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Question: By Simpson's rule the value of the interval \(\int_{1}^{6}{x\mspace{6mu} dx}\) on dividing, the inte...

By Simpson's rule the value of the interval ∫16x 6mudx\int_{1}^{6}{x\mspace{6mu} dx} on dividing, the interval into four equal parts is

A

16

B

16.5

C

17

D

17.5

Answer

17.5

Explanation

Solution

h=6−14=54=1.25h = \frac{6 - 1}{4} = \frac{5}{4} = 1.25

x0=a=1, 6mu 6mux1=x0+h=1+1.25=2.25x_{0} = a = 1,\mspace{6mu}\mspace{6mu} x_{1} = x_{0} + h = 1 + 1.25 = 2.25

x2=x0+2h=1+2(1.25)=3.50x_{2} = x_{0} + 2h = 1 + 2(1.25) = 3.50

x3=x0+3h=1+3(1.25)=4.75x_{3} = x_{0} + 3h = 1 + 3(1.25) = 4.75

x4=x0+4h=1+4(1.25)=6.0x_{4} = x_{0} + 4h = 1 + 4(1.25) = 6.0

By Simpson's rule, ∫abf(x)dx=∫16xdx\int_{a}^{b}{f(x)}dx = \int_{1}^{6}{xd}x

=1.253[(y0+y4)+4(y1+y3)+2(y2)]=1.253[(1+6)+4(2.25+4.75)+2(3.5)]=1.253[7+28+7]=17.5= \frac{1.25}{3}\left\lbrack (y_{0} + y_{4}) + 4(y_{1} + y_{3}) + 2(y_{2}) \right\rbrack = \frac{1.25}{3}\left\lbrack (1 + 6) + 4(2.25 + 4.75) + 2(3.5) \right\rbrack = \frac{1.25}{3}\lbrack 7 + 28 + 7\rbrack = 17.5