Question

${\int }_0^{\pi /2}(x-[\cos x\left]\right)dx=$ ____________ where $\left[t\right]=$greatest integer less or equal to t.

• A. $\frac{{\pi }^2}{8}-1$
• B. $\frac{{\pi }^2}{8}-\frac{\pi }{8}$
• C. $\frac{{\pi }^2}{8}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A)

$\frac{{\pi }^2}{8}-1$

${\int }_0^{\pi /2}(x-[cosx\left]\right)dx=(\frac{x^2}{2}-[sinx]{)}_0^{\pi /2}$

$=\frac{1}{2}(\frac{\pi }{2}-0{)}^2-[sin\left(\frac{\pi }{2}\right)-sin\left(0\right)]$

$=\frac{{\pi }^2}{8}-[1-0]$

$=\frac{{\pi }^2}{8}-1$ since, [1.0]=1