Q.30 - Calculus

Question

For each function f and interval [a, b] in Exercises 27–33, use the given approximation method to approximate the signed area between the graph of f and the x-axis on [a, b]. Determine whether each of your approximations is likely to be an over-approximation or an under-approximation of the actual area.   

$f\left(x\right)=1-2^x, \left[a, b\right]=\left[-3, 1\right] \mathrm{and} n=8$

a) left sum                     (b) right sum 

Step-by-Step Solution

Answer

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

$$\begin{gathered} a) {\int }_{-3}^{1}1-2^{x}dx=1.74 \\ b) {\int }_{-3}^{1}1-2^{x}dx=0.80 \end{gathered}$$

Part (a) Step 1. Given information.

We have given, 

$f\left(x\right)=1-2^x, \left[a, b\right]=\left[-3, 1\right] \mathrm{and} n=8$

Part (a) Step 2. Concept used.

Left endpoint Riemann sum formula:  

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx\hspace{0.17em}\approx \Delta x\left(f\right(x_0)+f(x_1)+f(x_2)+...+f(x_{n-1}\left)\right) \\ Where, \\ \Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n} \end{gathered}$$

Right endpoint Riemann sum formula: 

${\int }_a^{b}f\left(x\right)dx\hspace{0.17em}\approx \Delta x\left(f\right(x_1)+f(x_2)+f(x_3)+...+f(x_n\left)\right)$

Part (a) Step 3. Explanation.

We have given,

$f\left(x\right)=1-2^x, \left[a, b\right]=\left[-3, 1\right] \mathrm{and} n=8$

So length of the subintervals is, 

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=\frac{1}{2}$

So dividing the interval [-3, 1] in to the 8 subintervals with length $\frac{1}{2}$ is,

$\left[-3, -\frac{5}{2}\right], \left[-\frac{5}{2}, -2\right],\left[-2, -\frac{3}{2}\right], \left[-\frac{3}{2}, -1\right],\left[-1, -\frac{1}{2}\right], \left[-\frac{1}{2},0\right], \left[0, \frac{1}{2}\right],\left[\frac{1}{2}, 1\right]$

End points are:

$-3, -\frac{5}{2},-2, -\frac{3}{2},-1,-\frac{1}{2},0,\frac{1}{2}$

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

$f(-3)=\frac{7}{8}, f\left(-\frac{5}{2}\right)=1-\frac{\sqrt{2}}{8},f(-2)=\frac{3}{4}, f\left(-\frac{3}{2}\right)=1-\frac{\sqrt{2}}{4},f\left(-1\right)=\frac{1}{2}, f\left(-\frac{1}{2}\right)=1-\frac{\sqrt{2}}{2}, f\left(0\right)=0,f\left(\frac{1}{2}\right)= 1-\sqrt{2}$

Using left end point formula, 

$$\begin{gathered} {\int }_{-3}^{1}1-2^{x}dx\approx \frac{1}{2}\left(f\left(x_0\right)+f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)+f\left(x_5\right)+f\left(x_6\right)+f\left(x_7\right)\right) \\ \approx \frac{1}{2}\left[f(-3)+ f\left(-\frac{5}{2}\right)+f(-2)+ f\left(-\frac{3}{2}\right)+f\left(-1\right)+f\left(-\frac{1}{2}\right)+ f\left(0\right)+f\left(\frac{1}{2}\right)\right] \\ \approx \frac{1}{2}\left[\frac{7}{8}+1-\frac{\sqrt{2}}{8}+\frac{3}{4}+1-\frac{\sqrt{2}}{4}+\frac{1}{2}+1-\frac{\sqrt{2}}{2}+0+1-\sqrt{2}\right] \\ \approx 1.74 \end{gathered}$$

Part (a) Step 4. Conclusion.

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

${\int }_{-3}^{1}1-2^{x}dx=1.74$

Part (b) Step 1. Explanation.

We have given,

$f\left(x\right)=1-2^x, \left[a, b\right]=\left[-3, 1\right] \mathrm{and} n=8$

Length of the subintervals is, 

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=\frac{1}{2}$

So dividing the interval [2, 3] in to the 4 subintervals with length $\frac{1}{2}$ is,

$\left[-3, -\frac{5}{2}\right], \left[-\frac{5}{2}, -2\right],\left[-2, -\frac{3}{2}\right], \left[-\frac{3}{2}, -1\right],\left[-1, -\frac{1}{2}\right], \left[-\frac{1}{2},0\right], \left[0, \frac{1}{2}\right],\left[\frac{1}{2}, 1\right]$

Right end points are: 

$-\frac{5}{2},-2, -\frac{3}{2},-1,-\frac{1}{2},0,\frac{1}{2} \mathrm{and} 1$

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

$f\left(-\frac{5}{2}\right)=1-\frac{\sqrt{2}}{8},f(-2)=\frac{3}{4}, f\left(-\frac{3}{2}\right)=1-\frac{\sqrt{2}}{4},f\left(-1\right)=\frac{1}{2}, f\left(-\frac{1}{2}\right)=1-\frac{\sqrt{2}}{2}, f\left(0\right)=0,f\left(\frac{1}{2}\right)= 1-\sqrt{2} and f\left(1\right)=-1$

Using the right end points Riemann sum formula,

$$\begin{gathered} {\int }_{-3}^{1}1-2^{x}dx\approx \frac{1}{2} \left[f\left(-\frac{5}{2}\right)+f(-2)+ f\left(-\frac{3}{2}\right)+f\left(-1\right)+ f\left(-\frac{1}{2}\right)+f\left(0\right)+f\left(\frac{1}{2}\right)+f\left(1\right)\right] \\ \approx \frac{1}{2}\left[1-\frac{\sqrt{2}}{8}+\frac{3}{4}+1-\frac{\sqrt{2}}{4}+\frac{1}{2}+1-\frac{\sqrt{2}}{2}+0+1-\sqrt{2}-1\right] \\ \approx 0.80 \end{gathered}$$

Part (b) Step 2. Conclusion.

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

${\int }_{-3}^{1}1-2^{x}dx=0.80$