac{3\pi }{2}}^{2\pi }0dt)+{\int }_{2n\pi }^{x}0dt$
$\therefore {\int }_0^x\left[\cos t\right]dt=-n\pi$