If I = sum_{k=1}^{98} int_k^{k+1} frac{k+1}{x | vedclass

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(C) We are given $I = \sum_{k=1}^{98} \int_{k}^{k+1} \frac{k+1}{x(x+1)} dx$.For $x \in [k, k+1]$,we have $k \le x \le k+1$.This implies $\frac{1}{k+1}

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$B, D$

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(C) We are given $I = \sum_{k=1}^{98} \int_{k}^{k+1} \frac{k+1}{x(x+1)} dx$.For $x \in [k, k+1]$,we have $k \le x \le k+1$.This implies $\frac{1}{k+1} \le \frac{1}{x} \le \frac{1}{k}$.Multiplying by $(k+1)$,we get $\frac{k+1}{k+1} \le \frac{k+1}{x} \le \frac{k+1}{k}$.Dividing by $(x+1)$,we get $\frac{1}{x+1} \le \frac{k+1}{x(x+1)} \le \frac{k+1}{k(x+1)}$.Integrating from $k$ to $k+1$:$\int_{k}^{k+1} \frac{1}{x+1} dx Summing from $k=1$ to $98$:$\sum_{k=1}^{98} (\ln(k+2) - \ln(k+1)) The left side is $\ln(100) - \ln(2) = \ln(50)$.Since $\frac{k+1}{k} = 1 + \frac{1}{k}$,the right side is $\sum_{k=1}^{98} (1 + \frac{1}{k}) \ln(\frac{k+2}{k+1})$.Using the property $\ln(1+x) Also,comparing with $\frac{49}{50}$,we find $I > \frac{49}{50}$.Thus,$I  \frac{49}{50}$ are both true.

If $I = \sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} dx$,then which of the following is true?

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