Question

Repeat Exercise 11 for the function f shown above at the right, on the interval [−2, 2].

Short answer

Part (a): Area = 0.54

Part (b): Area = 0

Part (c): Area = 0

Step-by-step solution

Step 1. Given information is:

A function f plotted on the graph on [-2,2]

Part (a) Step 1. Left Sum Approximation

Given function is f(x) = 3cos(3x)

$$\begin{gathered} \mathrm{The}\mathrm{left}\mathrm{sum}\mathrm{approximation}\mathrm{for}\mathrm{area}\mathrm{of}\mathrm{this}\mathrm{region}\mathrm{with}\mathrm{n}=8\mathrm{can}\mathrm{be}\mathrm{formulated}\mathrm{as}: \\ \begin{array}{r}\mathrm{\Delta }x=\frac{2-(-2)}{8}=\frac{1}{2};x_k=-2+\frac{1}{2}k;f\left(x_{k-1}\right)=-2+\frac{1}{2}(k-1)\\ \sum _{k=1}^8 f\left(x_{k-1}\right)\mathrm{\Delta }x=\\ =\frac{1}{2}\left[f\left(-2+\frac{1}{2}(1-1)\right)+f\left(-2+\frac{1}{2}(2-1)\right)+f\left(-2+\frac{1}{2}(3-1)\right)\right.\\ =+f\left(-2+\frac{1}{2}(4-1)\right)+f\left(-2+\frac{1}{2}(5-1)\right)+f\left(-2+\frac{1}{2}(6-1)\right)\\ \left.=+f\left(-2+\frac{1}{2}(7-1)\right)+f\left(-2+\frac{1}{2}(8-1)\right)\right]\\ =\frac{1}{2}\left[f(-2)+f\left(\frac{-3}{2}\right)+f(-1)+f\left(\frac{-1}{4}\right)+f\left(0\right)+f\left(\frac{1}{2}\right)+f\left(1\right)+f\left(\frac{3}{2}\right)\right]\\ =\frac{1}{2}[2.88-0.63-2.97+2.20+3.00+0.21-2.97-0.63]\\ =\frac{1}{2}\left(1.08\right)\\ =0.54\end{array} \end{gathered}$$

Part (b) Step 1. Signed Area

$$\begin{gathered} \mathrm{The}\mathrm{area}\mathrm{on}\mathrm{the}\mathrm{interval}[-2,-1.5],[-0.5,0.5]\mathrm{and}[1.5,2]\mathrm{are}\mathrm{positive}. \\ \mathrm{The}\mathrm{area}\mathrm{on}\mathrm{the}\mathrm{interval}[-1.5,0.5],[0.5,1.5]\mathrm{are}\mathrm{negative}. \\ \mathrm{Area}={\int }_{-2}^{-1.5}3\cos \left(3\mathrm{x}\right)+{\int }_{-1.5}^{-0.5}3\cos \left(3\mathrm{x}\right)+{\int }_{-0.5}^{0.5}3\cos \left(3\mathrm{x}\right)+{\int }_{0.5}^{1.5}3\cos \left(3\mathrm{x}\right)+{\int }_{1.5}^{2}3\cos \left(3\mathrm{x}\right) \\ \mathrm{Area}=[-0.63-2.88]+[0.21+0.63]+[0.21-0.21]+[-0.63-0.21]+[2.88+0.63] \\ \mathrm{Area}=0 \end{gathered}$$

Part (c) Step 1. Absolute Area

$$\begin{gathered} \mathrm{The}\mathrm{absolute}\mathrm{function}\mathrm{is}\mathrm{given}\mathrm{below}: \\ \left|\mathrm{f}\left(\mathrm{x}\right)\right|=\left\{\begin{array}{ll}3\cos \left(3\mathrm{x}\right)& -2\le \mathrm{x}\le -1.5\\ -3\cos \left(3\mathrm{x}\right)& -1.5\le \mathrm{x}\le -0.5\\ 3\cos \left(3\mathrm{x}\right)& -0.5\le \mathrm{x}\le 0.5\\ -3\cos \left(3\mathrm{x}\right)& 0.5\le \mathrm{x}\le 1.5\\ 3\cos \left(3\mathrm{x}\right)& 1.5\le \mathrm{x}\le 2\end{array}\right. \\ \mathrm{Area}={\int }_{-2}^{-1.5}3\cos \left(3\mathrm{x}\right)-{\int }_{-1.5}^{-0.5}3\cos \left(3\mathrm{x}\right)+{\int }_{-0.5}^{0.5}3\cos \left(3\mathrm{x}\right)-{\int }_{0.5}^{1.5}3\cos \left(3\mathrm{x}\right)+{\int }_{1.5}^{2}3\cos \left(3\mathrm{x}\right) \\ \mathrm{Area}=[-0.63-2.88]-[0.21+0.63]+[0.21-0.21]-[-0.63-0.21]+[2.88+0.63] \\ \mathrm{Area}=0 \end{gathered}$$