1. Tardigrade
  2. Question
  3. Mathematics
  4. Statement-I: displaystyle∑ r =0 n -1 (1/ n )(√( r / n )+1)<∫ limits01(√ x +1) dx < displaystyle∑ r =1 n (1/ n )(√( r / n )+1), n ∈ N. because Statement-II: If f( x ) is continuous and increasing in [0, 1], then displaystyle∑ r =0 n -1 (1/ n ) f(( r / n ))<∫ limits01 f( x ) dx < displaystyle∑ r =1 n (1/ n ) f(( r / n )), where n ∈ N

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Q. Statement-I : $ \displaystyle\sum_{ r =0}^{ n -1} \frac{1}{ n }\left(\sqrt{\frac{ r }{ n }}+1\right)<\int\limits_0^1(\sqrt{ x }+1) dx < \displaystyle\sum_{ r =1}^{ n } \frac{1}{ n }\left(\sqrt{\frac{ r }{ n }}+1\right), n \in N$. because
Statement-II : If $f( x )$ is continuous and increasing in [0, 1], then $ \displaystyle\sum_{ r =0}^{ n -1} \frac{1}{ n } f\left(\frac{ r }{ n }\right)<\int\limits_0^1 f( x ) dx < \displaystyle\sum_{ r =1}^{ n } \frac{1}{ n } f\left(\frac{ r }{ n }\right)$, where $n \in N$

Integrals

Solution:

Correct answer is (a) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.