How do you find the area using the trapezoidal approximation method, given #(x²-6x+9) dx#, on the interval [0,3] with n=3?

Question

Aurora Alvarado · Accepted Answer

# int_0^3 \ x^2-6x+9 \ dx ~~ 9.5 #

We have:

# y = x^2-6x+9 #

We want to estimate #int \ y \ dx# over the interval #[0,3]# with #3# strips; thus:

# Deltax = (3-0)/3 = 1#

Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;

Trapezium Rule

# A = int_0^3 \ x^2-6x+9 \ dx #
# \ \ \ ~~ 1/2 * { 9 - 0 + 2*(4 + 1) } #
# \ \ \ = 0.5 * { 9 + 2*(5) } #
# \ \ \ = 0.5 * { 9 + 10 } #
# \ \ \ = 0.5 * 19 #
# \ \ \ = 9.5 #

Actual Value

For comparison of accuracy:
# A = int_0^3 \ x^2-6x+9 \ dx #

# \ \ \ = [x^3/3-3x^2+9x]_0^3 #
# \ \ \ = 9-27+27 #
# \ \ \ = 9 #

Adam Adkins · Answer

undefined To find the area using the trapezoidal approximation method with $ n = 3 $ on the interval $[0,3]$ for the function $ x^2 - 6x + 9 $, follow these steps:

1. Divide the interval $[0,3]$ into $ n $ subintervals. Since $ n = 3 $, there will be $ 3 $ subintervals of equal width. The width of each subinterval will be $ \Delta x = \frac{3-0}{3} = 1 $.

2. Compute the function values at the endpoints of each subinterval and at each interior point. Since $ f(x) = x^2 - 6x + 9 $, evaluate $ f(x) $ at $ x = 0, 1, 2, $ and $ 3 $.

3. Use the trapezoidal rule formula to calculate the area for each trapezoid. The formula is:

$$
A = \frac{1}{2} \cdot (\text{{base}}_1 + \text{{base}}_2) \cdot \text{{height}}
$$

where the bases are the function values at the endpoints of each subinterval, and the height is the width of each subinterval ($ \Delta x $).

4. Sum up the areas of all the trapezoids to find the total approximate area under the curve.

Following these steps, you'll approximate the area under the curve $ x^2 - 6x + 9 $ on the interval $[0,3]$ using the trapezoidal approximation method with $ n = 3 $.