Question Statement

Evaluate ${\int }_0^{\frac{\pi }{2}}\sin xdx$ using the Trapezoidal Rule and Simpson's $\frac{1}{3}$ Rule by taking $n=6$.

---

Background and Explanation

This problem requires numerical integration techniques to approximate the area under the curve $y=\sin x$ from $0$ to $\frac{\pi }{2}$. The Trapezoidal Rule approximates the area using trapezoids, while Simpson's $\frac{1}{3}$ Rule uses parabolic arcs for better accuracy, requiring an even number of subintervals ($n=6$ satisfies this).

---

Solution

Setup and Parameters

First, we identify the given parameters by comparing ${\int }_0^{\frac{\pi }{2}}\sin xdx$ with the standard form ${\int }_a^{b}f(x)dx$:

$a=0,b=\frac{\pi }{2},f(x)=\sin x$

Step 1: Calculate Step Size $h$ and Grid Points $x_i$

The step size is: $h=\frac{b-a}{n}=\frac{\frac{\pi }{2}-0}{6}=\frac{\pi }{12}$

The grid points are calculated as $x_i=x_{i-1}+h$:

$\begin{array}{rl}{x}_0& =0\\ x_1& =0+\frac{\pi }{12}=\frac{\pi }{12}\\ x_2& =\frac{\pi }{12}+\frac{\pi }{12}=\frac{\pi }{6}\\ x_3& =\frac{\pi }{6}+\frac{\pi }{12}=\frac{\pi }{4}\\ x_4& =\frac{\pi }{4}+\frac{\pi }{12}=\frac{\pi }{3}\\ x_5& =\frac{\pi }{3}+\frac{\pi }{12}=\frac{5\pi }{12}\\ x_6& =\frac{5\pi }{12}+\frac{\pi }{12}=\frac{\pi }{2}\end{array}$

Step 2: Calculate Function Values $f(x_i)=\sin (x_i)$

Evaluating the sine function at each grid point:

$\begin{array}{rl}f(x_0)& =\sin (0)=0\\ f(x_1)& =\sin \left(\frac{\pi }{12}\right)\approx 0.2588\\ f(x_2)& =\sin \left(\frac{\pi }{6}\right)=0.5\\ f(x_3)& =\sin \left(\frac{\pi }{4}\right)\approx 0.7071\\ f(x_4)& =\sin \left(\frac{\pi }{3}\right)\approx 0.8660\\ f(x_5)& =\sin \left(\frac{5\pi }{12}\right)\approx 0.9659\\ f(x_6)& =\sin \left(\frac{\pi }{2}\right)=1\end{array}$

Step 3: Apply the Trapezoidal Rule

The Trapezoidal Rule formula is: ${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[f(x_0)+2{\sum }_{i=1}^{n-1}f(x_i)+f(x_n)\right]$

Substituting the values: $\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\sin xdx& \approx \frac{\pi /12}{2}\left[f(x_0)+2\left(f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)\right)+f(x_6)\right]\\ & \approx \frac{\pi }{24}\left[0+2(0.2588+0.5+0.7071+0.8660+0.9659)+1\right]\\ & \approx \frac{\pi }{24}\left[0+2(3.2978)+1\right]\\ & \approx \frac{\pi }{24}(6.5956+1)\\ & \approx \frac{\pi }{24}(7.5956)\\ & \approx 0.9943\end{array}$

Step 4: Apply Simpson's $\frac{1}{3}$ Rule

Simpson's $\frac{1}{3}$ Rule formula (for even $n$) is: ${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[f(x_0)+4{\sum }_{\text{odd }i}f(x_i)+2{\sum }_{\text{even }i}f(x_i)+f(x_n)\right]$

Substituting the values (note: $h/3=\pi /36$): $\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\sin xdx& \approx \frac{\pi }{36}\left[f(x_0)+4\left(f(x_1)+f(x_3)+f(x_5)\right)+2\left(f(x_2)+f(x_4)\right)+f(x_6)\right]\\ & \approx \frac{\pi }{36}\left[0+4(0.2588+0.7071+0.9659)+2(0.5+0.8660)+1\right]\\ & \approx \frac{\pi }{36}\left[0+4(1.9318)+2(1.366)+1\right]\\ & \approx \frac{\pi }{36}(7.7272+2.732+1)\\ & \approx \frac{\pi }{36}(11.4592)\\ & \approx 1.000004\end{array}$

---

Key Formulas or Methods Used

• Step Size: $h=\frac{b-a}{n}$
• Trapezoidal Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{2}\left[y_0+2(y_1+y_2+\cdots +y_{n-1})+y_n\right]$
• Simpson's $\frac{1}{3}$ Rule: ${\int }_a^{b}f(x)dx\approx \frac{h}{3}\left[y_0+4(y_1+y_3+\cdots )+2(y_2+y_4+\cdots )+y_n\right]$ (where $n$ is even)

---

Summary of Steps

1. Identify parameters: Extract $a=0$, $b=\frac{\pi }{2}$, $n=6$, and calculate $h=\frac{\pi }{12}$
2. Generate grid: Compute $x_0$ through $x_6$ using $x_i=x_{i-1}+h$
3. Evaluate function: Calculate $f(x_i)=\sin (x_i)$ for all grid points
4. Apply Trapezoidal Rule: Use $\frac{h}{2}[y_0+2(\text{sum of middle terms})+y_6]$ to get $\approx 0.9943$
5. Apply Simpson's Rule: Use $\frac{h}{3}[y_0+4(\text{sum of odd indices})+2(\text{sum of even indices})+y_6]$ to get $\approx 1.000004$