Question

Taking number of intervals n = 3, the value of the integral
\[ \int_0^{0.3} e^{-x^2}\, dx \]
using Trapezoidal method, is ______ (rounded off to two decimal places).

Hint:

Use h = 0.1, evaluate e^(-x^2) at x = 0, 0.1, 0.2, 0.3, and apply the trapezoidal rule I is about (h/2)[f0 + 2f1 + 2f2 + f3].

Answer 1

Correct Answer: 0.27

Step 1: Set up the trapezoidal rule.
We need to estimate
\[ I = \int_0^{0.3} e^{-x^2}\, dx \]
using \(n = 3\) intervals. With the interval from \(0\) to \(0.3\) split into 3 equal parts, the step size is
\[ h = \frac{0.3 - 0}{3} = 0.1 \]

Step 2: List the nodes and function values.
The nodes are \(x_0 = 0\), \(x_1 = 0.1\), \(x_2 = 0.2\), \(x_3 = 0.3\).
Evaluate \(f(x) = e^{-x^2}\) at each node:
\[ f(x_0) = e^{-0^2} = e^{0} = 1 \]
\[ f(x_1) = e^{-(0.1)^2} = e^{-0.01} \approx 0.990050 \]
\[ f(x_2) = e^{-(0.2)^2} = e^{-0.04} \approx 0.960789 \]
\[ f(x_3) = e^{-(0.3)^2} = e^{-0.09} \approx 0.913931 \]

Step 3: Apply the trapezoidal rule formula.
The composite trapezoidal rule for \(n\) intervals is
\[ I \approx \frac{h}{2}\left[f(x_0) + 2\sum_{i=1}^{n-1} f(x_i) + f(x_n)\right] \]
Here, the interior points are \(x_1\) and \(x_2\), so
\[ I \approx \frac{0.1}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3)\right] \]

Step 4: Substitute the function values.
\[ I \approx 0.05\left[1 + 2(0.990050) + 2(0.960789) + 0.913931\right] \]
\[ I \approx 0.05\left[1 + 1.980100 + 1.921578 + 0.913931\right] \]
\[ I \approx 0.05 \times 5.815609 \]
\[ I \approx 0.290780 \]

Step 5: Round off the answer.
Rounded to two decimal places,
\[ I \approx 0.29 \]

Final Answer:
\[ \boxed{0.29} \]