Answer:
Given function is,
[tex]y=20x-x^2[/tex]To approximate the area under the curve in the interval [0, 20] by dividing the area into the given numbers of rectangles.
1) To use five rectangles to approximate the area under the curve.
we get that,
width of each rectangle is, (where a and b are end points of the interval)
[tex]=\frac{b-a}{n}=\frac{20-0}{5}[/tex][tex]=4\text{ units}[/tex]width of each rectangle is 4 units.
we get the x values as, 4,8,12,16 and 20
We can now calculate the height of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:
when x=4 we get,
[tex]\begin{gathered} y=20(4)-4^2 \\ y_1=64 \end{gathered}[/tex]x=8 we get,
[tex]\begin{gathered} y=20(8)-8^2 \\ y_2=96 \end{gathered}[/tex]x=12 we get
[tex]\begin{gathered} y=20(12)-12^2 \\ y_3=96 \end{gathered}[/tex]x=16 we get,
[tex]\begin{gathered} y=20(16)-16^2 \\ y_4=64 \end{gathered}[/tex]x=20 we get,
[tex]\begin{gathered} y_5=20(20)-20^2 \\ y_5=0 \end{gathered}[/tex]Area under the curve using 5 rectangle is,
[tex]=\text{width of each rectangle}\times\text{ sum of the height of the rectangle}[/tex][tex]=4\times(64+96+96+64+0)[/tex][tex]=1280[/tex]Area under the curve using 5 rectangle is 1280 sq.units.
Use 10 rectangles to approximate the area under the curve.
width of each rectangle is, (where a and b are end points of the interval)
[tex]=\frac{b-a}{n}=\frac{20-0}{10}[/tex][tex]=2\text{ units}[/tex]width of each rectangle is 2 units.
we get the x values as, 2,4,6,8,10,12,14,16,18 and 20
We can now calculate the height of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:
when x=2 we get,
[tex]y_1=20(2)-2^2=36[/tex]when x=4 we get,
[tex]\begin{gathered} y=20(4)-4^2 \\ y_2_{}=64 \end{gathered}[/tex]x=6 we get,
[tex]y_3=20(6)-6^2=84[/tex]x=8 we get,
[tex]\begin{gathered} y=20(8)-8^2 \\ y_4=96 \end{gathered}[/tex]x=10 we get,
[tex]y_5=20(10)-10^2=100[/tex]x=12 we get
[tex]\begin{gathered} y=20(12)-12^2 \\ y_6=96 \end{gathered}[/tex]x=14 we get,
[tex]\begin{gathered} y=20(14)-14^2 \\ y_7=84 \end{gathered}[/tex]x=16 we get,
[tex]\begin{gathered} y=20(16)-16^2 \\ y_8=64 \end{gathered}[/tex]x=18 we get,
[tex]\begin{gathered} y=20(18)-18^2 \\ y_9=36 \end{gathered}[/tex]x=20 we get,
[tex]\begin{gathered} y_5=20(20)-20^2 \\ y_5=0 \end{gathered}[/tex]Area under the curve using 5 rectangle is,
[tex]=\text{width of each rectangle}\times\text{ sum of the height of the rectangle}[/tex][tex]\begin{gathered} =2\times(36+64+84+96+100+96+84+64+36+0) \\ =1320 \end{gathered}[/tex]Area under the curve in the interval [0, 20] by dividing the area into 10 number of rectangles is 1320 sq units.
