Let f- be a continuous function in -0-1- then lim n - j -0 n 1- n f j - n i-A- -1 - 21 f x dx-01 f x dxC- 1-2-01 - 2 f x dxD- -01 - 2 f x dx

The correct option is B ∫^1_0 f(x) dxlim_n→∞∑_j=0^n1n f ( jn ) 1n→ dx , jn→ x, upper limit , lim_n →∞n/n =1, nbsp;lower limit = lim_n →∞0/n = 0 ∴∫^1_0 f(x ...

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Standard XII

Mathematics

Properties of Inequalities

Let f, be a c...

Question

Let $f$ , be a continuous function in $[0, 1]$ , then $lim n \to \infty n \sum j = 0 \frac{1}{n} f (\frac{j}{n})$ is

\frac{1}{2} 1 / 2 \int 0 f (x) d x

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1 \int 1 / 2 f (x) d x

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1 \int 0 f (x) d x

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1 / 2 \int 0 f (x) d x

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Solution

The correct option is C $1 \int 0 f (x) d x$
$lim n \to \infty n \sum j = 0 \frac{1}{n} f (\frac{j}{n}) \frac{1}{n} \to d x, \frac{j}{n} \to x, upper limit, lim n \to \infty \frac{n}{n} = 1, lower limit = lim n \to \infty \frac{0}{n} = 0 ∴ 1 \int 0 f (x) d x$

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Similar questions

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and twice differentiable in

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lim x \to 0 \frac{1}{x^{2}} x \int 0 f (t) d t

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{lim}_{n \to 0} \frac{1 + n}{1 + n^{2} + n^{4}}

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Let f, be a continuous function in [0,1], then limn→∞n∑j=01nf(jn) is

The correct option is C 1∫0f(x)dxlimn→∞n∑j=01nf(jn)1n→dx,jn→x, upper limit,limn→∞nn=1, lower limit=limn→∞0n=0∴1∫0f(x)dx

Let $f$ , be a continuous function in $[0, 1]$ , then $lim n \to \infty n \sum j = 0 \frac{1}{n} f (\frac{j}{n})$ is

The correct option is C $1 \int 0 f (x) d x$
$lim n \to \infty n \sum j = 0 \frac{1}{n} f (\frac{j}{n}) \frac{1}{n} \to d x, \frac{j}{n} \to x, upper limit, lim n \to \infty \frac{n}{n} = 1, lower limit = lim n \to \infty \frac{0}{n} = 0 ∴ 1 \int 0 f (x) d x$