Q. 25 - Calculus

Question

For each function $f$ and interval $\left[a,b\right]$ in Exercises 23–25, use at least eight rectangles to approximate (a) the signed area and (b) 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.

$f\left(x\right)=1-e^x, \left[-1,3\right]$

Step-by-Step Solution

Answer

a. The signed area is $\approx 11.1974$.

b. The absolute area is $\approx 12.229$.

Step 1. Given Information

The function is,

$f\left(x\right)=1-e^x$

The interval is  $\left[-1,3\right]$.

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

Where,$∆x=\frac{b-a}{n}$

             $x_k=a+k∆x$

Now,

$$\begin{gathered} ∆x=\frac{3+1}{8} \\ =\frac{4}{8} \\ =\frac{1}{2} \end{gathered}$$

So, 

$x_k=-1+k\frac{1}{2}$

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

So, 

$$\begin{gathered} x_{k-1}=-1+\left(k-1\right)\frac{1}{2} \\ =-1+\frac{k}{2}-\frac{1}{2} \\ =\frac{k}{2}-\frac{3}{2} \\ =\frac{k-3}{2} \end{gathered}$$

The left sum is,

$$\begin{gathered} \sum _{k=1}^8\left(1-e^{\frac{k-3}{2}}\right)\frac{1}{2} \\ =\frac{1}{2}\left(1-e^{-1}\right)+\frac{1}{2}\left(1-e^{-\frac{1}{2}}\right)+\frac{1}{2}\left(1-e^0\right)+ \\ \frac{1}{2}\left(1-e^{\frac{1}{2}}\right)+\frac{1}{2}\left(1-e^1\right)+\frac{1}{2}\left(1-e^{\frac{3}{2}}\right)+\frac{1}{2}\left(1-e^2\right)+\frac{1}{2}\left(1-e^{\frac{5}{2}}\right) \\ \approx -11.1974 \end{gathered}$$

Therefore, the signed area is $\approx -11.1974$.

Part (b). Step 2. 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). Step 3. Calculation

The absolute area is,

$$\begin{gathered} {\int }_{-1}^{3}f\left(x\right)dx \\ ={\int }_{-1}^3\left|\left(1-e^x\right)\right|dx \\ =-{\left[x\right]}_{-1}^3+{\left[e^x\right]}_{-1}^3 \\ =-3-1+e^3-e^{-1} \\ \approx 12.229 \end{gathered}$$

Therefore, the absolute area is $\approx 12.229$.