By John Gilbert (auth.)

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We write . Js . g((t + ~t) 2 - t 2 ) = g(t + &/2)<5t. Thus, ~s/ ~t = g(t + bt/2) and v(t) . ~s I. = I tm - = tm &~o & &~o g(t + ~t/2) = gt We now consider the geometric problem of finding the slope of the tangent to the graph of a function fat a point x. We approximate the tangent at the point P, whose coordinates are (x, f(x)) by the chord joining P with the point Q whose coordinates are (x + ~x, f(x + ~x)) for a small ~x. 1, is f(x + ~x)- f(x) ~X We obtain the slope of the tangent by taking the limit of this slope as ~x--+0.

1 Calculate J6 x 2 dx. We divide the interval of integration into n subintervals I,= [i ~ 1, ~l 49 GUIDE TO MATHEMATICAL METHODS y 0 n n n Fig. 5, and set X; = ~ (corresponding n to the end of/;), so that f(x;) = height f(x;) on base I; is An= n i2 2 . The total area of then rectangles R; of n n n i2 1 1 n L f(x;)bx = L 2-n = 3n L i ;~t ;~tn 2 ;~t The latter sum is just the sum of the squares of the first n natural numbers, whose value you may have come across before as 7;n(n + 1)(2n + 1). Using this, we obtain An= which we rewrite as 1( 1+ (n + 1)(2n + 1) 6n 2 ~) ( 2 + ~).

X 2 dx = t. This method of calculating areas was first used by Archimedes and so predates the development of calculus by many centuries. Notice that if, in the above, f(x;) is negative, then the rectangle R; will be below the x axis. Its area is - f(x;)bx, since areas are normally taken as positive. For integrals;however, we adopt the convention that any contribution from below the axis will be negative; this is the case if we take f(x;)bx as the contribution of the rectangle R; to the area An.