Let f(0)=1, f(0.5)=frac{5}{4}, f(1)=2, f(1.5) | vedclass

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(A) Given $a=0, b=2$ and $n=4$ intervals. The step size $h = \frac{b-a}{n} = \frac{2-0}{4} = 0.5$. The values are $y_0=f(0)=1, y_1=f(0.5)=\frac{5}{4},

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$\frac{14}{3}$

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(A) Given $a=0, b=2$ and $n=4$ intervals. The step size $h = \frac{b-a}{n} = \frac{2-0}{4} = 0.5$. The values are $y_0=f(0)=1, y_1=f(0.5)=\frac{5}{4}, y_2=f(1)=2, y_3=f(1.5)=\frac{13}{4}, y_4=f(2)=5$. According to Simpson's $1/3$ rule: $\int_a^b f(x) dx = \frac{h}{3} [ (y_0 + y_4) + 4(y_1 + y_3) + 2(y_2) ]$ Substituting the values: $\int_0^2 f(x) dx = \frac{0.5}{3} [ (1 + 5) + 4(\frac{5}{4} + \frac{13}{4}) + 2(2) ]$ $= \frac{0.5}{3} [ 6 + 4(\frac{18}{4}) + 4 ]$ $= \frac{0.5}{3} [ 6 + 18 + 4 ]$ $= \frac{0.5}{3} [ 28 ] = \frac{14}{3}$.

Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule, $\int_0^2 f(x) dx$ is equal to

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