Answer
El área aproximada bajo la curva de la función \( f(x) = x^{3} + 2x^{2} + 1 \) desde \( x = 0 \) hasta \( x = 2 \) utilizando una suma de Riemann con 4 subintervalos es **15.75 unidades cuadradas**.
Solution
Alright, I need to calculate the area under the curve of the function \( f(x) = x^{3} + 2x^{2} + 1 \) over the interval from \( x_{1} = 0 \) to \( x_{2} = 2 \) using Riemann sums. Let's break this down step by step.
First, I recall that a Riemann sum is a method for approximating the area under a curve by dividing the area into small rectangles and summing their areas. The more rectangles I use, the better the approximation will be. For this problem, I'll use the right Riemann sum, which means I'll evaluate the function at the right endpoint of each subinterval.
Let's start by determining the width of each subinterval. The total interval is from 0 to 2, so the width \( \Delta x \) is:
\[
\Delta x = \frac{2 - 0}{n} = \frac{2}{n}
\]
where \( n \) is the number of subintervals. For simplicity, I'll choose \( n = 4 \), which means there will be 4 rectangles.
Now, let's find the right endpoints of each subinterval:
1. \( x_1 = 0 + \Delta x = \frac{2}{4} = 0.5 \)
2. \( x_2 = 0.5 + \Delta x = 1.0 \)
3. \( x_3 = 1.0 + \Delta x = 1.5 \)
4. \( x_4 = 1.5 + \Delta x = 2.0 \)
Next, I'll evaluate the function \( f(x) \) at each of these right endpoints:
1. \( f(0.5) = (0.5)^3 + 2(0.5)^2 + 1 = 0.125 + 0.5 + 1 = 1.625 \)
2. \( f(1.0) = (1)^3 + 2(1)^2 + 1 = 1 + 2 + 1 = 4 \)
3. \( f(1.5) = (1.5)^3 + 2(1.5)^2 + 1 = 3.375 + 4.5 + 1 = 8.875 \)
4. \( f(2.0) = (2)^3 + 2(2)^2 + 1 = 8 + 8 + 1 = 17 \)
Now, I'll calculate the area of each rectangle by multiplying the height \( f(x_i) \) by the width \( \Delta x \):
1. \( \text{Area}_1 = 1.625 \times 0.5 = 0.8125 \)
2. \( \text{Area}_2 = 4 \times 0.5 = 2 \)
3. \( \text{Area}_3 = 8.875 \times 0.5 = 4.4375 \)
4. \( \text{Area}_4 = 17 \times 0.5 = 8.5 \)
Finally, I'll sum these areas to get the total approximation of the area under the curve:
\[
\text{Total Area} = 0.8125 + 2 + 4.4375 + 8.5 = 15.75
\]
But wait, let me double-check my calculations to ensure accuracy. Recalculating:
- \( f(0.5) = 1.625 \)
- \( f(1.0) = 4 \)
- \( f(1.5) = 8.875 \)
- \( f(2.0) = 17 \)
- \( \Delta x = 0.5 \)
- \( \text{Total Area} = 0.8125 + 2 + 4.4375 + 8.5 = 15.75 \)
Everything seems to add up correctly. Therefore, the approximate area under the curve \( f(x) = x^{3} + 2x^{2} + 1 \) from \( x = 0 \) to \( x = 2 \) using a right Riemann sum with 4 subintervals is 15.75 square units.
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