Let's solve step-by-step: ### **Given Table:** | \( x \) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|----|----|----|----|----| | \( y \) | 0 | 2 | 2.5 | 2.3 | 2.1 | 1.5 | 0.5 | ### **Required:** The x-coordinate of the center of gravity (\( \bar{x} \)) is given by: \[ \bar{x} = \frac{\int_0^6 x y \, dx}{\int_0^6 y \, dx} \] Use Simpson’s Rule for both integrals. --- ## **Step 1: Simpson’s Rule Formula** For an interval \([a, b]\) divided into \( n \) **equal subintervals** (\( n \) even): \[ \int_a^b f(x) \, dx \approx \frac{h}{3} \left[ f_0 + 4f_1 + 2f_2 + 4f_3 + 2f_4 + \ldots + 4f_{n-1} + f_n \right] \] where \( h = \frac{b-a}{n} \). Here: \( a = 0 \), \( b = 6 \), \( n = 6 \) (since 7 points), \( h = 1 \). --- ## **Step 2: Compute Area \( A = \int_0^6 y \, dx \)** Let \( f(x) = y \): - \( y_0 = 0 \) - \( y_1 = 2 \) - \( y_2 = 2.5 \) - \( y_3 = 2.3 \) - \( y_4 = 2.1 \) - \( y_5 = 1.5 \) - \( y_6 = 0.5 \) Plug into Simpson’s Rule: \[ A = \frac{1}{3} [ y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6 ] \] Calculate: - \( y_0 = 0 \) - \( 4y_1 = 4 \times 2 = 8 \) - \( 2y_2 = 2 \times 2.5 = 5 \) - \( 4y_3 = 4 \times 2.3 = 9.2 \) - \( 2y_4 = 2 \times 2.1 = 4.2 \) - \( 4y_5 = 4 \times 1.5 = 6 \) - \( y_6 = 0.5 \) Sum: \( 0 + 8 + 5 + 9.2 + 4.2 + 6 + 0.5 = 32.9 \) So, \[ A = \frac{1}{3} \times 32.9 = 10.97 \] --- ## **Step 3: Compute \( \int_0^6 x y \, dx \)** Let \( f(x) = x y \): | \( x \) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|----|----|----|----|----| | \( y \) | 0 | 2 | 2.5 | 2.3 | 2.1 | 1.5 | 0.5 | | \( x y \) | 0 | 2 | 5 | 6.9 | 8.4 | 7.5 | 3 | Now, Simpson’s Rule: \[ \int_0^6 x y \, dx = \frac{1}{3} [ (0) + 4(2) + 2(5) + 4(6.9) + 2(8.4) + 4(7.5) + 3 ] \] - \( 4 \times 2 = 8 \) - \( 2 \times 5 = 10 \) - \( 4 \times 6.9 = 27.6 \) - \( 2 \times 8.4 = 16.8 \) - \( 4 \times 7.5 = 30 \) Sum: \( 0 + 8 + 10 + 27.6 + 16.8 + 30 + 3 = 95.4 \) So, \[ \int_0^6 x y \, dx = \frac{1}{3} \times 95.4 = 31.8 \] --- ## **Step 4: Compute \( \bar{x} \):** \[ \bar{x} = \frac{31.8}{10.97} = 2.90 \] --- ## **Final Answer:** \[ \boxed{\bar{x} = 2.90} \] --- ### **Summary Table** | Integral | Value | |---|---| | \( \int_0^6 y \, dx \) | 10.97 | | \( \int_0^6 x y \, dx \) | 31.8 | | \( \bar{x} \) | 2.90 | --- Let me know if you need further explanation or steps!

Let's solve the problem step-by-step. **Given Data:** - The x-coordinates are 0, 1, 2, 3, 4, 5, 6. - Corresponding y-values are 0, 2, 2.5, 2.3, 2.1, 1.5, 0.5. **Goal:** Find the x-coordinate of the centroid (\( \bar{x} \)) of the area bounded by the curve, the end ordinates, and the x-axis. The formula for \( \bar{x} \) is: \[ \bar{x} = \frac{\int_0^6 x y \, dx}{\int_0^6 y \, dx} \] We will evaluate both integrals using Simpson's Rule, which is suitable here because the data is given at equally spaced points and the number of intervals \( n = 6 \) (which is even). ### Step 1: Calculate the area \( A = \int y \, dx \) Using Simpson's Rule: \[ A \approx \frac{h}{3} \left[ y_0 + 4 y_1 + 2 y_2 + 4 y_3 + 2 y_4 + 4 y_5 + y_6 \right] \] where \( h = 1 \). Plugging in the y-values: - \( y_0 = 0 \) - \( y_1 = 2 \) - \( y_2 = 2.5 \) - \( y_3 = 2.3 \) - \( y_4 = 2.1 \) - \( y_5 = 1.5 \) - \( y_6 = 0.5 \) Compute the sum inside the brackets: \[ 0 + 4 \times 2 + 2 \times 2.5 + 4 \times 2.3 + 2 \times 2.1 + 4 \times 1.5 + 0.5 \] \[ = 0 + 8 + 5 + 9.2 + 4.2 + 6 + 0.5 = 32.9 \] Therefore, \[ A \approx \frac{1}{3} \times 32.9 = 10.97 \] ### Step 2: Calculate \( \int x y \, dx \) Construct the values of \( x y \): - At \( x=0 \): \( 0 \times 0 = 0 \) - At \( x=1 \): \( 1 \times 2 = 2 \) - At \( x=2 \): \( 2 \times 2.5 = 5 \) - At \( x=3 \): \( 3 \times 2.3 = 6.9 \) - At \( x=4 \): \( 4 \times 2.1 = 8.4 \) - At \( x=5 \): \( 5 \times 1.5 = 7.5 \) - At \( x=6 \): \( 6 \times 0.5 = 3 \) Applying Simpson's Rule: \[ \int_0^6 x y \, dx \approx \frac{h}{3} \left[ x y_0 + 4 x y_1 + 2 x y_2 + 4 x y_3 + 2 x y_4 + 4 x y_5 + x y_6 \right] \] Sum inside brackets: \[ 0 + 4 \times 2 + 2 \times 5 + 4 \times 6.9 + 2 \times 8.4 + 4 \times 7.5 + 3 \] \[ = 0 + 8 + 10 + 27.6 + 16.8 + 30 + 3 = 95.4 \] Thus, \[ \int_0^6 x y \, dx \approx \frac{1}{3} \times 95.4 = 31.8 \] ### Step 3: Calculate \( \bar{x} \) Now, compute: \[ \bar{x} = \frac{31.8}{10.97} \approx 2.90 \] ### **Final answer:** \[ \boxed{\bar{x} \approx 2.90} \] This is the x-coordinate of the centroid of the area bounded by the curve and the x-axis.