Lecture 3: The Fundamental Theorem of Calculus

Today: The Fundamental Theorem of Calculus, Leibniz Integral Rule, Mean Value Theo- rem

We defined the definite integral and introduced the Evaluation Theorem to help us evaluate definite integrals. In today’s class, we will explore the relation between two central ideas of calculus: differentiation and integration.

The Fundamental Theorem of Calculus

The fundamental theorem of calculus deals with functions of the form x F (x) = f(t) dt, Za where f(x) is continuous on the interval [a, b] and x is taken on [a, b]. Recall that the definite integral is always a number that depends on the integrand, the upper and lower limits, but does not depend on the variable t that we integrate over. If x is fixed, the value of the function F (x) is also fixed, and if we let x vary, the function F (x), defined as the definite integral above, would vary with x.

Example. Let f(x) = x + 1. Define x F (x) = f(t) dt. Z0 (1) Sketch the graph of f(x) over the interval [0, 3]. (2) Evaluate F (0), F (1), F (2), and F (3). (3) Find a formula for F (x).

(4) Calculate F 0(x).

The graph of f(x) is a straight line.

0 1 2 3 4 4

3 3

2 2

1 1

0 0 0 1 2 3

19 0 First, F (0) = 0 f(t) dt = 0. The values of F (x) at x = 1, 2, 3 could be computed by area of trapezoids: R 1 (1 + 2) 1 3 F (1) = f(t) dt = · = , 2 2 Z0 2 (1 + 3) 2 F (2) = f(t) dt = · = 4, 2 Z0 3 (1 + 4) 3 15 F (3) = f(t) dt = · = . 2 2 Z0 To find a formula for F (x), we could compute F (x) directly using the Evaluation Theorem:

x t2 x x2 F (x) = (t + 1) dt = + t = + x, 0 2 0 2 Z h i and its derivative is

F 0(x) = x + 1.

Exercise. Let a be a fixed real number, and let f be a continuous function. Use the Evaluation Theorem to find the derivative of the function x F (x) = f(t) dt. Za

Let’s assume is an antiderivative of f, that is 0(x) = f(x). By the Evaluation Theorem, F F we have x x F (x) = f(x) dx = (t) = (x) (a). a F a F − F Z h i Notice that (a) is a constant, so it would vanish after we take one derivative. Therefore, F

F 0(x) = 0(x) 0 = f(x). F −

The result of this exercise leads us to the first part of the Fundamental Theorem of Calculus.

Theorem 10 (The Fundamental Theorem of Calculus, Part 1). If f is continuous on [a, b], then the function F defined by x F (x) = f(t) dt Za on the interval [a, b] is an antiderivative of f.

Proof. The proof of this theorem is surprisingly easy. It only uses the following facts about definite integrals: (1) If f is continuous on [a, b], then f is integrable on [a, b]. This is Theorem 5.

20 (2) If f is continuous on [a, b] and c [a, b], then ∈ b c b f(t) dt = f(t) dt + f(t) dt. Za Za Zc This is the interval concatenation property of Theorem 7. (3) If m f(t) M on [a, b], then ≤ ≤ b (b a)m f(t) dt (b a)M. − ≤ ≤ − Za This is a corollary of the comparison property (Theorem 8). We start by taking a point x in the open interval (a, b). Then we take h > 0 small enough such that x + h is also in the open interval (a, b). By definition of F , we have

x+h x x+h F (x + h) F (x) = f(t) dt f(t) dt = f(t) dt. − − Za Za Zx So the difference quotient

F (x + h) F (x) 1 x+h − = f(t) dt. h h Zx Because f is continuous on the closed interval [x, x + h], the Extreme Value Theorem tells us that f attains its minimum and maximum on [x, x + h], i.e., there exist numbers u and v in [x, x + h] such that m = f(u), M = f(v), and m f(t) M for all t in [x, x + h]. By ≤ ≤ the comparison property, we get

x+h f(u)h f(t) dt f(v)h. ≤ ≤ Zx Divide the inequalities through by h, we get

1 x+h f(u) f(t) dt f(v). ≤ h ≤ Zx Thus we have the upper and lower bounds on the difference quotient F (x + h) F (x) f(u) − f(v). ≤ h ≤ Now take h 0. Then u x and v x since both u and v lie on the interval [x, x + h]. → → → Thus lim f(u) = lim f(u) = f(x), and lim f(v) = lim f(v) = f(x). h 0+ u x h 0+ v x → → → → It follows from the Squeeze Theorem that F (x + h) F (x) lim − = f(x). h 0+ h → 21 A similar argument can give the result

F (x + h) F (x) lim − = f(x). h 0 h → − By definition of the derivative, we obtain the desired result:

F (x + h) F (x) F 0(x) = lim − = f(x). h 0 → h

The complete Fundamental Theorem of Calculus also includes the Evaluation Theorem as its Part 2.

