Question

Find the integral of $\left[{\tan }^{-1}x\right]dx$ from $0$ to $102$, where $\left[x\right]$ is the largest integer not exceeding $x$.

Answer 1

Find the integral of the given function

Given integration: ${\int }_0^{102}\left[{\tan }^{-1}x\right]dx$

${\int }_0^{102}\left[{\tan }^{-1}x\right]dx={\int }_0^{\tan \left(1\right)}\left[{\tan }^{-1}xdx\right]+{\int }_{\tan \left(1\right)}^{102}\left[{\tan }^{-1}xdx\right]$

As the greatest integer value of ${\tan }^{-1}x$ in the interval $[0,\tan 1)$ is $0$ and the greatest integer value of ${\tan }^{-1}x$ in the interval $[\tan 1,102]$ is $1$.

So,

$$\begin{gathered} {\int }_0^{102}\left[{\tan }^{-1}x\right]dx={\int }_0^{\tan \left(1\right)}\left[0\right]+{\int }_{\tan \left(1\right)}^{102}\left[1\right] \\ =0+[102–(\tan 1)] \\ =[102–(\tan 1)] \\ \therefore {\int }_0^{102}\left[{\tan }^{-1}x\right]dx=[102–(\tan 1)] \end{gathered}$$

Hence, the integral of $\left[{\tan }^{-1}x\right]dx$ from $0$ to $102$ is $[102–(\tan 1)]$.