Question

By trapezoidal rule, approximate value of \(\int_0^6 \frac{dx}{1+x^2}\)

• A. 1.3128
• B. 1.4108
• C. 1.4218
• D. None of these

Hint:

For trapezoidal rule, equal spacing simplifies calculations.

Answer 1

The Correct Option is B

Concept: Trapezoidal rule: \[ \int_a^b f(x)\,dx \approx \frac{h}{2}\left[y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\right] \]

Step 1: Take step size
Let \(h = 1\), so points: \[ x = 0,1,2,3,4,5,6 \]

Step 2: Compute values
\[ y = \frac{1}{1+x^2} \] \[ y_0 = 1,\ y_1 = \frac{1}{2},\ y_2 = \frac{1}{5},\ y_3 = \frac{1}{10},\ y_4 = \frac{1}{17},\ y_5 = \frac{1}{26},\ y_6 = \frac{1}{37} \]

Step 3: Substitute
\[ \int \approx \frac{1}{2}\left[y_0 + y_6 + 2(y_1+y_2+y_3+y_4+y_5)\right] \] \[ = \frac{1}{2}\left[1 + \frac{1}{37} + 2\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\frac{1}{26}\right)\right] \]

Step 4: Approximate
\[ \approx \frac{1}{2}(1.027 + 2 \times 0.691) = \frac{1}{2}(1.027 + 1.382) = \frac{2.409}{2} = 1.4108 \] Conclusion: \[ 1.4108 \]