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» » » » Post: Re: Definite Integration

Print at May 18, 2013 11:38:42 AM

Posted by pskinner at Jan 1, 2008 10:45:48 AM

Re: Definite Integration

i still can't figure it out. I was trying the first way you explained how to do it. But I didn't understand how you said the endpoints (-b/2), (b/2), (-a/2) and (a/2). If I place the trapezoid on de first quadrant how can I have negative values such as -a or -b?? I'm working on it because I have to turn it in wednesday.

The coordinates (-b/2,0) and (b/2,0), which are on the x-axis, represent the base of the trapezoid of length b.

The coordinates (-a/2,h) and (a/2,h), which are on a line parallel to the x-axis and h units above the x-axis, represent the top of the trapezoid. So, the trapezoid is in the second quadrant as well as the first quadrant. Further, from the diagram, you can see that the area to the left of the y-axis is the same as the area to the right of the y-axis. So, the total area is twice the area in the first quadrant.

If you want to place everything in the first quadrant, then you need to label the points: (0,0) and (0,b) for the base, and ((b-a)/2,h) and (a+(b-a)/2,h).
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Principal Skinner