Question

In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use graphing skills in addition to the knowledge gathered in this unit. Sketch the graph of the function y = 20x − x2, and approximate the area under the curve in the interval [0, 20] by dividing the area into the given numbers of rectangles.Use five rectangles to approximate the area under the curve.Use 10 rectangles to approximate the area under the curve.Calculate the area under the curve using rectangles as their number becomes arbitrarily large (tends to infinity

Answer 1

Given function is,

`y=20x-x^2`

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)

`=(b-a)/(n)=(20-0)/(5)`
`=4\text{ units}`

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,

`\begin{gathered} y=20(4)-4^2 \\ y_1=64 \end{gathered}`

x=8 we get,

`\begin{gathered} y=20(8)-8^2 \\ y_2=96 \end{gathered}`

x=12 we get

`\begin{gathered} y=20(12)-12^2 \\ y_3=96 \end{gathered}`

x=16 we get,

`\begin{gathered} y=20(16)-16^2 \\ y_4=64 \end{gathered}`

x=20 we get,

`\begin{gathered} y_5=20(20)-20^2 \\ y_5=0 \end{gathered}`

Area under the curve using 5 rectangle is,

`=\text{width of each rectangle}*\text{ sum of the height of the rectangle}`
`=4*(64+96+96+64+0)`
`=1280`

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)

`=(b-a)/(n)=(20-0)/(10)`
`=2\text{ units}`

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,

`y_1=20(2)-2^2=36`

when x=4 we get,

`\begin{gathered} y=20(4)-4^2 \\ y_2_{}=64 \end{gathered}`

x=6 we get,

`y_3=20(6)-6^2=84`

x=8 we get,

`\begin{gathered} y=20(8)-8^2 \\ y_4=96 \end{gathered}`

x=10 we get,

`y_5=20(10)-10^2=100`

x=12 we get

`\begin{gathered} y=20(12)-12^2 \\ y_6=96 \end{gathered}`

x=14 we get,

`\begin{gathered} y=20(14)-14^2 \\ y_7=84 \end{gathered}`

x=16 we get,

`\begin{gathered} y=20(16)-16^2 \\ y_8=64 \end{gathered}`

x=18 we get,

`\begin{gathered} y=20(18)-18^2 \\ y_9=36 \end{gathered}`

x=20 we get,

`\begin{gathered} y_5=20(20)-20^2 \\ y_5=0 \end{gathered}`

Area under the curve using 5 rectangle is,

`=\text{width of each rectangle}*\text{ sum of the height of the rectangle}`
`\begin{gathered} =2*(36+64+84+96+100+96+84+64+36+0) \\ =1320 \end{gathered}`

Area under the curve in the interval [0, 20] by dividing the area into 10 number of rectangles is 1320 sq units.