91影视

The function is,

$f\left(x\right)=\cos x$

The interval is$\left[-2\mathrm{蟺},2\mathrm{蟺}\right]$.

The objective is to find the signed area with at least eight rectangles. The left-sum defined for $n$rectangles on $\left[a,b\right]$is $\underset{k=1}{\overset{n}{鈭�}}f\left(x_{k-1}\right)鈭�x$.

Where, $鈭�x=\frac{b-a}{n}$

localid="1648708502892" $x_k=a+k鈭�x$

Now,

$$\begin{gathered} 鈭�x=\frac{2\mathrm{蟺}+2\mathrm{蟺}}{8} \\ =\frac{4\mathrm{蟺}}{8} \\ =\frac{\mathrm{蟺}}{2} \end{gathered}$$

So,

$x_k=-2\mathrm{蟺}+\mathrm{k}\frac{\mathrm{蟺}}{8}$

In the left sum, $x_{k-1}$is the left most point of the interval $\left[x_{k-1},x_k\right]$.

So,

$x_{k-1}=-2\mathrm{蟺}+\left(\mathrm{k}-1\right)\frac{\mathrm{蟺}}{8}$

The left sum is,

$$\begin{gathered} \underset{k=1}{\overset{8}{鈭�}}\left(\cos \left(-2\mathrm{蟺}+\left(\mathrm{k}-1\right)\frac{\mathrm{蟺}}{8}\right)\right)\frac{\mathrm{蟺}}{2} \\ =\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(1-1\right)\frac{\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(2-1\right)\frac{\mathrm{蟺}}{8}\right)+ \\ \frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(3-1\right)\frac{\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(4-1\right)\frac{\mathrm{蟺}}{8}\right) \\ +\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(5-1\right)\frac{\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(6-1\right)\frac{\mathrm{蟺}}{8}\right)+ \\ \frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(7-1\right)\frac{\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\left(8-1\right)\frac{\mathrm{蟺}}{8}\right) \\ =\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{\mathrm{蟺}}{4}\right) \\ +\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{3\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{\mathrm{蟺}}{2}\right) \\ +\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{5\mathrm{蟺}}{8}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{3\mathrm{蟺}}{4}\right)+\frac{\mathrm{蟺}}{2}\cos \left(-2\mathrm{蟺}+\frac{7\mathrm{蟺}}{8}\right) \\ =0 \end{gathered}$$

Therefore, the signed area is $0$.

The objective is to find the absolute area between the graphs of the function from $x=a$to $x=b$.

The graph of the function is,

The absolute area is,

$$\begin{gathered} {鈭�}_{-2\mathrm{蟺}}^{2\mathrm{蟺}}f\left(x\right)dx \\ ={鈭�}_{-2\mathrm{蟺}}^{2\mathrm{蟺}}\cos xdx \\ ={鈭�}_{-2\mathrm{蟺}}^{\mathrm{蟺}}-\cos xdx+{鈭�}_{-\mathrm{蟺}}^0\cos xdx+{鈭�}_0^{\mathrm{蟺}}-\cos xdx+{鈭�}_{\mathrm{蟺}}^{2\mathrm{蟺}}\cos xdx \\ =-{\left[\sin x\right]}_{-2\mathrm{蟺}}^{-\mathrm{蟺}}+{\left[\sin x\right]}_{-\mathrm{蟺}}^0-{\left[\sin x\right]}_0^{\mathrm{蟺}}+{\left[\sin x\right]}_{\mathrm{蟺}}^{2\mathrm{蟺}} \\ =0+0+0+0 \\ =0 \end{gathered}$$

Therefore, the absolute area is$0$.