lim _{n rightarrow infty} frac{1}{n} sum_{r=0 | vedclass

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Question

(B) આપેલ લક્ષ $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ છે.અંશ અને છેદને $n^2$ વડે ભાગતા:$L = \lim

Vedclass · Accepted Answer

$\frac{1}{2} 	an ^{-1}(4)$

Vedclass · Answer

(B) આપેલ લક્ષ $L = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ છે.અંશ અને છેદને $n^2$ વડે ભાગતા:$L = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{1}{1+4(\frac{r}{n})^2}$.નિશ્ચિત સંકલનની વ્યાખ્યા મુજબ,$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=0}^{kn-1} f(\frac{r}{n}) = \int_{0}^{k} f(x) dx$.અહીં,$f(x) = \frac{1}{1+4x^2}$ અને ઉપરની સીમા $k = \lim_{n ightarrow \infty} \frac{2n-1}{n} = 2$ છે.તેથી,$L = \int_{0}^{2} \frac{1}{1+(2x)^2} dx$.સૂત્ર $\int \frac{1}{1+a^2x^2} dx = \frac{1}{a} 	an^{-1}(ax) + C$ નો ઉપયોગ કરતા:$L = \left[ \frac{1}{2} 	an^{-1}(2x) ight]_{0}^{2}$.$L = \frac{1}{2} 	an^{-1}(2 	imes 2) - \frac{1}{2} 	an^{-1}(2 	imes 0)$.$L = \frac{1}{2} 	an^{-1}(4) - 0 = \frac{1}{2} 	an^{-1}(4)$.

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ ની કિંમત શોધો:

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