Definite Integration Question 3
Question 3
Objective Questions II
(One or more than one correct option)
Show Answer
Answer:
Correct Answer: 3. (b)
Solution:
- Let
Taking log on both sides, we get
$$ \begin{aligned} & \log l=\lim {n \rightarrow \infty} \frac{1}{n} \log 1+\frac{1}{n} 1+\frac{2}{n} \ldots 1+\frac{2 n}{n} \ & \Rightarrow \quad \log l=\lim {n \rightarrow \infty} \frac{1}{n} \ & \log 1+\frac{1}{n}+\log 1+\frac{2}{n}+\ldots+\log 1+\frac{2 n}{n} \ & \Rightarrow \quad \log l=\lim {n \rightarrow \infty} \frac{1}{n} \sum{r=1}^{2 n} \log 1+\frac{r}{n} \ & \Rightarrow \quad \log l=\int{0}^{2} \log (1+x) d x \ & \Rightarrow \quad \log l=\log (1+x) \cdot x-\int \frac{1}{1+x} \cdot x d x \ & \Rightarrow \quad \log l=[\log (1+x) \cdot x]{0}^{2}-\int_{0}^{2} \frac{x+1-1}{1+x} d x \ & \Rightarrow \quad \log l=2 \cdot \log 3-\int_{0}^{2} 1-\frac{1}{1+x} d x \ & \Rightarrow \quad \log l=2 \cdot \log 3-[x-\log |1+x|]_{0}^{2} \ & \Rightarrow \quad \log l=2 \cdot \log 3-[2-\log 3] \ & \Rightarrow \quad \log l=3 \cdot \log 3-2 \ & \Rightarrow \quad \log l=\log 27-2 \ & \therefore \quad l=e^{\log 27-2}=27 \cdot e^{-2}=\frac{27}{e^{2}} \end{aligned} $$
