MHB How Do You Calculate the Integral of 1/Floor(x^2) from 1 to 2?

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To evaluate the integral of 1/Floor(x^2) from 1 to 2, it is essential to recognize the behavior of the floor function within the specified limits. For x in the interval [1, 2], x^2 ranges from 1 to 4, leading to different constant values for the floor function: it equals 1 for x in [1, sqrt(2)) and 2 for x in [sqrt(2), 2]. The integral can be split into two parts: from 1 to sqrt(2) and from sqrt(2) to 2, allowing for straightforward integration of the constant values. The final result of the integral is computed by summing the contributions from both segments. The solution demonstrates the importance of understanding piecewise functions in integration.
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Evaluate $\displaystyle \int_{1}^{2} \dfrac{1}{\left\lfloor{x^2}\right\rfloor}\,dx$ where $\left\lfloor{u}\right\rfloor$ denotes the greatest integer less than or equal to $u$.

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Congratulations to the following members for their correct solutions::)

1. Olok
2. laura123
3. MarkFL
4. Euge
5. kaliprasad

Solution from Euge:
Partition the interval $[1,2)$ into the subintervals $[1,\sqrt{2})$, $[\sqrt{2}, \sqrt{3})$, and $[\sqrt{3}, 2)$, in which $\lfloor x^2 \rfloor$ has values $1$, $2$, and $3$, respectively. Then

$$\int_1^2 \frac{dx}{\lfloor x^2 \rfloor} = \int_1^{\sqrt{2}} \frac{dx}{1} + \int_{\sqrt{2}}^{\sqrt{3}} \frac{dx}{2} + \int_{\sqrt{3}}^2 \frac{dx}{3} = \sqrt{2} - 1 + \frac{1}{2}(\sqrt{3} - \sqrt{2}) + \frac{1}{3}(2 - \sqrt{3})$$

$$ = \frac{1}{6}(6\sqrt{2} - 6 + 3\sqrt{3} - 3\sqrt{2} + 4 - 2\sqrt{3}) = \frac{1}{6}(3\sqrt{2} + \sqrt{3} - 2).$$
 
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