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But this just says that F (x) = f (x) What is the practical significance of this calculation Suppose that we wish to calculate the area under the curve f , above the x-axis, and between x = a and x = b Obviously this area is F (b) F (a) See Figure 410 But we also b know that that area is a f (x) dx We conclude therefore that

(Continued)

f (x) dx = F (b) F (a)

-= 2

= = = =

383 1291 -1405 3315

y = f (x)

degrees inches inches rpm To avoid cam undercut, roc > 05600 To avoid cam undercut, roc < -05600 <10,000 recommended by McGill

F(b) _ F (a)

( 13, Cam System Dynamics Response)

The city dump can accommodate the trash from 38 thousand p ople, but at 200 thousand the curve becomes progressively more positive and the tras problem progressively worse

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Since the explicit relationship between the vectors p(0) and x(1) is not known, the in uence function matrix, Px, cannot easily be calculated One approach is to take partial derivatives of Eqs (1348) and (1349) with respect to the initial value of the costate vector, p(0) and interchange the order of differentiation (since continuity with respect to p(0) and t is assumed) The resulting two-point boundary-value problem is derived as follows x(t ) = f ( x(t ), p(t ),t ) (t ) = g( x(t ), p(t ),t ) p 2H 2H Px (p1 ( 0),t ) = Px (p1 ( 0),t ) + 2 Pp (p1 ( 0),t ) p x p 2 H 2H 1 Pp (p1 ( 0),t ) = - 2 Px (p1 ( 0),t ) + Pp (p ( 0),t ) x x p with the following initial conditions x ( 0 ) = x0 p( 0) = p1 ( 0) Px (p1 ( 0),t ) = 0 Pp ( P1 ( 0),t ) = 1 where x1 (t ) p1 ( 0) D Px (p1 ( 0),t ) = x5 (t ) p1 ( 0) Pp (p1 ( 0),t ) p(t ) p( 0) x1 (t ) p5 ( 0) x(t ) p( 0) x5 (t ) p5 ( 0)

Finally, if G is any other antiderivative for f then G(x) = F (x) + C Hence G(b) G(a) = [F (b) + C] [F (a) + C] = F (b) F (a) = That is the content of the Fundamental Theorem of Calculus

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Without saying so explicitly, we have implicitly assumed in our discussion of area in the last section that our function f is positive, that is, its graph lies about the x-axis But of course many functions do not share that property We nevertheless would like to be able to calculate areas determined by such functions, and to calculate the corresponding integrals This turns out to be simple to do Consider the function y = f (x) shown in Figure 411 It is negative on the interval [a, b] and positive on the interval [b, c] Suppose that we wish to calculate the shaded area as in Figure 412 We can do so by breaking the problem into pieces Of course, because f 0, the area between x = b and x = c is given by c the integral b f (x) dx, just as we have discussed in the last section But our

2H 2H p x = p x ij i j pi (t ) p(t ) p(t ) = p ( 0) , etc ij j Each subsequent new approximation for the vector, p(j+1)(0), is given by the iteration formula of Eq (1350) This approach to the solution of the two-point boundary-value problem is called the variation of extremals, also known as the generalized NewtonRaphson algorithm It is a second-order gradient technique and hence a good initial approximation for the costate vector, p(0)(0), is usually needed for convergence

.

discussions do not apply directly to the area between x = a and x = b What we can do is instead consider the function g = f Its graph is shown in Figure 413 Of course g is a positive function, except at the endpoints a and b; and the area under g ---between x = a and x = b---is just the same as the shaded area between x = a and x = b in Figure 414 (refer also to Figure 412) That area is

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