Question

The interval $[1,7]$ is divided into $6$ equal inter- vals and the value of ${\int }_1^7(x^2+x+1)dx$ using Trapezoidal and Simpson's rules are respectively ${\Delta }_1$ and ${\Delta }_2$

then which of the following is correct?

• A. ${\Delta }_1={\Delta }_2$
• B. ${\Delta }_1>{\Delta }_2$
• C. ${\Delta }_1<{\Delta }_2$
• D. ${\Delta }_1-{\Delta }_2=2$

Answer 1

The correct option is B ${\Delta }_1>{\Delta }_2$

Interval $h=\frac{b-a}{n}$
Here $n=6$ $b=7$ and $a=1$
Therefore $h=1$.
$f\left(1\right)=3=y_0$
$f\left(2\right)=7=y_1$
$f\left(3\right)=13=y_2$
$f\left(4\right)=21=y_3$
$f\left(5\right)=31=y_4$
$f\left(6\right)=43=y_5$
$f\left(7\right)=57=y_6$.
Using Simpson's rule, we get
${△}_2=\frac{h}{3}[y_0+2[y_2+y_4]+4[y_1+y_3+y_5]+y_6]$
$=\frac{1}{3}[3+2[13+31]+4[7+21+43]+57]$
$=144$
Using Trapezoidal rule, we get
${△}_1=\frac{h}{2}[y_0+2(y_1+y_2...y_5)+y_6]$
$=145$
Hence
${△}_1>{△}_2$.