Question

The estimate of ${\int }_{0.5}^{1.5}\frac{dx}{x}$ obtained using Simpson's rule with three-point function evaluation exceeds the exact value by

• A. $0.235$
• B. $0.068$
• C. $0.024$
• D. $0.012$

Answer 1

The correct option is D $0.012$

Exact value $={\int }_{0.5}^{1.5}\frac{dx}{x}dx=1.0986$
Using Simpson's rule in three-point form,
$h=\frac{b-a}{2}$
$=\frac{1.5-0.5}{2}$
$=0.5$
So,

$x$	$0.5$	$1$	$1.5$
$y$	$2$	$1$	$0.67$

$\int \frac{1}{x}dx=\frac{h}{3}\times [y_0+y_2+4y_1]$
$=\frac{0.5}{3}[2+0.67+4\times 1$]
$=1.1116$ (Approx. value)
It exceeds exact value by $0.012$