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Question

Let f, be a continuous function in [0,1], then limnnj=01nf(jn) is

A
121/20f(x)dx
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B
11/2f(x)dx
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C
10f(x)dx
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D
1/20f(x)dx
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Solution

The correct option is C 10f(x)dx
limnnj=01nf(jn)1ndx,jnx, upper limit,limnnn=1, lower limit=limn0n=010f(x)dx

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