Question

$\underset{0}{\overset{102}{\int }}\left[{\tan }^{-1}x\right]dx$ is equal to
(where $[\cdot ]$ denotes the greatest integer function)

• A. $102-\tan 1$
• B. $101$
• C. $102+\tan 1$
• D. $102-\frac{\pi }{4}$

Answer 1

The correct option is A $102-\tan 1$

For, $0\le x\le \tan 1\Rightarrow 0\le {\tan }^{-1}x<1$ and $\tan 1\le x\le 102$
$\Rightarrow 1\le {\tan }^{-1}x<2$
$\Rightarrow \underset{0}{\overset{102}{\int }}\left[{\tan }^{-1}x\right]dx=\underset{0}{\overset{\tan 1}{\int }}0\cdot dx+\underset{\tan 1}{\overset{102}{\int }}1\cdot dx={x|}_{\tan 1}^{102}=102-\tan 1$