(b) (10) use simpson’s rule with n = 6 to approximate ∫₀³ e^(t²) dt.

Explanation:

Step1: Recall Simpson's Rule formula

Simpson's Rule is given by $S_n=\frac{\Delta x}{3}[f(x_0) + 4f(x_1)+2f(x_2)+4f(x_3)+\cdots+2f(x_{n - 2})+4f(x_{n - 1})+f(x_n)]$, where $\Delta x=\frac{b - a}{n}$, and $x_i=a + i\Delta x$ for $i = 0,1,\cdots,n$. Here, $a = 0$, $b = 3$, and $n = 6$.

Step2: Calculate $\Delta x$

$\Delta x=\frac{b - a}{n}=\frac{3-0}{6}=\frac{1}{2}$.

Step3: Determine the $x_i$ values

$x_0 = 0$, $x_1=0+\frac{1}{2}=\frac{1}{2}$, $x_2 = 0 + 2\times\frac{1}{2}=1$, $x_3=0 + 3\times\frac{1}{2}=\frac{3}{2}$, $x_4=0 + 4\times\frac{1}{2}=2$, $x_5=0 + 5\times\frac{1}{2}=\frac{5}{2}$, $x_6=0+6\times\frac{1}{2}=3$.

Step4: Calculate $f(x_i)$ values

$f(x)=e^{x^{2}}$. So, $f(x_0)=e^{0^{2}} = 1$, $f(x_1)=e^{(\frac{1}{2})^{2}}=e^{\frac{1}{4}}$, $f(x_2)=e^{1^{2}}=e$, $f(x_3)=e^{(\frac{3}{2})^{2}}=e^{\frac{9}{4}}$, $f(x_4)=e^{2^{2}}=e^{4}$, $f(x_5)=e^{(\frac{5}{2})^{2}}=e^{\frac{25}{4}}$, $f(x_6)=e^{3^{2}}=e^{9}$.

Step5: Apply Simpson's Rule

$S_6=\frac{\frac{1}{2}}{3}[1 + 4e^{\frac{1}{4}}+2e+4e^{\frac{9}{4}}+2e^{4}+4e^{\frac{25}{4}}+e^{9}]$ $=\frac{1}{6}[1 + 4e^{\frac{1}{4}}+2e+4e^{\frac{9}{4}}+2e^{4}+4e^{\frac{25}{4}}+e^{9}]$. Using a calculator: $e^{\frac{1}{4}}\approx1.284$, $e\approx2.718$, $e^{\frac{9}{4}}\approx9.488$, $e^{4}\approx54.598$, $e^{\frac{25}{4}}\approx331.117$, $e^{9}\approx8103.084$. $S_6=\frac{1}{6}[1+4\times1.284 + 2\times2.718+4\times9.488+2\times54.598+4\times331.117+8103.084]$ $=\frac{1}{6}[1 + 5.136+5.436 + 37.952+109.196+1324.468+8103.084]$ $=\frac{1}{6}[9586.272]$ $= 1597.712$.

Answer:

$1597.712$