The trapezoidal sum method estimates the area under a curve by dividing the interval into $n$ subintervals, calculating the function's value at each endpoint, and approximating each subinterval's area using a trapezoid. Here's the step-by-step solution for the given problem:

Problem Breakdown:

• Function: $f(x) = 2^x + 3$
• Interval: $[0, 2]$
• Number of subintervals: $n = 4$
• Subinterval width ($\Delta x$):
  $\Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5$

Subintervals:

The subintervals are: [ [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2] ] The corresponding $x$-values are: $x_0 = 0, \; x_1 = 0.5, \; x_2 = 1, \; x_3 = 1.5, \; x_4 = 2$

Function Values:

We evaluate $f(x) = 2^x + 3$ at each of these points: $f(0) = 2^0 + 3 = 4, \; f(0.5) = 2^{0.5} + 3 \approx 4.414, \; f(1) = 2^1 + 3 = 5$ $f(1.5) = 2^{1.5} + 3 \approx 5.828, \; f(2) = 2^2 + 3 = 7$

Trapezoidal Sum Formula:

The trapezoidal sum is: $A \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right]$ Substitute the values: $A \approx \frac{0.5}{2} \left[ 4 + 2(4.414) + 2(5) + 2(5.828) + 7 \right]$ $A \approx 0.25 \left[ 4 + 8.828 + 10 + 11.656 + 7 \right]$ $A \approx 0.25 \cdot 41.484$ $A \approx 10.371$

Final Answer:

The approximate area under the graph is: $A \approx 10.37$

Let me know if you want more details or explanations!

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Related Questions:

1. How does the trapezoidal sum compare with the midpoint or Simpson's rule for this problem?
2. What is the exact integral of $f(x) = 2^x + 3$ over $[0, 2]$?
3. How does increasing the number of partitions ($n$) affect the accuracy of the trapezoidal sum?
4. What is the error bound for the trapezoidal sum in this case?
5. How would the calculation change if the function were $f(x) = 2^x - 3$ instead?

Tip: Always increase $n$ for better accuracy in numerical integration methods.