Question

Evaluate integral of a = 0 and b =π\2 sin(t)dt by applying Simpsons rule with four equal intervals

Answer 1

$$\displaystyle\int_{a}^{b}f(t)dt$$

$$\approx\dfrac{\Delta t}{3}\big(f(t_0)+4f(t_1)+2f(t_2)+4f(t_3)$$

$$+2f(t_4)+...+4f(t_{n-2})+2f(t_{n-1})+f(t_n)\big)$$

$$\Delta t=\dfrac{b-a}{n}$$

We have that $f(t)=\sin t, a=0, b=\pi/2, n=4.$

Therefore

$$\Delta t=\dfrac{\pi/2-0}{4}=\pi/8$$

Divide the interval $[0, \pi/2]$  into $n=4$ subintervals of the length $Δt=π/8$ with the following endpoints: $a=0, \pi/8, \pi/4, 3\pi/8, \pi/2=b.$

 Evaluate the function at these endpoints

$$f(t_0)=f(0)=\sin(0)=0$$

$$4f(t_1)=4f(\pi/8)=4\sin(\pi/8)$$

$$≈1.530733729460359$$

$$2f(t_2)=2f(\pi/4)=2\sin(\pi/4)=\sqrt{2}$$

$$\approx1.414213562373095$$

$$4f(t_3)=4f(3\pi/8)=4\sin(3\pi/8)$$

$$≈3.695518130045147$$

$$f(t_4)=f(\pi/2)=\sin(\pi/2)=1$$

Therefore

$$\displaystyle\int_{0}^{\pi/2}\sin t dt$$

$$\approx(\pi/12)(0+1.530733729460359$$

$$+1.414213562373095$$

$$+3.695518130045147+1)$$

$$\approx1.000134584974194$$

$$\displaystyle\int_{0}^{\pi/2}\sin t dt\approx1.000134584974194$$