Area between curves

This is a parabolic function:[br][math]f\left(x\right)=x^2[/math][br]What is its graph?[br][math]...[/math][br][math]...[/math][br][math]x=-2\longrightarrow f\left(-2\right)=\left(-2\right)^2=4[/math][br][math]x=-1\longrightarrow f\left(-1\right)=\left(-1\right)^2=1[/math][br][math]x=0\longrightarrow f\left(0\right)=\left(0\right)^2=0[/math][br][math]x=1\longrightarrow f\left(1\right)=\left(1\right)^2=1[/math][br][math]...[/math][br][math]...[/math]

If you get infinite points you will have the graph.[br]We don't have infinite time to draw infinite points. However, we have Geogebra!!

[b]How we can get the area between the graph and the x-axis in a closed interval?[/b]

How do you think that we can measure the area? Write your answer.

We can use rice grains to measure the area. [b]Look at the next photo![br][br][img]http://www.danipartal.net/imagenes-cuestionario/parabola-con-arroz-area.jpg[/img][/b]

How do you can measure a mathematical area with rice grains?

[list=1][*]Print the graphic on paper and spread glue over the rectangle ([b]Rectangle Area: 8 units square[/b]).[/*][*]Throw alleatory rice grains over the rectangle.[/*][*]Repeat the process until you have finished with all the grains.[/*][*]Wait a while for the glue to dry.[/*][*]Count the number of grains over the region between the function and the x-axis [b](success number: F)[/b]. Add to F the number of grains situated over the rest of the rectangle [b](total number: N)[/b].[/*][*]If N is a big number, we can assume that the relative frecuency F/N tend to the probability ([b]P[/b]) of finding a rice grain on the area created by the function and the x-axis, in the closed interval [0,2].[/*][*]Multiply [math]Area\times P[/math] to get the solution area.[/*][*]What is your result?[/*][/list]

Geogebra can create bunch of rice grains to measure the area!!

If Start = 0, Finish = 2 and maxPoints = 10000, what value does the probability tend to? Write your answer.

Challenge: Design a Geogebra simulation with a parabolic function, a closed interval and 500 points to estimate the area

[b][i]Hints for the task![/i][/b][br]Function:[br] [math]f\left(x\right)=x^2[/math][br]Interval: [br] [0,2][br]Image of the interval: [br] [0,4][br]Slider[b] [i]numPoints:[/i][/b] [br] from 1 to 500 [br]Random horizontal coordinate of a point from x=0,001 to x=2: [br] [b][i]aleaHor = AleatorioEntre(1,2000)/1000[/i][/b][br]Random vertical coordinate of a point from y=0,001 to y=4: [br] [b][i]aleaVer = AleatorioEntre(1,4000)/1000[/i][/b][br]Create a list of random horizontal coordinates: [br] [b][i]coordHor =[/i] [i]Secuencia(aleaHor, i , 1, 500)[br][/i][/b]Create a list of random vertical coordinates:[br] [b][i]coordVer =[/i] [i]Secuencia(aleaVer, i , 1, 500)[br][/i][/b]Create a list of points:[br] [i][b]points = Secuencia((coordHor(i), coordVer(i)),i,1,500)[/b][/i][b][br][br][color=#980000]Watch out! [/color][/b][color=#980000]Compare the vertical coordinate of each point with the image of the function generated by the horizontal coordinate: [br][i][b]count = Secuencia(Si(y(points(i)) < f(x(points(i))), true, false), i, 1, numPoints)[/b][br][/i]How do we know the number of points that belong to the solution area?[br]Commands: Secuencia() and Si(condition, then, else)[br]We need to compare the vertical coordinate of each point with the image of each horizontal coordinate in the function.[/color][br][i][i][i][br]Probability: [br] [b]prob = Suma(count)/500[/b][br]Solution area:[br] [b]areaSolution = 8·prob[/b][/i][/i][/i]