Question

The approximate value of ${\int }_1^3\frac{dx}{2+3x}$ using Simpson's Rules and dividing the interval [1, 3] into two equal parts is

• A. $\frac{1}{3}\log \left(\frac{11}{5}\right)$
• B. $\frac{107}{110}$
• C. $\frac{29}{110}$
• D. $\frac{116}{440}$

Answer 1

The correct option is D $\frac{116}{440}$

${\int }_a^{b}f\left(x\right)dx$
$=\frac{b-a}{6}\left[f\right(a)+4f(\frac{a+b}{2})+f(b\left)\right]$ ... when interval is divided into two equal parts.
Hence
${\int }_1^3\frac{1}{2+3x}dx$
$=\frac{2}{6}[\frac{1}{5}+\frac{4}{8}+\frac{1}{11}]$
$=\frac{2}{6}(\frac{28}{40}+\frac{1}{11})$
$=\frac{2}{6}\left(\frac{308+40}{440}\right)$
$=\frac{348}{440}\times \frac{2}{6}$
$=\frac{116}{440}$