The correct option is D π4 n →∞Lt∑_r=0^n 1nn^2+r^2 #10; ⇒n →∞Lt∑_r=0^n 11n #10;( 11+ ( r^2n^2 ) ) #10;We can write in the form of: #10; ...

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Definite Integral as Limit of Sum

n →∞Lt∑r=0n-1...

Question

$L t n \to \infty n - 1 \sum r = 0 \frac{n}{n^{2} + r^{2}}$

1

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π

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\frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is D $\frac{π}{4}$
$L t n \to \infty \sum_{r = 0}^{n - 1} \frac{n}{n^{2} + r^{2}}$
$\Rightarrow L t n \to \infty \sum_{r = 0}^{n - 1} \frac{1}{n} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{1 + (\frac{r^{2}}{n^{2}})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
We can write in the form of:
$L t n \to \infty \sum_{r = 0}^{n - 1} \frac{1}{n} f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$

$f (\frac{r}{n}) = \frac{1}{1 + {(\frac{r}{n})}^{2}}; f (x) = \frac{1}{1 + x^{2}}$
$\int_{0}^{1} \frac{1}{1 + x^{2}} d x = t a n^{- 1} x |_{0}^{1} = \frac{π}{4}$

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