91影视

We have given,

$f\left(x\right)=\sin \left(x\right),[a,b]=[0,蟺]$and n = 3

The trapezoidal rule uses trapezoids to approximate the area:

$$\begin{gathered} {鈭�}_a^{b}f\left(x\right)dx鈥�鈮�\frac{螖x}{2}\left(f\right(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n\left)\right) \\ Where, \\ 螖鈥�x鈥�=鈥�\frac{b-a}{n} \end{gathered}$$

Where, n is the total number of subintervals.

Upper Riemann sum formula:

${鈭�}_a^{b}f\left(x\right)dx鈥�鈮�螖x\left(f\right(x_1)+f(x_2)+f(x_3)+...+f(x_n\left)\right)$

From the given information in step 1. in part (a),

$螖鈥�x=\frac{\mathrm{蟺}-0}{3}=\frac{\mathrm{蟺}}{3}$

Length of the subintervals is $\frac{蟺}{3}$.

So dividing the interval $[0,蟺]$in to the subintervals with length $\frac{蟺}{3}$is,

$\left[0,\frac{蟺}{3}\right],\left[\frac{蟺}{3},\frac{2蟺}{3}\right],\left[\frac{2蟺}{3},\mathrm{蟺}\right]$

So end points are: $0,\frac{蟺}{3},\frac{2蟺}{3},\mathrm{蟺}$

Now, just evaluating the function at the left endpoints of the subintervals,

$sin\left(0\right)=0,sin\left(\frac{蟺}{3}\right)=\frac{\sqrt{3}}{2},sin\left(\frac{2蟺}{3}\right)=\frac{\sqrt{3}}{2}\mathrm{and}sin\left(蟺\right)=0$

Using trapezoidal formula,

localid="1648579359642" $$\begin{gathered} {鈭�}_{鈥�0}^{蟺}\sin \left(x\right)dx鈮�\frac{蟺}{6}[0+2脳\frac{\sqrt{3}}{2}+2脳\frac{\sqrt{3}}{2}+0] \\ 鈮�\frac{\sqrt{3}蟺}{3} \\ 鈮�\frac{蟺}{\sqrt{3}} \end{gathered}$$

Hence, using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is,

${鈭�}_{鈥�0}^{蟺}\sin \left(x\right)dx鈮�\frac{蟺}{\sqrt{3}}$

Using given information from the part (a) in step 1,

$f\left(x\right)=\sin \left(x\right),[a,b]=[0,\mathrm{蟺}]$

Using upper Riemann sum formula,

$螖鈥�x=\frac{\mathrm{蟺}-0}{3}=\frac{\mathrm{蟺}}{3}$

So length of the intervals is $\frac{\mathrm{蟺}}{3}$.

Dividing the given interval in to the subintervals with length $\frac{\mathrm{蟺}}{3}$,

So intervals are: $\left[0,\frac{\mathrm{蟺}}{3}\right],\left[\frac{\mathrm{蟺}}{3},\frac{2\mathrm{蟺}}{3}\right],\left[\frac{2\mathrm{蟺}}{3},\mathrm{蟺}\right]$

Upper end points are:$\frac{\mathrm{蟺}}{3},\frac{2\mathrm{蟺}}{3},\mathrm{蟺}$

Now just evaluating the functions for those endpoints,

$\mathrm{f}\left(\frac{\mathrm{蟺}}{3}\right)=\frac{\sqrt{3}}{2},\mathrm{f}\left(\frac{2\mathrm{蟺}}{3}\right)=\frac{\sqrt{3}}{2}\mathrm{and}\mathrm{f}\left(\mathrm{蟺}\right)=0$

Using upper Riemann sum formula,

$$\begin{gathered} {鈭�}_0^{蟺}\sin \left(x\right)dx鈮�\frac{\mathrm{蟺}}{3}\left[\mathrm{f}\left(\frac{\mathrm{蟺}}{3}\right)+\mathrm{f}\left(\frac{2\mathrm{蟺}}{3}\right)+\mathrm{f}\left(\mathrm{蟺}\right)\right] \\ 鈮�\frac{\mathrm{蟺}}{3}\left[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}+0\right] \\ 鈮�\frac{\mathrm{蟺}}{\sqrt{3}} \end{gathered}$$

Hence, using approximation method to approximate the signed area between the graph of f and the x-axis on [a, b] is,

${鈭�}_0^{蟺}\sin \left(x\right)dx=\frac{蟺}{\sqrt{3}}$