 INTEGRALS 5. 5.3 The Fundamental Theorem of Calculus INTEGRALS In this section, we will learn about: The Fundamental Theorem of Calculus and its significance.

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INTEGRALS 5

5.3 The Fundamental Theorem of Calculus INTEGRALS In this section, we will learn about: The Fundamental Theorem of Calculus and its significance.

The Fundamental Theorem of Calculus (FTC) is appropriately named.  It establishes a connection between the two branches of calculus—differential calculus and integral calculus. FUNDAMENTAL THEOREM OF CALCULUS

FTC Differential calculus arose from the tangent problem. Integral calculus arose from a seemingly unrelated problem—the area problem.

Newton’s mentor at Cambridge, Isaac Barrow (1630–1677), discovered that these two problems are actually closely related.  In fact, he realized that differentiation and integration are inverse processes. FTC

The FTC gives the precise inverse relationship between the derivative and the integral. FTC

It was Newton and Leibniz who exploited this relationship and used it to develop calculus into a systematic mathematical method.  In particular, they saw that the FTC enabled them to compute areas and integrals very easily without having to compute them as limits of sums—as we did in Sections 5.1 and 5.2 FTC

The first part of the FTC deals with functions defined by an equation of the form where f is a continuous function on [a, b] and x varies between a and b. Equation 1 FTC

 Observe that g depends only on x, which appears as the variable upper limit in the integral.  If x is a fixed number, then the integral is a definite number.  If we then let x vary, the number also varies and defines a function of x denoted by g(x). FTC

If f happens to be a positive function, then g(x) can be interpreted as the area under the graph of f from a to x, where x can vary from a to b.  Think of g as the ‘area so far’ function, as seen here. FTC Figure 5.3.1, p. 314

If f is the function whose graph is shown and, find the values of: g(0), g(1), g(2), g(3), g(4), and g(5).  Then, sketch a rough graph of g. Example 1 FTC Figure 5.3.2, p. 314

First, we notice that: FTC Example 1

From the figure, we see that g(1) is the area of a triangle: Example 1 FTC Figure 5.3.3a, p. 314

To find g(2), we add to g(1) the area of a rectangle: Example 1 FTC Figure 5.3.3b, p. 314

We estimate that the area under f from 2 to 3 is about 1.3. So, Example 1 FTC Figure 5.3.3c, p. 314

Thus, FTC Example 1 Figure 5.3.3d, p. 314Figure 5.3.3e, p. 314

We use these values to sketch the graph of g.  Notice that, because f(t) is positive for t < 3, we keep adding area for t < 3.  So, g is increasing up to x = 3, where it attains a maximum value.  For x > 3, g decreases because f(t) is negative. Example 1 FTC Figure 5.3.4, p. 314

If we take f(t) = t and a = 0, then, using Exercise 27 in Section 5.2, we have: FTC

Notice that g’(x) = x, that is, g’ = f.  In other words, if g is defined as the integral of f by Equation 1, g turns out to be an antiderivative of f—at least in this case. FTC

If we sketch the derivative of the function g, as in the first figure, by estimating slopes of tangents, we get a graph like that of f in the second figure.  So, we suspect that g’ = f in Example 1 too. FTC Figure 5.3.4, p. 314 Figure 5.3.2, p. 314

To see why this might be generally true, we consider a continuous function f with f(x) ≥ 0.  Then, can be interpreted as the area under the graph of f from a to x. FTC Figure 5.3.1, p. 314

To compute g’(x) from the definition of derivative, we first observe that, for h > 0, g(x + h) – g(x) is obtained by subtracting areas.  It is the area under the graph of f from x to x + h (the gold area). FTC Figure 5.3.5, p. 315

For small h, you can see that this area is approximately equal to the area of the rectangle with height f(x) and width h: So, FTC Figure 5.3.5, p. 315

Intuitively, we therefore expect that:  The fact that this is true, even when f is not necessarily positive, is the first part of the FTC (FTC1). FTC

FTC1 If f is continuous on [a, b], then the function g defined by is continuous on [a, b] and differentiable on (a, b), and g’(x) = f(x).

If x and x + h are in (a, b), then Proof FTC1

So, for h ≠ 0, Proof—Equation 2 FTC1

For now, let us assume that h > 0.  Since f is continuous on [x, x + h], the Extreme Value Theorem says that there are numbers u and v in [x, x + h] such that f(u) = m and f(v) = M.  m and M are the absolute minimum and maximum values of f on [x, x + h]. Proof FTC1 Figure 5.3.6, p. 316

By Property 8 of integrals, we have: That is, Proof FTC1

Since h > 0, we can divide this inequality by h: FTC1 Proof

Now, we use Equation 2 to replace the middle part of this inequality:  Inequality 3 can be proved in a similar manner for the case h < 0. Proof—Equation 3 FTC1

Now, we let h → 0. Then, u → x and v → x, since u and v lie between x and x + h.  Therefore, and because f is continuous at x. Proof FTC1

