To find the approximate area under the curve $y = x^2 + 1$ between $x = 0$ and $x = 2$, we can use the trapezoidal rule with $n = 4$ intervals.

The trapezoidal rule formula is given by:

$A \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right]$

Where:

• $f(x) = x^2 + 1$
• $a = 0$ and $b = 2$, the limits of integration
• $n = 4$, the number of subintervals
• $\Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5$

Now, we calculate the points:

$x_0 = 0, \quad x_1 = 0.5, \quad x_2 = 1, \quad x_3 = 1.5, \quad x_4 = 2$

Evaluating the function at each of these points:

$f(x_0) = f(0) = 0^2 + 1 = 1$ $f(x_1) = f(0.5) = (0.5)^2 + 1 = 0.25 + 1 = 1.25$ $f(x_2) = f(1) = 1^2 + 1 = 2$ $f(x_3) = f(1.5) = (1.5)^2 + 1 = 2.25 + 1 = 3.25$ $f(x_4) = f(2) = 2^2 + 1 = 4 + 1 = 5$

Now applying the trapezoidal rule formula:

$A \approx \frac{0.5}{2} \left[ 1 + 2(1.25) + 2(2) + 2(3.25) + 5 \right]$ $A \approx 0.25 \left[ 1 + 2.5 + 4 + 6.5 + 5 \right]$ $A \approx 0.25 \times 19$ $A \approx 4.75$

Thus, the approximate area is 4.75, but the choices given are smaller. Based on these, the closest option would be 3.75.

Do you need more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the result change if $n$ was increased?
2. What is the exact area using integration instead of approximation?
3. How does the trapezoidal rule compare to Simpson's rule?
4. Can you approximate the area using the midpoint rule?
5. What happens if the curve was different, like $y = x^3$?

Tip: The more intervals ($n$) you use in numerical methods like the trapezoidal rule, the more accurate the approximation becomes.