Question

let ( f(x)=4 + 2x^{2} ).
(a) estimate the area under the graph of ( f ), the ( x )-axis, and the lines ( x=-1 ) and ( x = 2 ) using three rectangles and a right endpoints.
( r_{3}= )
improve your estimate by using six rectangles.
( r_{6}= )

Explanation:

Step1: Determine the width of each subinterval

The interval is from \( x = -1 \) to \( x = 2 \), so the length of the interval is \( 2 - (-1)=3 \). For \( n \) rectangles, the width \( \Delta x=\frac{3}{n} \).

Step2: Find the right endpoints for \( R_3 \)

For \( n = 3 \), \( \Delta x=\frac{3}{3} = 1 \). The subintervals are \([-1,0]\), \([0,1]\), \([1,2]\). The right endpoints are \( x_1 = 0 \), \( x_2 = 1 \), \( x_3 = 2 \).
Now, calculate \( f(x) \) at these endpoints:
\( f(0)=4 + 2(0)^2=4 \)
\( f(1)=4 + 2(1)^2=6 \)
\( f(2)=4 + 2(2)^2=12 \)
The area \( R_3=\Delta x(f(0)+f(1)+f(2))=1\times(4 + 6+12)=22 \)

Step3: Find the right endpoints for \( R_6 \)

For \( n = 6 \), \( \Delta x=\frac{3}{6}=0.5 \). The subintervals are \([-1,-0.5]\), \([-0.5,0]\), \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), \([1.5,2]\). The right endpoints are \( x_1=-0.5 \), \( x_2 = 0 \), \( x_3=0.5 \), \( x_4 = 1 \), \( x_5=1.5 \), \( x_6 = 2 \)
Calculate \( f(x) \) at these endpoints:
\( f(-0.5)=4+2(-0.5)^2=4 + 2\times0.25 = 4.5 \)
\( f(0)=4+2(0)^2 = 4 \)
\( f(0.5)=4+2(0.5)^2=4 + 2\times0.25=4.5 \)
\( f(1)=4+2(1)^2 = 6 \)
\( f(1.5)=4+2(1.5)^2=4 + 2\times2.25=8.5 \)
\( f(2)=4+2(2)^2=12 \)
The area \( R_6=\Delta x(f(-0.5)+f(0)+f(0.5)+f(1)+f(1.5)+f(2))=0.5\times(4.5 + 4+4.5+6+8.5+12) \)
First, sum the values inside the parentheses: \( 4.5+4 = 8.5 \), \( 8.5+4.5 = 13 \), \( 13+6 = 19 \), \( 19+8.5 = 27.5 \), \( 27.5+12 = 39.5 \)
Then, \( R_6=0.5\times39.5 = 19.75 \)

Answer:

For \( R_3 \): \(\boldsymbol{22}\)
For \( R_6 \): \(\boldsymbol{19.75}\)