From Equation 3 and the Squeeze Theorem, we conclude that: Proof—Equation 4 FTC1

Using Leibniz notation for derivatives, we can write the FTC1 as when f is continuous.  Roughly speaking, Equation 5 says that, if we first integrate f and then differentiate the result, we get back to the original function f. Equation 5 FTC1

Find the derivative of the function  As is continuous, the FTC1 gives: Example 2 FTC1

FRESNEL FUNCTION For instance, consider the Fresnel function  It is named after the French physicist Augustin Fresnel (1788–1827), famous for his works in optics.  It first appeared in Fresnel’s theory of the diffraction of light waves.  More recently, it has been applied to the design of highways. Example 3

FRESNEL FUNCTION The FTC1 tells us how to differentiate the Fresnel function: S’(x) = sin(πx 2 /2)  This means that we can apply all the methods of differential calculus to analyze S. Example 3

The figure shows the graphs of f(x) = sin(πx 2 /2) and the Fresnel function  A computer was used to graph S by computing the value of this integral for many values of x. Example 3 FRESNEL FUNCTION Figure 5.3.7, p. 317

It does indeed look as if S(x) is the area under the graph of f from 0 to x (until x ≈ 1.4, when S(x) becomes a difference of areas). Example 3 FRESNEL FUNCTION Figure 5.3.7, p. 317

The other figure shows a larger part of the graph of S. Example 3 FRESNEL FUNCTION Figure 5.3.7, p. 317 Figure 5.3.8, p. 317

If we now start with the graph of S here and think about what its derivative should look like, it seems reasonable that S’(x) = f(x).  For instance, S is increasing when f(x) > 0 and decreasing when f(x) < 0. Example 3 FRESNEL FUNCTION Figure 5.3.7, p. 317

Find  Here, we have to be careful to use the Chain Rule in conjunction with the FTC1. Example 4 FTC1

Let u = x 4. Then, Example 4 FTC1

FTC2 If f is continuous on [a, b], then where F is any antiderivative of f, that is, a function such that F’ = f.

FTC2 Let We know from the FTC1 that g’(x) = f(x), that is, g is an antiderivative of f. Proof

FTC2 If F is any other antiderivative of f on [a, b], then we know from Corollary 7 in Section 4.2 that F and g differ by a constant: F(x) = g(x) + C for a < x < b. Proof—Equation 6

FTC2 However, both F and g are continuous on [a, b]. Thus, by taking limits of both sides of Equation 6 (as x → a + and x → b - ), we see it also holds when x = a and x = b. Proof

FTC2 If we put x = a in the formula for g(x), we get: Proof

FTC2 So, using Equation 6 with x = b and x = a, we have: Proof

FTC2 Evaluate the integral  The function f(x) = x 3 is continuous on [-2, 1] and we know from Section 4.9 that an antiderivative is F(x) = ¼x 4.  So, the FTC2 gives: Example 5

FTC2  Notice that the FTC2 says that we can use any antiderivative F of f.  So, we may as well use the simplest one, namely F(x) = ¼x 4, instead of ¼x 4 + 7 or ¼x 4 + C. Example 5

FTC2 We often use the notation So, the equation of the FTC2 can be written as:  Other common notations are and.

FTC2 Find the area under the parabola y = x 2 from 0 to 1.  An antiderivative of f(x) = x 2 is F(x) = (1/3)x 3.  The required area is found using the FTC2: Example 6

FTC2 Find the area under the cosine curve from 0 to b, where 0 ≤ b ≤ π/2.  Since an antiderivative of f(x) = cos x is F(x) = sin x, we have: Example 7

FTC2 In particular, taking b = π/2, we have proved that the area under the cosine curve from 0 to π/2 is sin(π/2) =1. Example 7 Figure 5.3.9, p. 319

FTC2 What is wrong with this calculation? Example 8

FTC2 To start, we notice that the calculation must be wrong because the answer is negative but f(x) = 1/x 2 ≥ 0 and Property 6 of integrals says that when f ≥ 0. Example 9

FTC2 The FTC applies to continuous functions.  It can’t be applied here because f(x) = 1/x 2 is not continuous on [-1, 3].  In fact, f has an infinite discontinuity at x = 0.  So, does not exist. Example 9

FTC Suppose f is continuous on [a, b]. 1.If, then g’(x) = f(x). 2., where F is any antiderivative of f, that is, F’ = f.

INVERSE PROCESSES We noted that the FTC1 can be rewritten as:  This says that, if f is integrated and then the result is differentiated, we arrive back at the original function f.

INVERSE PROCESSES As F’(x) = f(x), the FTC2 can be rewritten as:  This version says that, if we take a function F, first differentiate it, and then integrate the result, we arrive back at the original function F.  However, it’s in the form F(b) - F(a).

SUMMARY Before it was discovered—from the time of Eudoxus and Archimedes to that of Galileo and Fermat—problems of finding areas, volumes, and lengths of curves were so difficult that only a genius could meet the challenge.

SUMMARY Now, armed with the systematic method that Newton and Leibniz fashioned out of the theorem, we will see in the chapters to come that these challenging problems are accessible to all of us.

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