第4.2节 微积分学的基本定理

4.2 FUNDAMENTAL THEOREM OF CALCULUS

In this section we shall state five basic theorems about the integral, culminating in the Fundamental Theorem of Calculus. Right now we can only approximate a definite integral by the laborious computation of a finite Riemann sum. At the end of this section we will be in a position easily to compute exact values for many definite integrals. The key to the method is the Fundamental Theorem. Our first theorem shows that we are free to choose any positive infinitesimal we wish for dx in the definite integral.

THEOREM 1

Given a continuous function f on [a, b] and two positive infinitesimals dx and du, the definite integrals with respect to dx and du are the same,

From now on when we write a definite integral _____f(x)dx, it is understood that dx is a positive infinitesimal. By theorem 1, it doesn’t matter which infinitesimal.

The proof of Theorem 1 is based on the following intuitive idea. Figure 4.2.1 shows the two Riemann sums _____f(x)dx and ___ f(u) du. We see from the figure that the difference ______f(x) dx -_____ f(u) du is a sum of rectangles of infinitesimal height. These difference rectangles all lie between the horizontal lines y= -ε and y=ε, where ε is the largest height. Thus -ε(b-a) ____ _____ f(x) dx -____f(u) du≤ ε (b -a). Taking standard parts,

Figure 4.2.1

Theorem 1 shows that whenever Δx is positive infinitesimal, the Riemann sum is infinitely close to the definite integral,

This fact can also be expressed in terms of limits. It shows that the Riemann sum approaches the definite integral as Δx approaches 0 from above, in symbols

Given a continuous function f on an interval I, Theorem I shows that the definite integral is a real function of two variables a and b,

We now formally define the area as the definite integral shown in Figure 4.2.2.

Figure 4.2.2

DEFINITION

If f is continuous and f(x) ≥0 on[a,b], the area of the region below the curve y=f(x) from a to b is defined as the definite integral:

The next two theorems give basic properties of the integral.

THEOREM 2 (The Rectangle Property)

Suppose f is continuous and has minimum value m and maximum value M on a closed interval [a, b]. Then

That is, the area of the region under the curve is between the area of the rectangle whose height is the minimum value of and the area of the rectangle whose height is the maximum value of in the interval [a,b].

The Extreme Value Theorem is needed to show that the minimum value m and maximum value M exist. The rectangle of height m is called the inscribed rectangle of the region, and the rectangle of height M is called the circumscribed rectangle. From Figure 4.2.3, we see that the inscribed rectangle is a subset of the region under the curve, which is in turn a subset of the circumscribed rectangle. The Rectangle Property says that the area of the region is between the areas of the inscribed and circumscribed rectangles.

Figure 4.2.3 the Rectangle Property

PROOF By Theorem 1, any positive infinitesimal may be chosen for dx. Let us choose a positive infinite hyperinteger H and let dx=(b-a) /H. Then dx evenly divides b-a; that is, the interval [a, b] is divided into H subintervals of exactly the same length dx. Then

For each x, we have m≤ f(x)≤M. Adding up and taking standard parts, we obtain the required formula.

One useful consequence of the Rectangle Property is that the integral of a positive function is positive and the integral of a negative function is negative:

The definite integral of a negative function f(x)= -g(x) from a to b is just the negative of the area of the region above the curve and below the x axis.

This is because

f(x) dx = -g(x)dx,

(See Figure 4.2.4.)

Figure 4.2.4

THEOREM 3 (The Addition Property)

Suppose f is continuous on an interval I. Then for all a, b, c in I,

This property is illustrated in Figure 4.2.5 for the case a<b<c. The Addition Property holds even if the points a, b,c are in some other order on the real line, such as c<a<b.

Figure 4.2.5

PROOF First suppose that a<b<c. Choose a dx that evenly divides the first interval length b -a. This simplifies our computation because it makes b a partition point, b=a+H dx. Then, as Figure 4.2.6 suggests,

Taking standard parts we have the desired formula

Figure 4.2.6

To illustrate the other cases, we prove the Addition Property when

c<a<b. The previous case gives

Since reversing the endpoints changes the sign of the integral,

and the desired formula

follows.

