Questions: Question 19 Mark this question Use a Riemann sum with 4 rectangles of equal width to approximate the area between y=x^3+2 and the x-axis on the interval [-1,1]. Use the righthand endpoint of each subinterval. 4.5 units ^2 9 units ^2 8.0625 units ^2 3.5 units ^2

Solution Steps

To approximate the area under the curve \( y = x^3 + 2 \) on the interval \([-1, 1]\) using a Riemann sum with 4 rectangles of equal width, we will:

1. Divide the interval \([-1, 1]\) into 4 equal subintervals.
2. Calculate the width of each subinterval.
3. Use the right-hand endpoint of each subinterval to determine the height of each rectangle.
4. Sum the areas of the rectangles to approximate the total area.

We are given the function \( y = x^3 + 2 \) and the interval \([-1, 1]\). We need to approximate the area under this curve using a Riemann sum with 4 rectangles of equal width.

The interval \([-1, 1]\) is divided into 4 equal subintervals. The width of each subinterval is: \[ \text{width} = \frac{1 - (-1)}{4} = \frac{2}{4} = 0.5 \]

The right-hand endpoints of each subinterval are: \[ x_1 = -1 + 0.5 = -0.5, \quad x_2 = -0.5 + 0.5 = 0, \quad x_3 = 0 + 0.5 = 0.5, \quad x_4 = 0.5 + 0.5 = 1 \]

Using the right-hand endpoints, we calculate the height of each rectangle: \[ f(x_1) = f(-0.5) = (-0.5)^3 + 2 = -0.125 + 2 = 1.875 \] \[ f(x_2) = f(0) = 0^3 + 2 = 2 \] \[ f(x_3) = f(0.5) = (0.5)^3 + 2 = 0.125 + 2 = 2.125 \] \[ f(x_4) = f(1) = 1^3 + 2 = 1 + 2 = 3 \]

The area of each rectangle is the height times the width. Summing these areas gives: \[ \text{Area} = \sum_{i=1}^{4} f(x_i) \cdot \text{width} = (1.875 \cdot 0.5) + (2 \cdot 0.5) + (2.125 \cdot 0.5) + (3 \cdot 0.5) \] \[ = 0.9375 + 1 + 1.0625 + 1.5 = 4.5 \]

Final Answer