Question

By the application of Simpson's one-third rule for numerical integration with two subintervals, the value of ${\int }_0^1\frac{dx}{1+x}$ is

• A. $\frac{17}{36}$
• B. $\frac{17}{25}$
• C. $\frac{25}{36}$
• D. $\frac{17}{24}$

Answer 1

The correct option is C $\frac{25}{36}$

Since, the given integration is divided into two subintervals

ie $h=\frac{1-0}{2}=\frac{1}{2}$

$\therefore$ ${\int }_0^1\frac{1}{1+x}dx=\frac{h}{3}[(y_0+y_2)+4\left\{y_1\right\}]$

At $x=0,y=1$

$x=\frac{1}{2},y_1=\frac{2}{3}$

and $x=1,y_2=\frac{1}{2}$

${\int }_0^1\frac{1}{1+x}dx=\frac{h}{3}[(y_0+y_2)+4\left\{y_1\right\}]$

${\int }_0^1\frac{1}{1+x}dx=\frac{1}{2.3}[(1+\frac{1}{2})+4\left(\frac{2}{3}\right)]=\frac{1}{6}[\frac{3}{2}+\frac{8}{3}]=\frac{25}{36}$