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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Area with Geogebra

Hi;

This question came up in another thread.

Find the area bounded by the x-axis, and the curve y = 1 - (8/x^2) from x = 4 to x = 8. If the straight line passing through the point (a,k) and parallel to the y-axis divides the area A into two parts in the ratio 4:5, find the value of a.

1)In the input bar enter 1-8 / x^2. f(x) =1-8 / x^2 will appear in the algebra pane and the graph will be plotted.

2)Use the move graphics view button to position the graph.

3)Enter in the input bar integral(f,4,8). Geogebra will shade the appropriate area and put a = 3 which is the area in the algebra pane.

4)Create a slider using the slider tool. Call it b and give it Min = 4 and Max = 8 with an increment of .01.

5)Create  a point on the curve by entering in the input bar (b, f(b)). Label that point A. See the picture.

6)Draw a perpendicular line through A. This will of course be parallel with the y axis.
Notice that when you move b that A travels right and left along f.

7)Input integral(f,4,b) in the input bar. Color the area under the curve that is left of the perpendicular line a dark blue. See the second picture. d in the algebra pane is the area of the blue region. Now use the slider until d is1.33333 or as close as you can get it.

I got b = 6 and that is the answer to the question, a = 6.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
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