Question

From the given data value of $\int_{1}^{2} \frac{1}{x} dx$ using Simpson’s 1/3rd rule is

• A. 0.06932
• B. 0.6932
• C. 6.932
• D. 0.006932

Hint:

Simpson’s 1/3rd rule requires an even number of subintervals; verify the given \( f(x) \) values match the function, and compute carefully to match the exact integral.

Answer 1

The Correct Option is B

Step 1: Understand Simpson’s 1/3rd rule. 
Simpson’s 1/3rd rule approximates the integral \( \int_a^b f(x) \, dx \) over an interval \([a, b]\) with \( n \) subintervals (where \( n \) is even). The formula is: \[ \int_a^b f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{\text{odd } i} f(x_i) + 2 \sum_{\text{even } i} f(x_i) + f(x_n) \right], \] where \( h = \frac{b - a}{n} \), and \( x_i = a + i h \). Here, \( a = 1 \), \( b = 2 \), and the data points are given at \( x = 1, 1.25, 1.5, 1.75, 2.0 \), so there are 4 subintervals (\( n = 4 \)). 
Step 2: Identify the data and compute \( h \). 
\( x_0 = 1 \), \( x_1 = 1.25 \), \( x_2 = 1.5 \), \( x_3 = 1.75 \), \( x_4 = 2.0 \),
\( f(x_0) = 1 \), \( f(x_1) = 1.25 \), \( f(x_2) = 1.75 \), \( f(x_3) = 1.5 \), \( f(x_4) = 0.5 \),
\( h = \frac{2 - 1}{4} = 0.25 \).
Note: The given \( f(x) \) values do not match \( f(x) = \frac{1}{x} \). For \( f(x) = \frac{1}{x} \), the values should be:
\( f(1) = 1 \),
\( f(1.25) = \frac{1}{1.25} = 0.8 \),
\( f(1.5) = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667 \),
\( f(1.75) = \frac{1}{1.75} = \frac{4}{7} \approx 0.5714 \),
\( f(2) = \frac{1}{2} = 0.5 \).
The provided \( f(x) \) values (1, 1.25, 1.75, 1.5, 0.5) seem incorrect for \( f(x) = \frac{1}{x} \). However, since the problem states “from the given data,” we use the provided values and note the discrepancy. Let’s proceed with the given data and then verify the expected integral. 
Step 3: Apply Simpson’s 1/3rd rule with the given data. 
With \( n = 4 \): \[ \int_{1}^{2} f(x) \, dx \approx \frac{h}{3} \left[ f(x_0) + 4(f(x_1) + f(x_3)) + 2(f(x_2)) + f(x_4) \right], \] \[ = \frac{0.25}{3} \left[ 1 + 4(1.25 + 1.5) + 2(1.75) + 0.5 \right], \] \[ = \frac{0.25}{3} \left[ 1 + 4(2.75) + 2(1.75) + 0.5 \right], \] \[ = \frac{0.25}{3} \left[ 1 + 11 + 3.5 + 0.5 \right] = \frac{0.25}{3} \cdot 16 = \frac{0.25 \cdot 16}{3} = \frac{4}{3} \approx 1.3333. \] This result does not match any of the options, indicating the given \( f(x) \) values are inconsistent with \( f(x) = \frac{1}{x} \). 
Step 4: Recompute with the correct \( f(x) = \frac{1{x} \).} 
Using the correct values for \( f(x) = \frac{1}{x} \):
\( f(x_0) = 1 \),
\( f(x_1) = 0.8 \),
\( f(x_2) = \frac{2}{3} \approx 0.6667 \),
\( f(x_3) = \frac{4}{7} \approx 0.5714 \),
\( f(x_4) = 0.5 \),
\[ \int_{1}^{2} \frac{1}{x} \, dx \approx \frac{0.25}{3} \left[ 1 + 4(0.8 + 0.5714) + 2(0.6667) + 0.5 \right], \] \[ = \frac{0.25}{3} \left[ 1 + 4(1.3714) + 2(0.6667) + 0.5 \right], \] \[ = \frac{0.25}{3} \left[ 1 + 5.4856 + 1.3334 + 0.5 \right] = \frac{0.25}{3} \cdot 8.319 \approx \frac{0.25 \cdot 8.319}{3} \approx 0.69325. \] The exact value of \( \int_{1}^{2} \frac{1}{x} \, dx = \ln x \big|_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \approx 0.69315 \), so our approximation is very close. 
Step 5: Evaluate the options. 
(1) 0.06932: Incorrect, as the integral is approximately 0.6932. Incorrect.
(2) 0.6932: Correct, as this matches the computed value and the exact value of \( \ln 2 \). Correct.
(3) 6.932: Incorrect, as the integral is approximately 0.6932. Incorrect.
(4) 0.006932: Incorrect, as the integral is approximately 0.6932. Incorrect.