Question

The value of the integral: $\underset{1}{\overset{3}{\int }}\left[x\right]\cos \left(\frac{\pi }{2}\right(x-\left[x\right]\left)\right)dx$
where $\left[x\right]$ denotes the largest integer not exceeding $x$ is :

• A. $6$
• B. $\frac{1}{6}$
• C. $\frac{\pi }{6}$
• D. $\frac{6}{\pi }$

Answer 1

The correct option is D

$\frac{6}{\pi }$

${\int }_1^3\left[x\right]\cos \left(\frac{\pi }{2}\right(x-\left[x\right])dx$

$={\int }_1^2\cos \left(\frac{\pi }{2}\right(x-1\left)\right)dx+{\int }_2^{3}2\cos \left(\frac{\pi }{2}\right(x-2\left)\right)dx$

$={\left[\frac{\sin \left(\frac{\pi }{2}\right(x-1\left)\right)}{\frac{\pi }{2}}\right]}_1^2+2{\left[\frac{\sin \left(\frac{\pi }{2}\right(x-2\left)\right)}{\frac{\pi }{2}}\right]}_2$

$=\frac{2}{\pi }[1-0]+\frac{4}{\pi }[1-0]$

$=\frac{6}{\pi }$