Question

Repeat Exercise 11, using the lower sum approximation and the values$m_k$

Short answer

1. The required values are $M_1=M_2=1$
2. The value are $M_1=\frac{2}{3},M_2=0,M_3=\frac{4}{3}$
   
3. The values are $M_1=\frac{1}{2},M_2=0,M_3=0,M_4=\frac{3}{2}$.

   
   

Step-by-step solution

Part (a) Step 1: Given information

The function is,

$f\left(x\right)=(x-1{)}^2$

Part (a) Step 2: Calculation

The aim is to calculate the lower sum approximation for the graph's area and the x-axis from $x=0$to width="36">x=2and to list the values of width="22">Mkutilised for rectangles with $n=2$.

The lower sum defined for n rectangles on $[a,b]$is $\sum _{k=1}^{n}f\left(M_k\right)\Delta x$.

Where, localid="1659709086168" $M_k$is the minimum value of height chosen in the interval $\left[x_{k-1},x_k\right]$,

$\Delta x=\frac{b-a}{n},x_k=a+k\Delta x.$

Think about the illustration below,

The minimum values $M_k$ occurs at $x=1$.

Therefore, the values are $M_1=M_2=1$.

Part (b) Step 1: Given information

The function is,

$f\left(x\right)=(x-1{)}^2$

Part (b) Step 2: Calculation

The objective is to list the values of $M_k$used for $n=3$rectangles while calculating the lower sum approximation for the area of the graph and the x-axis from $x=0$to $x=2$.

The lower sum defined for n rectangles on $[a,b]$is $\sum _{k=1}^{n}f\left(M_k\right)\Delta x$-

Where, $M_k$is the minimum value of height chosen in the interval $\left[x_{k-1},x_k\right]$,

$\Delta x=\frac{b-a}{n},x_k=a+k\Delta x$

Consider the following figure,

The minimum values $M_k$occurs at $x=\frac{2}{3},x=0$and $x=\frac{4}{3}$.

Therefore, the values are $M_1=\frac{2}{3},M_2=0,M_3=\frac{4}{3}$.

Part (c) Step 1: Given information

The function is, $f\left(x\right)=(x-1{)}^2$

Part (c) Step 2: Calculation

The objective is to list the values of $M_k$ used for $n=4$ rectangles while calculating the lower sum approximation for the area of the graph and the $x$-axis from $x=0$ to $x=2$.

The lower-sum defined for $n$ rectangles on $[a,b]$ is $\sum _{k=1}^{n}f\left(M_k\right)\Delta x$.

Where, $M_k$ is the minimum value of height chosen in the interval $\left[x_{k-1},x_k\right]$,

$\Delta x=\frac{b-a}{n},x_k=a+k\Delta x.$

Let us consider the below figure

The minimum values $M_k$occurs at $x=\frac{1}{2},x=0,x=\frac{3}{2}$.

Hence, the values are $M_1=\frac{1}{2},M_2=0,M_3=0,M_4=\frac{3}{2}$.