Q.29 - 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.  

$f\left(x\right)=\sqrt{x-1}, [a, b]=[2, 3] \mathrm{and} n=4$

(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, 

$$\begin{gathered} a) {\int }_2^3\sqrt{x-1}dx\approx 1.17 \\ b) {\int }_2^3\sqrt{x-1}dx\approx 1.27 \end{gathered}$$

Part (a) Step 1. Given information.

We have given, 

$f\left(x\right)=\sqrt{x-1}, [a, b]=[2, 3] \mathrm{and} n=4$

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, 

$f\left(x\right)=\sqrt{x-1}, [a, b]=[2, 3] \mathrm{and} n=4$

So length of the subintervals is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=\frac{1}{4}$

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

$\left[2, \frac{9}{4}\right], \left[\frac{9}{4}, \frac{5}{2}\right], \left[\frac{5}{2},\frac{11}{4}\right], \left[\frac{11}{4},3\right]$

Left end points are: $2,\frac{9}{4},\frac{5}{2} \mathrm{and} \frac{11}{4}$

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

$f\left(2\right)=1, f\left(\frac{9}{4}\right)=\frac{\sqrt{5}}{2}, f\left(\frac{5}{2}\right)=\frac{\sqrt{6}}{2} \mathrm{and} f\left(\frac{11}{4}\right)=\frac{\sqrt{7}}{2}$

Using left end point formula,

$$\begin{gathered} {\int }_2^3\sqrt{x-1}dx\approx \frac{1}{4}\left[f\left(2\right)+ f\left(\frac{9}{4}\right)+f\left(\frac{5}{2}\right)+ f\left(\frac{11}{4}\right)\right] \\ \approx \frac{1}{4}\left[1+\frac{\sqrt{5}}{2}+\frac{\sqrt{6}}{2}+ \frac{\sqrt{7}}{2}\right] \\ \approx 1.17 \end{gathered}$$

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, 

${\int }_2^3\sqrt{x-1}dx\approx 1.17$

Part (b) Step 1. Explanation.

We have given,

$f\left(x\right)=\sqrt{x-1}, [a, b]=[2, 3] \mathrm{and} n=4$

Length of the subintervals is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}=\frac{1}{4}$

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

$\left[2, \frac{9}{4}\right], \left[\frac{9}{4}, \frac{5}{2}\right], \left[\frac{5}{2},\frac{11}{4}\right], \left[\frac{11}{4},3\right]$

Right end points are:

$\frac{9}{4},\frac{5}{2},\frac{11}{4} \mathrm{and} 3$

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

$f\left(\frac{9}{4}\right)=\frac{\sqrt{5}}{2}, f\left(\frac{5}{2}\right)=\frac{\sqrt{6}}{2}, f\left(\frac{11}{4}\right)=\frac{\sqrt{7}}{2} and f\left(3\right)=\sqrt{2}$

Using the right end point Riemann sum formula,

$$\begin{gathered} {\int }_2^3\sqrt{x-1}dx\approx \frac{1}{4} \left[f\left(\frac{9}{4}\right)+ f\left(\frac{5}{2}\right)+f\left(\frac{11}{4}\right)+ f\left(3\right)\right] \\ \approx \frac{1}{4}\left[\frac{\sqrt{5}}{2}+\sqrt{\frac{3}{2}}+\frac{\sqrt{7}}{2}+\sqrt{2}\right] \\ \approx 1.27 \end{gathered}$$

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, 

$\hspace{0.17em}{\int }_2^3\sqrt{x-1}dx\approx 1.27$