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The question was to find $$\int_{1}^{100} \lfloor \arctan x \rfloor dx $$. I hesitated because I learnt from illustrations in my book that when there is step up function, it is compulsory to break it at integral limits. Did that mean I've to break the limits $\int_{1}^{2} \cdots \int_{99}^{100}$? I always had problem what that means-$\arctan 100^\circ \quad \text{or}\quad \arctan 100$. If it was later, I felt very difficult in drawing the graph:( I went to wolfram alpha & there they showed the graph of the integrand as:

floor function of inverse of tangent http://www4c.wolframalpha.com/Calculate/MSP/MSP29821ec4face5cdfb21300003g6a31b5hg9d438f?MSPStoreType=image/gif&s=44&w=300.&h=188.&cdf=RangeControl

I couldn't understand why the graph is so. Then I saw the answer which was $ 100 - \tan 1$ . I couldn't comprehend how to to do it. Can anyone help me to evaluate this??

1 Answers1

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When we draw the Graph of $\tan^{-1}(x)\;,$ Then $\tan^{-1}(x)$ is an Strictly Increasing function.

Or Simply we can check it by Using Derivative Test. Means If $\displaystyle f(x) = \tan^{-1}(x)\;,$

Then $\displaystyle f'(x) = \frac{1}{1+x^2}>0$ for all $x\in \mathbb{R}$

Now in $1<x<\tan 1\;,$ Then $$\lfloor \tan ^{-1}(x)\rfloor = 0$$ and in $\tan 1<x<100\;,$ Then $$\lfloor \tan^{-1}(x)\rfloor = 1$$

Bcz $\displaystyle -\frac{\pi}{2}<\tan^{-1}(x)<\frac{\pi}{2}$ and Here $\displaystyle 1<\tan^{-1}(x)<\frac{\pi}{2}$

So we can break the Integral like $$\displaystyle \int_{1}^{100}\tan^{-1}(x)dx = \int_{1}^{\tan 1}0.dx+\int_{\tan 1}^{100}1dx = 0+ \left(100-\tan 1\right)$$

juantheron
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  • Sir,a bit confusion still persists: why have you broken the limits into 1 & $\tan 1$ ?? Have you done this by seeing the answer or some other reason?? I want to know if I am given such sort of step function, how would I divide the limits. Can you please explain? –  Aug 16 '15 at 07:28
  • bcz $\lfloor x \rfloor $ is an Integer quantity, and here $ 1 = \tan \left(\frac{\pi}{4}\right)<\tan (1)$So $\displaystyle \tan^{-1}\left(\tan \frac{\pi}{4}\right) = \frac{\pi}{4} \approx 0.8$ and $\tan^{-1}(\tan 1) = 1$ So in this Interval $\lfloor \tan^{-1}(x) \rfloor = 0$ and in $\tan (1)$ to $100;,$ We get $\lfloor \tan^{-1}(x)\rfloor = 1$ bcz $\displaystyle \tan^{-1}(x)< \frac{\pi}{2}\approx 1.57$ – juantheron Aug 16 '15 at 07:32