Question

The integral ${\int }_1^3\frac{1}{x}dx$, when evaluated by using Simpson's $1/3$ rule on two equal subintervales each of length 1, equal

• A. $1.000$
• B. $1.098$
• C. $1.111$
• D. $1.120$

Answer 1

The correct option is C $1.111$

By simpson's $1/3$ rule
${\int }_1^3\frac{1}{x}dx=\frac{h}{3}\left[\right(y_0+y_2)+4y_1]$
$y\left(x\right)=\frac{1}{x}$
$y_0=y\left(1\right)=1$
$y_1=y\left(2\right)=\frac{1}{2}$
$y_2=y_{(}3)=\frac{1}{3}$
${\int }_1^3\frac{1}{x}dx=\frac{1}{3}[(1+\frac{1}{3})+4\left(\frac{1}{2}\right)]$
$=\frac{10}{9}=1.111$