JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}} \sum_{\mathrm{r}=0}^{2 \mathrm{n}-1} \frac{\mathrm{n}^{2}}{\mathrm{n}^{2}+4 \mathrm{r}^{2}}\) is:
- A \(\frac{1}{2} \tan ^{-1}(2)\)
- B \(\frac{1}{2} \tan ^{-1}(4)\)
- C \(\tan ^{-1}(4)\)
- D \(\frac{1}{4} \tan ^{-1}(4)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \tan ^{-1}(4)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{L}=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}} \cdot \sum_{\mathrm{r}=0}^{2 \mathrm{n}-1} \frac{1}{1+4\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^{2}}\) \(\Rightarrow \mathrm{L}=\int_{0}^{2} \frac{1}{1+4 \mathrm{x}^{2}} \mathrm{dx}\)…
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