Question

Considering four sub-intervals, the value of ${\int }_0^4{2}^{x}dx$ by Simpson's rule is

• A. $\frac{64}{8}$
• B. $\frac{65}{3}$
• C. $\frac{62}{12}$
• D. $\frac{61}{8}$

Answer 1

Answer: (B)

$\frac{65}{3}$

Here, $a=0,b=4,n=4$

$h=\frac{b-a}{n}=\frac{4-0}{4}=1$

$x$ $0$ $1$ $2$ $3$ $4$

$y=2^x$

$1$ $2$ $4$ $8$ $16$

$y_0$ $y_1$ $y_2$ $y_3$ $y_4$

By Simpson's rule

${\int }_0^4{2}^{x}dx=\frac{h}{3}[y_0+y_4+4(y_1+y_3)+2y_2]$

$=\frac{1}{3}[1+16+4(2+8)+2(4\left)\right]$

$=\frac{1}{3}(17+40+8)=\frac{1}{3}\left(65\right)$