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» » » » Post: Re: Definite Integration

Print at May 23, 2013 12:06:24 PM

Posted by pskinner at Dec 27, 2007 8:53:59 AM

Re: Definite Integration

I got it for the Circle, What about the trapezoid. I Know I have to place it at a coordenate plane, I also know that I can divide it into two right triangle and a rectangle, but I don't know what else to do.

Well, it's not fully clear to me, but I'd place the bottom of the trapezoid on the x-axis, of length b, with the end points at -b/2 and b/2. Also, since the area doesn't vary once the dimensions a, b, and h are known, you can assume you can work with an isosceles trapezoid, so the top portion can be at the end points (-a/2,h) and (a/2,h). Continue the lines along the sides so they meet at (H,0), where H is unknown. You now have a large triangle with the trapezoid inside and the triangle and trapezoid share the same base. The equation of the hypotenuse of the large triangle in the first quadrant is y=(-H/b)x. To get the area of the 1/2 the trapezoid, you can now integrate from 0 to a, (-H/b)x-h plus the integral of (-H/b)x from a to b and that will give you the area of the trapezoid. Finally, you can solve for H before doing this by setting up a proportion between similar triangles: b is to a as H is to H-h.

It seems like a lot of work. Perhaps there is a simpler way, but it escapes me.
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Principal Skinner