Question

Solve ${\int }_0^{100}\left[{\tan }^{-1}x\right]dx$

• A. $100+tan1$
• B. $100-tan1$
• C. $tan1$
• D. $99+tan1$

Answer 1

The correct option is B $100-tan1$

We know

$\frac{-\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}$

$-1.57\le {\tan }^{-1}x<1.57$

$0\le {\tan }^{-1}x\le 1.57$

Now

$0\le {\tan }^{-1}x<1\left[{\tan }^{-1}x\right]=0$

$0\le x\le \tan x$

$1\le {\tan }^{-1}x<\frac{\pi }{2}$

$\tan 1\le x\le 100$

${\int }_0^{100}\left[{\tan }^{-1}x\right]dx$

${\int }_0^{\tan 1}0dx+{\int }_{\tan 1}^{100}1dx$

${\left[x\right]}_{\tan 1}^{100}$

$100-\tan 1$

Option B