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Archimedes’ Quadrature of the Parabola Consider a parabola given by the formula f ( x ) = a ! x 2 . Let b < c be given. Archimedes determined the area of figure bounded by the line joining the points ( b , a ! b 2 ) and ( c , a ! c 2 ) . By modern methods we would determine this by integration which we now proceed to do. Let g ( x ) = a ! c 2 " a ! b 2 c " b ( x " b ) + a ! b 2 , the linear function whose graph passes through the points ( b , a ! b 2 ) and ( c , a ! c 2 ) . Then one can easily compute that ( g ( x ) ! f ( x )) dx = a 6 ( c ! b ) 3 b c " . (1) Now consider the triangle whose vertex is on the graph of f where the slope of the tangent line is the same as the slope joining ( b , a ! b 2 ) and ( c , a ! c 2 ) . Again, using modern techniques one can determine the x –coordinate of that point by using the derivative. 2 ! a ! x 0 = a ! c 2 " a ! b 2 c " b x 0 = b + c 2 The area of the triangle determined by the vertices ( b , a ! b 2 ) , ( c , a ! c 2 ) , and x 0 , a ! x 0 2 ( ) is given by

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