Q.33 - 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. 

$f\left(x\right)={\left(x-2\right)}^2+1, \left[a,b\right]=\left[1,3\right]$ lower sum with

(a) n = 2 (b) n = 3 (c) n = 4 

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,   

$$\begin{gathered} a) {\int }_1^3{\left(x-2\right)}^2+1dx=3 \\ b) {\int }_1^3{\left(x-2\right)}^2+1dx=\frac{76}{27} \\ c) {\int }_1^3{\left(x-2\right)}^2+1dx=\frac{22}{8} \end{gathered}$$

Part (a) Step 1. Given information.

We have given,

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

Part (a) Step 2. Concept used.

Lower Riemann sum formula: 

${\int }_a^{b}f\left(x\right)dx\hspace{0.17em}\approx \Delta x\left(f\right(x_0)+f(x_1)+f(x_2)+...+f(x_{n-1}\left)\right)$

$\mathrm{where}\hspace{0.17em}\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{b-a}{n}$

Part (a) Step 3. Explanation.

We have,

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

Length of the subintervals of the interval [1 , 3] is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{3-1}{2}=1$

Dividing the interval [1, 3] in to the 2 subintervals with length 1,

$\left[1, 2\right],\left[2,3\right]$

Left end points are: 1 and 2

Just evaluating the function for those end points, 

$f\left(1\right)=2, f\left(2\right)=1$

Using lower sum formula,

$$\begin{gathered} {\int }_1^3{\left(x-2\right)}^2+1dx\approx 1·\left(f\left(x_0\right)+f\left(x_1\right)\right) \\ =1·\left(2+1\right) \\ =3 \end{gathered}$$

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,  

${\int }_1^3{\left(x-2\right)}^2+1dx=3$

Part (b) Step 1. Explanation.

We have given,

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

Length of the subintervals is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{3-1}{3}=\frac{2}{3}$

Dividing the interval [1, 3] in to the three subintervals whose length is $\frac{2}{3}$,

$\left[1, \frac{5}{3}\right],\left[\frac{5}{3}, \frac{7}{3}\right],\left[\frac{7}{3},3\right]$

Left end points are:

$1, \frac{5}{3},\frac{7}{3}$

Now just evaluating the functions for those left end points,

$f\left(1\right)=2, f\left(\frac{5}{3}\right)=\frac{10}{9}, f\left(\frac{7}{3}\right)=\frac{10}{9}$

Using left sum formula,

$$\begin{gathered} {\int }_1^3{\left(x-2\right)}^2+1dx\approx \frac{2}{3}\left(f\left(x_0\right)+f\left(x_1\right)+f\left(x_2\right)\right) \\ =\frac{2}{3}\left(2+\frac{10}{9}+\frac{10}{9}\right) \\ =\frac{76}{27} \end{gathered}$$

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,   

${\int }_1^3{\left(x-2\right)}^2+1dx=\frac{76}{27}$

Part (c) Step 1. Explanation

We have given,

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

Length of the intervals is,

$\Delta \hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}\frac{3-1}{4}=\frac{1}{2}$

Dividing the interval [1, 3] into the 4 subintervals with length $\frac{1}{2}$is,

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

Left end points are:

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

Just evaluating the function for those end points,

$f\left(1\right)=2, f\left(\frac{3}{2}\right)=\frac{5}{4}, f\left(2\right)=1, f\left(\frac{5}{2}\right)=\frac{5}{4}$

Using Riemann sum,

$$\begin{gathered} \hspace{0.17em}{\int }_1^3{\left(x-2\right)}^2+1dx\approx \frac{1}{2}\left(f\left(x_0\right)+f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\right) \\ =\frac{1}{2}\left(2+\frac{5}{4}+1+\frac{5}{4}\right) \\ =\frac{22}{8} \end{gathered}$$

Part (c) Step 2. Conclusion.

${\int }_1^3{\left(x-2\right)}^2+1dx=\frac{22}{8}$