# Thread: John Wallis Integration question

1. ## John Wallis Integration question

I have just been reading about John Wallis ( 17th Century English Mathematician) on wikipedia.
https://en.wikipedia.org/wiki/John_Wallis

Can anyone help with this:

the area enclosed between the curve y = xm, the axis of x, and any ordinate x = h, and he proved that the ratio of this area to that of the parallelogram on the same base and of the same height is 1/(m + 1),
Which parallelogram are we talking about here?

2. ## Re: John Wallis Integration question

Originally Posted by rodders
I have just been reading about John Wallis ( 17th Century English Mathematician) on wikipedia.
https://en.wikipedia.org/wiki/John_Wallis

Can anyone help with this:

the area enclosed between the curve y = xm, the axis of x, and any ordinate x = h, and he proved that the ratio of this area to that of the parallelogram on the same base and of the same height is 1/(m + 1),
Which parallelogram are we talking about here?
It looks to me like the word "rectangle" rather than "parallelogram" would be better here (of course, a rectangle [b]is[\b] a type of parallelogram). That is, the rectangle with vertices at (0, 0), (h, 0), ((h, h^m), and (0, h^m). That rectangle has length a and height a^m so area a^{m+1}. What is the are or the region bounded by the x-axis, y= x^m, and x= h?\

3. ## Re: John Wallis Integration question

That makes sense but presumably only between x=0 and x=h , rather than x=a to x=h ? Doesn't feel like much of a remarkable result??