Question

The approximate value of \(\int_1^5 x^2 dx\) using trapezoidal rule with \(n = 4\) is

• A. 41
• B. 41.5
• C. 41.75
• D. 42

Hint:

Exact value of \(\int_1^5 x^2 dx = [x^3/3]_1^5 = (125 - 1)/3 = 124/3 \approx 41.33\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Trapezoidal rule: \(\int_a^b f(x) dx \approx \frac{h}{2}[f(x_0) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f(x_n)]\)
Step 2: Detailed Explanation:
\(a = 1\), \(b = 5\), \(n = 4\), \(h = (5 - 1)/4 = 1\)
\(x_0 = 1\), \(x_1 = 2\), \(x_2 = 3\), \(x_3 = 4\), \(x_4 = 5\)
\(f(x) = x^2\)
\(f(1) = 1\), \(f(2) = 4\), \(f(3) = 9\), \(f(4) = 16\), \(f(5) = 25\)
Approx = \(\frac{1}{2}[1 + 2(4 + 9 + 16) + 25] = \frac{1}{2}[1 + 2(29) + 25] = \frac{1}{2}[1 + 58 + 25] = \frac{1}{2} \times 84 = 42\)
Step 3: Final Answer:
42.