Question

If the graph of $y=f\left(x\right)$ passes through $(0,0),(0.5,0.19),(1,0.34),(1.5,0.43),(2,0.47)$

then ${\int }_0^{2}f\left(x\right)dx$ by Simpson's rule is?

• A. $0.605$
• B. $0.61$
• C. $0.601$
• D. $0.615$

Answer 1

The correct option is A $0.605$

$f\left(0\right)=0=y_0$
$f\left(0.5\right)=0.19=y_1$
$f\left(1\right)=0.34=y_2$
$f\left(1.5\right)=0.43=y_3$
$f\left(2\right)=0.47=y_4$
Here interval $h=0.5$
Applying Simpson's rule, we get
$△=\frac{h}{3}[y_0+2[y_2]+4[y_1+y_3]+y_4]$
$=\frac{0.5}{3}[0+2(0.34)+4(0.43+0.19)+0.47]$
$=0.605$