Integral Calculus
86419
The value of \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}\) is equal to
1 \(5+\log _{\mathrm{e}}\left(\frac{3}{2}\right)\)
2 \(2-\log _{\mathrm{e}}\left(\frac{2}{3}\right)\)
3 \(3+2 \log _{\mathrm{e}}\left(\frac{2}{3}\right)\)
4 \(1+2 \log _{\mathrm{e}}\left(\frac{3}{2}\right)\)
Explanation:
(D) : \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n}\left(\frac{2 \frac{j}{n}-\frac{1}{n}+8}{\frac{2 j}{n}-\frac{1}{n}+4}\right)\)
\(=\int_{0}^{1} \frac{2 \mathrm{x}+8}{2 \mathrm{x}+4} \cdot \mathrm{dx}=\int_{0}^{1} \mathrm{dx}+\int_{0}^{1} \frac{4}{2 \mathrm{x}+4} \mathrm{dx}\)
\(=1+4 \cdot \frac{1}{2}[\ln (2 x+4)]_{0}^{1}=1+2[\ln 6-\ln 4]\)
\(=1+2 \ln \left(\frac{3}{2}\right)\)