91Ó°ÊÓ

The formula for the Trapezoid Rule is given by: \(T_{n}=\frac{\Delta x}{2}\left[f(x_{0})+2f(x_{1})+2f(x_{2})+\cdots+2f(x_{n-1})+f(x_{n})\right]\) where \(n\) is the number of subintervals, \(\Delta x= \frac{b-a}{n}\) is the width of each subinterval, and \(f(x_i)\) is the value of the function at \(x_i\). For our given integral, the function is \(f(x) = x^3\), the interval is \([1, 9]\), and we are asked to approximate using \(n = 2, 4, 8\) subintervals.

For \(n=2\), first find the width of each subinterval: \(\Delta x = \frac{9-1}{2} = 4\) Now, calculate the function values at the endpoints of the subintervals: \(f(x_0) = f(1) = 1^3 = 1\) \(f(x_1) = f(5) = 5^3 = 125\) \(f(x_2) = f(9) = 9^3 = 729\) Plug the values into the trapezoid rule formula: \(T_{2}=\frac{4}{2}\left[f(1)+2f(5)+f(9)\right] = \frac{4}{2}(1+2(125)+729) = 1716\)

For \(n=4\), find the width of each subinterval: \(\Delta x = \frac{9-1}{4} = 2\) Calculate the function values at the endpoints of the subintervals: \(f(x_0) = f(1) = 1^3 = 1\) \(f(x_1) = f(3) = 3^3 = 27\) \(f(x_2) = f(5) = 5^3 = 125\) \(f(x_3) = f(7) = 7^3 = 343\) \(f(x_4) = f(9) = 9^3 = 729\) Plug the values into the trapezoid rule formula: \(T_{4} = \frac{2}{2}\left[f(1)+2(f(3)+f(5)+f(7))+f(9)\right] = 1372\)

For \(n=8\), find the width of each subinterval: \(\Delta x = \frac{9-1}{8} = 1\) Calculate the function values at the endpoints of the subintervals: \(f(x_0) = f(1) = 1^3 = 1\) \(f(x_1) = f(2) = 2^3 = 8\) \(f(x_2) = f(3) = 3^3 = 27\) \(f(x_3) = f(4) = 4^3 = 64\) \(f(x_4) = f(5) = 5^3 = 125\) \(f(x_5) = f(6) = 6^3 = 216\) \(f(x_6) = f(7) = 7^3 = 343\) \(f(x_7) = f(8) = 8^3 = 512\) \(f(x_8) = f(9) = 9^3 = 729\) Plug the values into the trapezoid rule formula: \(T_{8} = \frac{1}{2}\left[f(1) + 2\sum_{i=1}^{7} f(x_i)+ f(9)\right] = \frac{1}{2}(1 + 2(8+27+64+125+216+343+512) + 729) = 5264\) The Trapezoid Rule approximations for the integral \(\int_{1}^{9} x^{3}d x\) using \(n=2,4,\) and \(8\) subintervals are: \(T_2 = 1716\) \(T_4 = 1372\) \(T_8 = 5264\)