Question

the instructions for the given integral have two parts, one for the trapezoidal rule and one for simpsons rule. complete the following parts.
int_{0}^{pi} sin t d t

1. using the trapezoidal rule complete the following.

a. estimate the integral with ( n = 4 ) steps and find an upper bound for ( left|e_{t}
ight| ).
( t approx square )
(round to four decimal places as needed.)

Explanation:

Step1: Calculate \(\Delta x\)

For the integral \(\int_{a}^{b}f(t)dt=\int_{0}^{\pi}\sin tdt\), \(a = 0\), \(b=\pi\), \(n = 4\).
\(\Delta x=\frac{b - a}{n}=\frac{\pi-0}{4}=\frac{\pi}{4}\)

Step2: Find the partition points

\(t_{0}=a = 0\), \(t_{1}=a+\Delta x=\frac{\pi}{4}\), \(t_{2}=a + 2\Delta x=\frac{\pi}{2}\), \(t_{3}=a+3\Delta x=\frac{3\pi}{4}\), \(t_{4}=a + 4\Delta x=\pi\)

Step3: Calculate \(f(t_{i})\)

\(f(t_{0})=\sin(0)=0\), \(f(t_{1})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\approx0.7071\), \(f(t_{2})=\sin(\frac{\pi}{2}) = 1\), \(f(t_{3})=\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}\approx0.7071\), \(f(t_{4})=\sin(\pi)=0\)

Step4: Apply the trapezoidal rule formula \(T=\frac{\Delta x}{2}[f(t_{0})+2f(t_{1})+2f(t_{2})+2f(t_{3})+f(t_{4})]\)

\(T=\frac{\pi/4}{2}[0 + 2\times0.7071+2\times1+2\times0.7071+0]\)
\(T=\frac{\pi}{8}(0 + 1.4142+2 + 1.4142+0)\)
\(T=\frac{\pi}{8}(4.8284)\approx\frac{3.1416}{8}\times4.8284\approx1.8961\)

Step5: Find the second - derivative of \(y = f(t)=\sin t\)

\(y'=\cos t\), \(y''=-\sin t\). The absolute value of the second - derivative \(|y''|=|-\sin t|\). On the interval \([0,\pi]\), \(|y''|\leq1\).
The error formula for the trapezoidal rule is \(|E_{T}|\leq\frac{(b - a)^{3}}{12n^{2}}M\), where \(M\) is the upper bound of \(|y''|\) on \([a,b]\).
Substitute \(a = 0\), \(b=\pi\), \(n = 4\), \(M = 1\)
\(|E_{T}|\leq\frac{(\pi-0)^{3}}{12\times4^{2}}\times1=\frac{\pi^{3}}{192}\approx\frac{31.006}{192}\approx0.1615\)

Answer:

\(T\approx1.8961\)