The approximate value of int_1^3 frac{dx}{2+3 | vedclass

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(C) Let $f(x) = \frac{1}{2+3x}$. We divide the interval $[1,3]$ into $n=2$ equal parts. $h = \frac{3-1}{2} = 1$. The points are $x_0 = 1, x_1 = 2, x_2

Vedclass · Accepted Answer

$\frac{29}{110}$

Vedclass · Answer

(C) Let $f(x) = \frac{1}{2+3x}$. We divide the interval $[1,3]$ into $n=2$ equal parts. $h = \frac{3-1}{2} = 1$. The points are $x_0 = 1, x_1 = 2, x_2 = 3$. $f(x_0) = f(1) = \frac{1}{2+3(1)} = \frac{1}{5} = 0.2$. $f(x_1) = f(2) = \frac{1}{2+3(2)} = \frac{1}{8} = 0.125$. $f(x_2) = f(3) = \frac{1}{2+3(3)} = \frac{1}{11} \approx 0.0909$. Using Simpson's rule: $\int_a^b f(x) dx \approx \frac{h}{3} [f(x_0) + 4f(x_1) + f(x_2)]$. $\int_1^3 f(x) dx \approx \frac{1}{3} [0.2 + 4(0.125) + 0.0909] = \frac{1}{3} [0.2 + 0.5 + 0.0909] = \frac{0.7909}{3} \approx 0.2636$. Comparing with the options,$\frac{29}{110} \approx 0.2636$.

The approximate value of $\int_1^3 \frac{dx}{2+3x}$ using Simpson's rule and dividing the interval $[1,3]$ into two equal parts is

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