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

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

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

FTC 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. 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 Equation 1 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.

If x is a fixed number, then the integral is a definite number. 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.

If f is the function whose graph is shown and FTC Example 1 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.

FTC Example 1 First, we notice that:

From the figure, we see that g(1) is the area of a triangle: FTC Example 1 From the figure, we see that g(1) is the area of a triangle:

To find g(2), we add to g(1) the area of a rectangle: FTC Example 1 To find g(2), we add to g(1) the area of a rectangle:

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

For t > 3, f(t) is negative. FTC Example 1 For t > 3, f(t) is negative. So, we start subtracting areas, as follows.

FTC Example 1 Thus,

We use these values to sketch the graph of g. FTC Example 1 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.

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 .

FTC 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 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 For small h, we see that this area is approximately equal to the area of the rectangle with height f(x) and width h; that is, So,

Intuitively, we therefore expect that: FTC 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).

Fundamental Theorem of Calculus Version 1 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).

FTC1 In words, the FTC1 says that the derivative of a definite integral with respect to its upper limit is the integrand evaluated at the upper limit.

Using Leibniz notation for derivatives, we can write the FTC1 as Equation 5 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.

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

FTC1 A formula of the form may seem like a strange way of defining a function. However, books on physics, chemistry, and statistics are full of such functions.

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

FTC1 Example 4 Let u = x4. Then,

FTC1 In Section 5.2, we computed integrals from the definition as a limit of Riemann sums and saw that this procedure is sometimes long and difficult. The second part of the FTC (FTC2), which follows easily from the first part, provides us with a much simpler method for the evaluation of integrals.

Fundamental Theorem of Calculus Version 2 If f is continuous on [a, b], then where F is any antiderivative of f, that is, a function such that F ’ = f.

FTC2 It is very surprising that , which was defined by a complicated procedure involving all the values of f(x) for a ≤ x ≤ b, can be found by knowing the values of F(x) at only two points, a and b.

At first glance, the theorem may be surprising. FTC2 At first glance, the theorem may be surprising. However, it becomes plausible if we interpret it in physical terms.

So, s is an antiderivative of v. FTC2 If v(t) is the velocity of an object and s(t) is its position at time t, then v(t) = s’(t). So, s is an antiderivative of v.

FTC2 In Section 5.1, we considered an object that always moves in the positive direction. Then, we guessed that the area under the velocity curve equals the distance traveled. In symbols, That is exactly what the FTC2 says in this context.

FTC2 Example 5 Evaluate the integral The function f(x) = ex is continuous everywhere and we know that an antiderivative is F(x) = ex. So, the FTC2 gives:

Notice that the FTC2 says that we can use any antiderivative F of f. Example 5 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) = ex, instead of ex + 7 or ex + C.

We often use the notation FTC2 We often use the notation So, the equation of the FTC2 can be written as:

Find the area under the parabola y = x2 from 0 to 1. FTC2 Example 6 Find the area under the parabola y = x2 from 0 to 1. An antiderivative of f(x) = x2 is F(x) = (1/3)x3. The required area is found using the FTC2:

FTC2 If you compare the calculation in Example 6 with the one in Example 2 in Section 5.1, you will see the FTC gives a much shorter method.

Evaluate An antiderivative of f(x) = 1/x is F(x) = ln |x|. FTC2 Example 7 Evaluate An antiderivative of f(x) = 1/x is F(x) = ln |x|. As 3 ≤ x ≤ 6, we can write F(x) = ln x.

FTC2 Example 7 Therefore,

Find the area under the cosine curve from 0 to b, where 0 ≤ b ≤ π/2. FTC2 Example 8 Find the area under the cosine curve from 0 to b, where 0 ≤ b ≤ π/2. As an antiderivative of f(x) = cos x is F(x) = sin x, we have:

FTC2 Example 8 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

What is wrong with this calculation? FTC2 Example 9 What is wrong with this calculation?

and Property 6 of integrals says that when f ≥ 0. FTC2 Example 9 To start, we notice that the calculation must be wrong because the answer is negative but f (x) = 1/x2 ≥ 0 and Property 6 of integrals says that when f ≥ 0.

The FTC applies to continuous functions. Example 9 The FTC applies to continuous functions. It cannot be applied here because f(x) = 1/x2 is not continuous on [-1, 3]. In fact, f has an infinite discontinuity at x = 0. In section 7.8 we will see that , does not exist.

We end this section by bringing together the two parts of the FTC. INVERSE PROCESSES We end this section by bringing together the two parts of the FTC.

Suppose f is continuous on [a, b]. 1. If , then g’(x) = f(x). 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.

We noted that the FTC1 can be rewritten as: 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.

As F’(x) = f(x), the FTC2 can be rewritten as: 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 is in the form F(b) - F(a).

INVERSE PROCESSES Taken together, the two parts of the FTC say that differentiation and integration are inverse processes. Each undoes what the other does.