The definite integral of a curve can be thought of as area even if the curve crosses the x-axis. The curve in Figure 4.2.7 is positive from a to b and negative from b to c, crossing the x-axis at b. The integral ___ f(x) dx is a positive number and the integral ___ f(x) dx is a negative number. By the Addition Property, the integral

is equal to the area from a to b minus the area from b to c. The definite integral ___ f(x) dx always gives the net area between the x-axis and the curve, counting areas above the x-axis as positive and areas below the x-axis as negative.

The definite integral __ f(t)dt is a real function of two variables u and v and does not depend on the dummy variable t. If we replace u by a constant a and v by the variable x, we obtain a real function of one variable x, given by

Our fourth theorem states that this new function is continuous.

Figure 4.2.7

THEOREM 4

Let f be continuous on an interval 1. Choose a point a in 1. Then the function F(x) defined by

is continuous on 1.

SKETCH OF PROOF Let c be in 1, and let x be infinitely close to c and between the end points of 1. By the Addition Property.

and

This is the area of the infinitely thin strip under the curve y=f(t) between t=x and t=c (see Figure 4.2.8). The strip has width Δx=c -x. By the Rectangle Property, its area is between m Δx and M Δx and hence is infinitely small. Therefore F(x) is infinitely close to F(c), and F is continuous on 1.

Figure 4.2.8

Our fifth theorem, the Fundamental Theorem of Calculus, shows that the definite integral can be evaluated by means of antiderivatives. The process of antidifferentiation is just the opposite of differentiation. To keep things simple, let I be an open interval, and assume that all functions mentioned have domain I.

DEFINITION

Let f and F be functions with domain I. If f is the derivative of F, then F is called an antiderivative of f.

For example, suppose a particle is moving upward along the y-axis with velocity v=f(t) and position y=F(t) at time t. The position y=F(t) is an antiderivative of the velocity v=f(t). We shall discuss antiderivatives in more detail in the next section. We are now ready for the Fundamental Theorem.

FUNDAMENTAL THEOREM OF CALCULUS

Suppose f is continuous on its domain, which is an open interval I.

( i ) For each point a in I, the definite integral of f from a to x considered as a function of x is an antiderivative of f. That is,

(ii) If F is any antiderivative of f, then for any two points (a, b) in I the definite integral of f from a to b is equal to the difference F(b) - F(a),

The Fundamental Theorem of Calculus is important for two reasons. First, it shows the relation between the two main notions of Calculus: the derivative, which corresponds to velocity, and the integral, which corresponds to area. It shows that differentiation and integration are “inverse” processes. Second, it gives a simple method for computing many definite integrals.

EXAMPLE 1

(a) Find ____ cdx. Since cx is an antiderivative of c,

(b) Find ___ xdx. __x² is an antiderivative of x. Thus

The above example gives the same result that we got before but is much simpler. We can easily go further.

EXAMPLE 2 Find ___ x² dx. x3/3 is an antiderivative of x² because

Therefore

This gives the area of the region under the curve y=x² between a and b

(Figure 4.2.9).

Figure 4.2.9

If a particle moves along the y-axis with continuous velocity v=f(t), the position y=F(t) is an antiderivative of the velocity, because v=dy/dt. The Fundamental Theorem of Calculus shows that the distance moved (the change in y) between times t=a and t=b is equal to the definite integral of the velocity,

EXAMPLE 3 A particle moves along the y-axis with velocity v=8t3 cm/sec. How far does it move between times t= -1 and t= 2 sec? The function G(t)=2t4 is an antiderivative of the velocity v = 8t3. Thus the definite integral is

EXAMPLE 4 Find ____ dt (Figure 4.2.10). The function ___ is defined and continuous on the half-open interval [0,∝). But to apply the Fundamental Theorem we need a function continuous on an open interval that contains the limit points 0 and 4. We therefore define

This function is continuous on the whole real line. In particular it is continuous at 0 because if t≈ 0 then f(t) ≈ 0. The function

Figure 4.2.10

is an antiderivative of f. Then

In the next section we shall develop some methods for finding antiderivatives. The antiderivative of a very simple function may turn out to be a “new” function which we have not yet given a name.

