Question

The value of $\underset{n\to \infty }{lim}{\Sigma }_{i=1}^{n}log(2+\frac{5i}{n})\frac{5}{n}$ is equal to?

• A. ${\int }_2^{5}ln\left(x\right)dx$
• B. ${\int }_3^{7}ln\left(x\right)dx$
• C. ${\int }_2^{7}ln\left(x\right)dx$
• D. ${\int }_2^{7}ln\left(x^2\right)dx$

Answer 1

The correct option is C ${\int }_2^{7}ln\left(x\right)dx$

Given : $\underset{n\to \infty }{lim}{\sum }_1^n{\log }_e(2+\frac{5i}{n})\frac{5}{n}$

$\Delta x=\frac{5p}{n}=\frac{7-2}{n}a=2b=7f\left(x\right)=\log xf\left(x^{\ast }\right)=\log [2+\frac{5i}{n}]\underset{n\to \infty }{lim}{\sum }_1^n{\log }_e(2+\frac{5i}{n})\frac{5}{n}={\int }_2^7{\log }_e\left(x\right)dx$

Hence the correct answer is ${\int }_2^7{\log }_e\left(x\right)dx$