Question

Let f (0) = 1, f(1) = 0.5, f(2) = 0.333, f(3) = 0.25, f(4) = 0.2, f(5) = 0.166 f(6) = 0.142 . The value of int 0 ^ 6 :f(x)dx by Simpson's three eighth rule is​

Answer 1

Step-by-step explanation:

(0) = 1, f(1) = 0.5, f(2) = 0.333, f(3) = 0.25, f(4) = 0.2, f(5) = 0.166 f(6) = 0.142 . The value of int 0 ^ 6

Answer 2

Answer:

1.58

Step-by-step explanation:

Simpson's three eighth rule gives the formula :

$\int\limits^b_a {f(x)} \, dx = \frac{3h}{8} [(y_{0}+y_{n}) + 3(y_{1}+y_{2}+y_{4}+...)+2(y_{3}+y_{6}+..)]$

where a & b are the limits of the integral and h is calculated by dividing (b-a) by no of parts integral is to be divided in

We are given that a=0 & b=6, therefore [0,6] is divided into 6 equal parts

Therefore,

$h=\frac{b-a}{6} = \frac{6-0}{6} =1$

h=1

$x_{0} = 0 , f(0)=1$

$x_{1} = 1 , f(1)=0.5$

$x_{2} = 2 , f(2)=0.333$

$x_{3} = 3 , f(3)=0.25$

$x_{4} = 4 , f(4)=0.2$

$x_{5} = 5 , f(5)=0.166$

$x_{6} = 6 , f(6)=0.142$

Now to finally solve the integral:

$\int\limits^b_a {f(x)} \, dx = \frac{3h}{8} [(y_{0}+y_{n}) + 3(y_{1}+y_{2}+y_{4}+...)+2(y_{3}+y_{6}+..)]$

$\int\limits^6_0 {f(x)} \, dx = \frac{3*1}{8} [(0+0.142) + 3(0.5+0.333+0.2+0.166)+2(0.25)]$

$\int\limits^6_0 {f(x)} \, dx = \frac{3}{8} [(0.142) + 3(1.199)+0.5]$

$\int\limits^6_0 {f(x)} \, dx = \frac{3}{8} [(0.142) + 3.597+0.5]$

$\int\limits^6_0 {f(x)} \, dx = 1.58$

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