approximate the definite integral $int_{5}^{9} sin(x^{2}) dx$ using the trapezoidal rule with $n = 8$. round…

Explanation:

Step1: Calculate the width of sub - intervals

The formula for the width $\Delta x$ of sub - intervals in the Trapezoidal Rule is $\Delta x=\frac{b - a}{n}$, where $a = 5$, $b = 9$, and $n = 8$. $\Delta x=\frac{9 - 5}{8}=\frac{4}{8}=0.5$

Step2: Determine the endpoints of sub - intervals

The endpoints $x_i=a + i\Delta x$, for $i = 0,1,\cdots,8$. So $x_0 = 5$, $x_1=5 + 0.5=5.5$, $x_2=5+2\times0.5 = 6$, $x_3=5 + 3\times0.5=6.5$, $x_4=5+4\times0.5 = 7$, $x_5=5+5\times0.5 = 7.5$, $x_6=5+6\times0.5 = 8$, $x_7=5+7\times0.5 = 8.5$, $x_8=5+8\times0.5 = 9$.

Step3: Evaluate the function at endpoints

Let $f(x)=\sin(x^{2})$. Then $f(x_0)=\sin(5^{2})=\sin(25)$, $f(x_1)=\sin(5.5^{2})=\sin(30.25)$, $f(x_2)=\sin(6^{2})=\sin(36)$, $f(x_3)=\sin(6.5^{2})=\sin(42.25)$, $f(x_4)=\sin(7^{2})=\sin(49)$, $f(x_5)=\sin(7.5^{2})=\sin(56.25)$, $f(x_6)=\sin(8^{2})=\sin(64)$, $f(x_7)=\sin(8.5^{2})=\sin(72.25)$, $f(x_8)=\sin(9^{2})=\sin(81)$.

Step4: Apply the Trapezoidal Rule formula

The Trapezoidal Rule formula is $T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n - 1})+f(x_n)]$. $T_8=\frac{0.5}{2}[\sin(25)+2\sin(30.25)+2\sin(36)+2\sin(42.25)+2\sin(49)+2\sin(56.25)+2\sin(64)+2\sin(72.25)+\sin(81)]$ Using a calculator to evaluate each sine - value and perform the arithmetic operations: $\sin(25)\approx0.423$, $\sin(30.25)\approx - 0.988$, $\sin(36)\approx - 0.588$, $\sin(42.25)\approx0.891$, $\sin(49)\approx - 0.954$, $\sin(56.25)\approx - 0.657$, $\sin(64)\approx0.502$, $\sin(72.25)\approx0.997$, $\sin(81)\approx0.988$ $T_8 = 0.25[0.423+2\times(- 0.988)+2\times(- 0.588)+2\times0.891+2\times(- 0.954)+2\times(- 0.657)+2\times0.502+2\times0.997+0.988]$ $T_8=0.25[0.423-1.976 - 1.176 + 1.782-1.908-1.314 + 1.004+1.994+0.988]$ $T_8=0.25[0.423+1.782+1.004+1.994+0.988-(1.976 + 1.176+1.908+1.314)]$ $T_8=0.25[6.191 - 6.374]$ $T_8=0.25\times(- 0.183)=-0.046$

Answer:

$-0.046$