Question

If $[·]$ represents the greatest integer function, then the value of $\left|\underset{0}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}\left[\left[x^2\right]-\cos x\right]dx\right|$ is

Answer 1

Apply integration property:

Since $\left[x+a\right],\text{If}a\in I\mathrm{then}\left[x\right]+a$

$I=\underset{0}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}\left(\left[x^2\right]dx+\left(-\cos x\right)\right)dx$

Check limit:

$\begin{array}{l}0<x<\sqrt{\frac{\pi }{2}}\\ 0<x^2<\frac{\pi }{2}\\ 0<x^2\le 1\text{and}1<x^2\le \frac{\pi }{2}\\ \left[x^2\right]=0,\text{and}\left[x^2\right]=1\end{array}$

$I=\underset{0}{\overset{1}{\int }}(0+[-\cos x])dx+\underset{0}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}1+[-\cos x]$

$\begin{array}{c}I=\underset{0}{\overset{1}{\int }}(0+[-\cos x])dx+\underset{0}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}1+[-\cos x]\\ =\underset{0}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}[-\cos x]dx+\underset{1}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}1\left[\begin{array}{l}\because -1\le \cos x\le 1\\ 0<x<\sqrt{\frac{\pi }{2}}\\ 0<\cos x<1\\ [-\cos x]=-1\end{array}\right]\\ =\underset{0}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}(-1)+\underset{1}{\overset{\sqrt{\frac{\pi }{2}}}{\int }}dx\\ ={[-x]}_0^{\sqrt{\frac{\pi }{2}}}+{\left[x\right]}_1^{\sqrt{\frac{\pi }{2}}}\\ =-\sqrt{\frac{\pi }{2}}+\sqrt{\frac{\pi }{2}}-1\\ =-1\\ \left|I\right|=1\end{array}$

Hence, the value of $\underset{\mathbf{0}}{\overset{\sqrt{\frac{\mathbf{\pi }}{\mathbf{2}}}}{\mathbf{\int }}}\mathbf{(}\left[x^2\right]\mathbf{d}\mathbf{x}\mathbf{+}\left(-cosx\right)\mathbf{)}\mathit{d}\mathit{x}\mathbf{=}\mathbf{1}$