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(C) We are given the integral $I = \int_1^M \{x\}^{[x]} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can write the integral as a sum of int

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$4$

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(C) We are given the integral $I = \int_1^M \{x\}^{[x]} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can write the integral as a sum of integrals over unit intervals: $I = \sum_{n=1}^{M-1} \int_n^{n+1} (x-n)^n dx$. Let $u = x-n$,then $du = dx$. The integral becomes $\int_0^1 u^n du = \left[ \frac{u^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1}$. Thus,$I = \sum_{n=1}^{M-1} \frac{1}{n+1} = \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{M}$. For $M=2$,$I = \frac{1}{2} = 0.5 For $M=3$,$I = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \approx 0.833 For $M=4$,$I = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6+4+3}{12} = \frac{13}{12} > 1$. Therefore,the smallest positive integer $M$ is $4$.

Column $I$	Column $II$
$(A) \int_{-1}^1 \frac{dx}{1+x^2}$	$(p) \frac{1}{2} \log \left(\frac{2}{3}\right)$
$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}}$	$(q) 2 \log \left(\frac{2}{3}\right)$
$(C) \int_2^3 \frac{dx}{1-x^2}$	$(r) \frac{\pi}{3}$
$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}}$	$(s) \frac{\pi}{2}$

For each real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$,and let $\{x\} = x - [x]$. Then the smallest positive integer $M$ for which $\int_1^M \{x\}^{[x]} dx > 1$ is

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