How did Archimedes discover the Quadrature of the parabola without the use of calculus?
If someone could please explain, I would be eternally grateful.
that makes much more sense- I can proceed :)
\[ A=((-1/e)2^2+(2^2+4/e)2-2e^2/2)-((-1/e)0^2+(0^2+4/4)(0)-2e^0/2)) \]
which simplifies to
\[ A=((-1/e)(4))+((8/e)(2)-2e^1)-(2e^0) \]
this is as far as i got :)
\( A=((-1/e)(4))+((8/e)(2)-2e)-(2) \)
Actually, because I do not do very hard maths, it is just 0<x<2
That is wayyy to complicated for me to understand. Desmos deformed it because my graphing calculator says different.
What I am doing is for my maths assignment:
So I had to find the equation of the normal for the point (2,e)
Which...
The area of two lines that I need to find is 2.36, however i need this in exact form. The lines are y=-x/2e+1/e+e the other line is y=e^x/2
Since y=-x/2e+1/e+e is on top it is the first function.
A=(the lower boundary is 0 and the top is 2) -x/2e+1/e+e-e^x/2
If you could please help!