Q. 24 - Calculus

Question

For each function $f$ and interval $\left[a,b\right]$in Exercises $23-25$, use at least eight rectangles to approximate $\left(a\right)$ the signed area and $\left(b\right)$ the absolute area between the graph of $f$ and the $x-axis$ from $x=a$ to $x=b$. Your work should include a graph of $f$ together with the rectangles that you used.

$f\left(x\right)=\cos x, \left[-2\mathrm{\pi },2\mathrm{\pi }\right]$

Step-by-Step Solution

Answer

a. The signed area is $0$.

b. The absolute area is $0$.

Step 1. GIven Information

The function is,

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

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

Part (a). Step 2. Calculation

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 $\sum _{k=1}^{n}f\left(x_{k-1}\right)∆x$.

Where, $∆x=\frac{b-a}{n}$

              $x_k=a+k∆x$

Now,

         $$\begin{gathered} ∆x=\frac{2\mathrm{\pi }+2\mathrm{\pi }}{8} \\ =\frac{4\mathrm{\pi }}{8} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

So,

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

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

So,

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

The left sum is,

$$\begin{gathered} \sum _{k=1}^8\left(\cos \left(-2\mathrm{\pi }+\left(\mathrm{k}-1\right)\frac{\mathrm{\pi }}{8}\right)\right)\frac{\mathrm{\pi }}{2} \\ =\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(1-1\right)\frac{\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(2-1\right)\frac{\mathrm{\pi }}{8}\right)+ \\ \frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(3-1\right)\frac{\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(4-1\right)\frac{\mathrm{\pi }}{8}\right) \\ +\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(5-1\right)\frac{\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(6-1\right)\frac{\mathrm{\pi }}{8}\right)+ \\ \frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(7-1\right)\frac{\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\left(8-1\right)\frac{\mathrm{\pi }}{8}\right) \\ =\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\right) \\ +\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{3\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{\mathrm{\pi }}{2}\right) \\ +\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{5\mathrm{\pi }}{8}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{3\mathrm{\pi }}{4}\right)+\frac{\mathrm{\pi }}{2}\cos \left(-2\mathrm{\pi }+\frac{7\mathrm{\pi }}{8}\right) \\ =0 \end{gathered}$$

Therefore, the signed area is $0$.

Part (b). Step2. Graph

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,

Part (b). Step2. Calculation

The absolute area is,

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

Therefore, the absolute area is $0$.