Question

The approximate value of ${\int }_0^4(x^2+1)dx$, taking 4 equal intervals and using Simpson's $\left(\frac{1}{3}\right)$rd rule is

• A. $\frac{76}{3}$
• B. $22$
• C. $\frac{56}{3}$
• D. $\frac{86}{3}$

Answer 1

Answer: (A)

$\frac{76}{3}$

Here $h=\frac{4-0}{4}=1$

$x$	$0$	$1$	$2$	$3$	$4$
$1+x^2$	$1$	$2$	$5$	$10$	$17$
	$y_0$	$y_1$	$y_2$	$y_3$	$y_4$

By Simpson's rule $\left(\frac{1}{3}\right)$rd rule
${\int }_0^4(x^2+1)dx=\frac{1}{3}\left[\right(1+17)+4(2+10)+2(5\left)\right]$
$=\frac{1}{3}[18+48+10]$
$=\frac{76}{3}$