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

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Question

(C) Given the integral $I = \int_1^3 \frac{dx}{2+3x}$. Here,$n = 2$ and the interval $[1, 3]$ is divided into $n=2$ equal parts. The step size $h = \f

Vedclass · Accepted Answer

$\frac{29}{110}$

Vedclass · Answer

(C) Given the integral $I = \int_1^3 \frac{dx}{2+3x}$. Here,$n = 2$ and the interval $[1, 3]$ is divided into $n=2$ equal parts. The step size $h = \frac{b-a}{n} = \frac{3-1}{2} = 1$. The values of $x$ are $x_0 = 1, x_1 = 2, x_2 = 3$. The corresponding values of $y = f(x) = \frac{1}{2+3x}$ are: $y_0 = f(1) = \frac{1}{2+3(1)} = \frac{1}{5} = 0.2$ $y_1 = f(2) = \frac{1}{2+3(2)} = \frac{1}{8} = 0.125$ $y_2 = f(3) = \frac{1}{2+3(3)} = \frac{1}{11} \approx 0.0909$ According to Simpson's rule: $\int_a^b f(x) dx \approx \frac{h}{3} [y_0 + 4y_1 + y_2]$. Substituting the values: $I \approx \frac{1}{3} [0.2 + 4(0.125) + 0.0909] = \frac{1}{3} [0.2 + 0.5 + 0.0909] = \frac{0.7909}{3} \approx 0.2636$. Comparing with the options,$\frac{29}{110} \approx 0.2636$. Thus,the correct option is $C$.

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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