EXAMPLE 5 The only way we can show that the function f(x) = ________ has an antiderivative is to take a definite integral

This is a “new” function that cannot be expressed in terms of algebraic, trigonometric, and exponential functions without calculus.

The Fundamental Theorem cal also be used to find the derivative of a function which is defined as a definite integral with a variable limit of integration.. This can be done without actually evaluating the integral.

EXAMPLE 6 let

and

EXAMPLE 7 Let y=

By the Chain Rule,

We conclude this section with a proof of the Fundamental Theorem of Calculus.

PROOF (i) Let F(x) be the area under the curve y=f(t) from a to x,

Imagine that the vertical line cutting the t-axis at x moves to the right as in Figure 4.2.11

Figure 4.2.11

We show that the rate of change of F(x) is equal to the length f(x) of the moving vertical line.

Suppose x increases by an infinitesimal amount Δx> 0. Then

is the area of an infinitely thin strip of width Δx and height infinitely close to f(x). By the Rectangle Property the area of the strip is between the inscribed and circumscribed rectangles (Figure 4.2.12),

m Δx≤ F(x+Δx) - F(x) ≤M Δx.

Dividing by Δx, m≤_________ ≤ M.

Since f is continuous at x, the values m and M are both infinitely close to f(x),

and therefore

The proof is similar when Δx<0. Hence F'(x) =f(x).

Figure 4.2.12

PROOF (ii) Let F(x) be any antiderivative of f. Then, by(i),

In Section 3.7 on curve sketching, we saw that every function with derivative zero is constant. Thus

for some constant C0. Then

PROBLEMS FOR SECTION 4.2

In Problems 1-14, find an antiderivative of the given function.

1 f(x) =8___ 2 f(x) = 4/__

3 f(t) =3t² + 1 4 f(x) =5x3

5 f(t) =4 －3t² 6 f(z) =2/z²

7 f(s) =7s -3 8 f(t) =t² + t-2

9 f(x) =(x -6)² 10 f(u) =(5u + 1)²

11 f(y) = y3/2 12 f(x) =2/x__

13 f(x) =|x| 14 f(t) = |2t -4 |

15 If F '(x) = x + x² for all x, find F(1) - F(-1).

16 If F '(x) = x4 for all x, find F(2) - F(1).

17 If F '(t) = t1/3 for all t, find F(8) - F(0).

Evaluate the definite integrals in Problems 18-22.

18 __________ 19__________

20 __________ 21__________

22__________

In Problems 23-27 an object moves along the y-axis. Given the velocity v, find how far the object moves between the given times t0 and t1.

23 v =2t + 5, t0 =0, t1 =2

24 v =4 －t, t0 =1, t1 =4

25 v =3, t0 =2, t1 =6

26 v =3t², t0 =1, t1 =3

27 v =10t -2, t0 =1, t1 =100

In Problems28-32, find the area of the region under the curve y=f(x) from a to b.

28 y =4 －x², a = －2, b=2

29 y = ______, a = －2, b=2

30 y =9x －x², a = 0, b=3

31 y =__ －x, a =0, b=1

32 y = 3x?. a =1, b=8

33 If F '(t) = t-1 for all t and F(0) =2, find F(2).

34 If F '(x) = 1－x² for all x and F(3) =5, find F(-1).

□35 Suppose F(x) and G(x) have continuous derivatives and F '(x)+G'(x)=0 for all x.

Prove that F(x)+G(x) is constant.

□36 Suppose F(x) and G(x) have continuous derivatives such that F '(x)≤G'(x) for all x.

Prove that F(b)- F(a)≤G(b)- G(a)

Where a<b.

□37 Prove that a function F(x) has a constant derivative if and only if F(x) is linear, i.e.,

of the form F(x)=ax +b.

□38 Prove that a function F(x) has a constant second derivative if and only if F(x) has the form

F(x) = ax² +bx + c.

□39 Suppose that F''(x) = G''(x) for all x. Prove that F(x) and G(x) differ by a linear function,

that is, G(x) = F(x) + ax + b for some real numbers a and b.