Chapter 11.2, Problem 11E

Solution

To find: The area of the region below the graph of $f\left(x\right)=10-x^2$ over the interval $\left[1,3\right]$

The area of the region below the graph of $f\left(x\right)=10-x^2$ over the interval $\left[1,3\right]$ is $32.5$

Given information:= graph of $f\left(x\right)=10-x^2$ over the interval $\left[1,3\right]$

Calculation:

Base of each rectangle is $0.5$ and the heights are the function value at the left-hand endpoints, so the area of the region is, $A\approx 4.5+4.785+5+4.785+4.5+3.875+3+1.875=32.5$

Subintervals	Base	Height	Area
$\left[-1,-0.5\right]$	$0.5$	$f\left(-1\right)=10-{\left(-1\right)}^2=9$	$\left(0.5\right)\left(9\right)=4.5$
$\left[-0.5,0\right]$	$0.5$	$f\left(-0.5\right)=10-{\left(-0.5\right)}^2=9.75$	$\left(0.5\right)\left(9.75\right)=4.875$
$\left[0,0.5\right]$	$0.5$	$f\left(0\right)=10-{\left(0\right)}^2=10$	$\left(0.5\right)\left(10\right)=5$
$\left[0.5,1\right]$	$0.5$	$f\left(0.5\right)=10-{\left(0.5\right)}^2=9.75$	$\left(0.5\right)\left(9.75\right)=4.875$
$\left[1,1.5\right]$	$0.5$	$f\left(1\right)=10-{\left(1\right)}^2=9$	$\left(0.5\right)\left(9\right)=4.5$
$\left[1.5,2\right]$	$0.5$	$f\left(1.5\right)=10-{\left(1.5\right)}^2=7.75$	$\left(0.5\right)\left(7.75\right)=3.875$
$\left[2,2.5\right]$	$0.5$	$f\left(2\right)=10-{\left(2\right)}^2=6$	$\left(0.5\right)\left(6\right)=3$
$\left[2.5,3\right]$	$0.5$	$f\left(2.5\right)=10-{\left(2.5\right)}^2=3.75$	$\left(0.5\right)\left(3.75\right)=1.875$