By Simpson rule taking `n=4,` the value of the integral `int _(0)^(1)(1)/(1+x^(2))dx` is equal to

Options

A. `0.788`
B. `0.781`
C. `0.785`
D. None of these

Explanation

To solve the integral \(\int_0^1 \frac{1}{1+x^2} \, dx\) using Simpson's rule with \(n=4\), we will follow these steps:

### Step 1: Determine the values of \(a\), \(b\), and \(n\)
Here, \(a = 0\), \(b = 1\), and \(n = 4\).

### Step 2: Calculate \(\Delta x\)
\[ \Delta x = \frac{b - a}{n} = \frac{1 - 0}{4} = \frac{1}{4} = 0.25 \]

### Step 3: Identify the \(x\) values
The \(x\) values at which we will evaluate the function are:
- \(x_0 = a = 0\)
- \(x_1 = a + \Delta x = 0 + 0.25 = 0.25\)
- \(x_2 = a + 2\Delta x = 0 + 2(0.25) = 0.5\)
- \(x_3 = a + 3\Delta x = 0 + 3(0.25) = 0.75\)
- \(x_4 = b = 1\)

### Step 4: Evaluate the function at these points
Now we evaluate \(f(x) = \frac{1}{1+x^2}\) at these points:
- \(f(x_0) = f(0) = \frac{1}{1+0^2} = 1\)
- \(f(x_1) = f(0.25) = \frac{1}{1+(0.25)^2} = \frac{1}{1+0.0625} = \frac{1}{1.0625} \approx 0.9412\)
- \(f(x_2) = f(0.5) = \frac{1}{1+(0.5)^2} = \frac{1}{1+0.25} = \frac{1}{1.25} = 0.8\)
- \(f(x_3) = f(0.75) = \frac{1}{1+(0.75)^2} = \frac{1}{1+0.5625} = \frac{1}{1.5625} \approx 0.64\)
- \(f(x_4) = f(1) = \frac{1}{1+1^2} = \frac{1}{2} = 0.5\)

### Step 5: Apply Simpson's Rule
Using Simpson's Rule:
\[ \int_a^b f(x) \, dx \approx \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4) \right] \]

Substituting the values:
\[ \int_0^1 f(x) \, dx \approx \frac{0.25}{3} \left[ 1 + 4(0.9412) + 2(0.8) + 4(0.64) + 0.5 \right] \]

Calculating the terms:
- \(4f(x_1) = 4 \times 0.9412 \approx 3.7648\)
- \(2f(x_2) = 2 \times 0.8 = 1.6\)
- \(4f(x_3) = 4 \times 0.64 = 2.56\)

Now summing these:
\[ 1 + 3.7648 + 1.6 + 2.56 + 0.5 = 9.4258 \]

Calculating the integral:
\[ \int_0^1 f(x) \, dx \approx \frac{0.25}{3} \times 9.4258 \approx \frac{2.35645}{3} \approx 0.78548 \]

### Final Result
Thus, the approximate value of the integral \(\int_0^1 \frac{1}{1+x^2} \, dx\) using Simpson's rule with \(n=4\) is approximately \(0.785\).