Question

${\int }_{-2}^3(x+1)dx=\underset{n\to \infty }{lim}{\Sigma }_1^n(\frac{5i}{n}-1)$

State whether the above statment is True or False?

• A. True
• B. False

Answer 1

The correct option is B False

Given : ${\int }_{-2}^3(x+1)dx=\underset{n\to \infty }{lim}{\sum }_1^n[\frac{5i}{n}-1]$

$a=-2b=3f\left(x\right)=(x+1)\Delta x=\frac{b-a}{n}=\frac{3-(-2)}{n}=\frac{5}{n}{x}^{\ast }=a+i\Delta x=-2+i\frac{5}{n}{\int }_{-2}^3(x+1)dx=\underset{n\to \infty }{lim}{\sum }_1^{n}f\left(x^{\ast }\right)\Delta x=\underset{n\to \infty }{lim}{\sum }_1^n[-2+\frac{5i}{n}+1]\left(\frac{5}{n}\right)=\underset{n\to \infty }{lim}{\sum }_1^n[\frac{5i}{n}-1]\left(\frac{5}{n}\right)$

Hence above statement is false