1550078481-Ordinary_Differential_Equations__Roberts_

118 Ordinary Differential Equations

FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS

If f ( x) is continuous on the interval [a , b] and if F' ( x) = f ( x) for all

x E [a , b], then

1 b f(x) dx = F(x) lb a= F(b) - F(a)

Problems which appear in calculus texts are chosen very carefully so that

required integrations can be performed- that is , so that the antiderivative

of the integrand can be written as an elementary function. In this respect,

problems which appear in calculus texts are somewhat artificial, because in

practice the required integration usually cannot be performed explicitly. Look

at the section on computing arc length in any calculus text and you will see

that very few examples and exercises are given. This is because it is difficult

to choose many functions y(x) such that the integral of Jl + [y'(x)]2 can be

written as an elementary function.

Unfortunately, most functions f(x) which arise in practice do not

have antiderivatives that can be written as elementary functions.

So we need to answer the following questions:

"How can we rewrite any definite integral J: f(x) dx as an initial value

problem?"

"How can we numerically solve the resulting initial value problem and

thereby compute a value for the definite integral?"

If f(x) is integrable on [a, b], we can symboli cally write the antiderivative

of f(x) on [a, b] as

(1) F(x) = 1x f(t) dt for x E [a, b].

Suppose F(x) is differentiable on [a, b]. Differentiating (1) with respect to

x, we find F'(x) = f(x ) for all x E [a, b]. So F(x) satisfies the differential

equation y' = f(x) on [a , b]. Evaluating F(x) at x = a, we get F(a) = 0.

Thus, the antiderivative F(x) of f(x) satisfies the initial value problem

(2) y' = f(x); y(a) = 0

on the interval [a , b] and the value of the definite integral of f(x) on [a , b] is

lb f(t) dt = y(b) = F(b).