Chapter 11.2, Problem 18E

Solution

a)

To find: Draw the graph of the function for x in the specified interval and to verify that the function is non-negative in that interval.

The graph of the function $f\left(x\right)=x^2+2$ in the interval $\left[0,6\right]$ is,

  

Given information:

  $f\left(x\right)=x^2+2$ in the interval $\left[0,6\right]$

Calculation:

  

Plotting the graph

  

b)

To find: Draw and shade the approximating rectangles for the RRAM and to find the area.

The approximating rectangles for the RRAM and the area is,

  $A\approx 103$

Given information:

  $f\left(x\right)=x^2+2$ in the interval $\left[0,6\right]$

Calculation:

The six sub-intervals are $\left[0,1\right],\left[1,2\right],\left[2,3\right],\left[3,4\right],\left[4,5\right],\left[5,6\right]$

For RRAM the upper right corners of the rectangles touch the graph of the function

  

Each rectangles has a base of 1 while the heights are function value at the right-hand endpoint.

  $A\approx 1+4+9+18+27+38=103$

Subintervals	Base	Height	Area
$\left[0,1\right]$	$1$	$f\left(1\right)=1^2+2=3$	$\left(1\right)\left(1\right)=1$
$\left[1,2\right]$	$1$	$f\left(2\right)=2^2+2=6$	$\left(1\right)\left(6\right)=6$
$\left[2,3\right]$	$1$	$f\left(3\right)=3^2+2=11$	$\left(11\right)\left(1\right)=11$
$\left[3,4\right]$	$1$	$f\left(4\right)=4^2+2=18$	$\left(1\right)\left(18\right)=18$
$\left[4,5\right]$	$1$	$f\left(5\right)=5^2+2=27$	$\left(1\right)\left(27\right)=27$
$\left[5,6\right]$	$1$	$f\left(6\right)=6^2+2=38$	$\left(1\right)\left(38\right)=38$

c)

To find: Draw and shade the approximating rectangles for the LRAM and to find the area.

The approximating rectangles for the LRAM and the area is

  $A\approx 67$

Given information:

  $f\left(x\right)=x^2+2$ in the interval $\left[0,6\right]$

Calculation:

For LRAM there is no rectangle in the interval $\left[0.1\right]$

  $A\approx 2+3+6+11+18+27=67$

Subintervals	Base	Height	Area
$\left[0,1\right]$	$1$	$f\left(1\right)=0^2+2=2$	$\left(1\right)\left(2\right)=2$
$\left[1,2\right]$	$1$	$f\left(2\right)=1^2+2=3$	$\left(1\right)\left(3\right)=3$
$\left[2,3\right]$	$1$	$f\left(2\right)=2^2+2=6$	$\left(1\right)\left(6\right)=6$
$\left[3,4\right]$	$1$	$f\left(3\right)=3^2+2=11$	$\left(11\right)\left(1\right)=11$
$\left[4,5\right]$	$1$	$f\left(4\right)=4^2+2=18$	$\left(1\right)\left(18\right)=18$
$\left[5,6\right]$	$1$	$f\left(5\right)=5^2+2=27$	$\left(1\right)\left(27\right)=27$

d)

To find: Average area of RRAM and.LRAM.

Average area of RRAM and.LRAM is $65$

Given information:

  $f\left(x\right)=x^2+2$ in the interval $\left[0,6\right]$

Calculation:

Using RRAM the area is $103$ and using LRAM the are is $67$ the average estimate of the area is

  $\frac{103+67}{2}=65$