Theorem 11 (The Fundamental Theorem of Calculus). Let f be a continuous function on the interval [a, b]. x (1) If F (x) = a f(t) dt, then F 0(x) = f(x). (2) b f(x) dx = (b) (a), where is any antiderivative of f. a R F − F F R Proof of Part 2. Consider the function F in Part 1. Then both F and are continuous F on [a, b] and differentiable on (a, b) with the same derivative F 0(x) = 0(x) = f(x). By a F corollary of the Mean Value Theorem for derivatives, we know that F and differs by a F constant C, i.e., (x) = F (x) + C. Thus F b b a f(x) dx = f(x) dx f(x) dx − Za Za Za = F (b) F (a) = (F (b) + C) (F (a) + C) = (b) (a) − − F − F

Remark. If we use Leibniz notation for derivatives, we may write the first part of the Fundamental Theorem as d x f(t) dt = f(x), dx Za where f is a continuous function. In words, this is saying that if we integrate f first and then differentiate the result, we get the original function f back.

Leibniz Integral Rule

We begin with some exercises regarding the Fundamental Theorem of Calculus.

22 Exercise. Find the derivative of the function x F (x) = et sin(t) dt. Z0 Because the integrand et sin(t) is continuous, we can apply Part 1 of the Fundamental The- orem and get t F 0(x) = e sin(t).

Exercise. Find the derivative of the function

x2 g(x) = cos(t) dt. Z1 Evaluate the definite integral

x2 g(x) = sin(t) = sin(x2) sin(1). 1 − h i Then use Chain Rule to evaluate its derivative

2 2 g0(x) = 2x cos(x ) 0 = 2x cos(x ) · − Exercise. Find the derivative of the function

cos(x) g(x) = (1 t2)9 dt. − Zsin(x) The integrand is a polynomial. In principle, we could expand and evaluate the integral term by term. But this process would be tedious. The following is a simpler process using the Chain Rule and Part 1 of the Fundamental Theorem. Let f(t) = (1 t2)9, u(x) = sin(x) and v(x) = cos(x). Then − cos(x) v(x) u(x) g(x) = (1 t2)9 dt = f(t) dt f(t) dt = F (v(x)) F (u(x)). − − − Zsin(x) Z0 Z0 Using the Chain Rule and Part 1 of the Fundamental Theorem, we get

g0(x) = v0(x) F 0(v(x)) u0(x) F 0(u(x)) = v0(x) f(v(x)) u0(x) f(u(x)) · − · · − · = sin(x) (1 cos2(x))9 cos(x) (1 sin2(x))9 − · − − · − = sin(x) (sin2(x))9 cos(x) (cos2(x))9 − · − · = sin19(x) cos19(x). − −

The Leibniz Integral Rule tells us how to differentiate a definite integral whose the upper and lower limits are differentiable functions.

23 Theorem 12 (Leibniz Integral Rule). Let f(t) be a continuous function, and let a(x) and b(x) be differentiable functions. Then d b(x) f(t) dt = b0(x) f(b(x)) a0(x) f(a(x)). dx · − · Za(x)

Average Value of a Function

Question. How do we take the average value of a function, where the function has infinitely many possible inputs, and possibly gives infinitely many outputs? For example, how to compute the average temperature of a particular day?

We know how to take the average value of a finite collection of numbers x1, . . . , xn: 1 n x = x ave n k Xk=1 In the old days, at small weather stations, people used to make four temperature measure- ments at 2 am, 8 am, 2 pm, and 8 pm, then compute the average value of these four data points to get the average temperature of the day. Of course, this is a crude estimate of the true average. We can improve the estimate by taking more data points. For example, we can take a total of 24 measurements every hour on the hour, and compute the average of these data points. We can further improve by taking a measurement every minute, resulting in a total of 1440 data points. The process described above is very similar to the limiting process of taking Riemann sums, and in fact, it is closely related to Riemann sums. Let’s say the function T (x) defined on the interval [a, b] describes the temperature of a certain period of time. Start by dividing the interval [a, b] into n subintervals of equal length, each with length ∆x = (b a)/n. The − we pick the right endpoints x1, . . . , xn of each subinterval and calculate the average 1 n ∆x n 1 n f(x ) = f(x ) = f(x ) ∆x n k b a k b a k · Xk=1 − Xk=1 − Xk=1 If we let n , the right hand side above would become → ∞ 1 n 1 b lim f(xk) ∆x = f(x) dx. n →∞ b a · b a a − Xk=1 − Z Thus we have the following definition. Definition. The average value of an integrable function f on the interval [a, b] is 1 b f = f(x) dx ave b a − Za

24 Exercise. Find the average values of the function f(x) = sin(x) on the intervals [0, π/2], [0, π], and [0, 2π], respectively.

1 π/2 2 π/2 2 2 sin(x) dx = cos(x) = ( cos(π/2) + cos(0)) = . π/2 π − 0 π − π Z0 1 π 1 h iπ 1 2 sin(x) dx = cos(x) = ( cos(π) + cos(0)) = . π 0 π − 0 π − π Z 2π h i 1 1 2π 1 sin(x) dx = cos(x) = ( cos(2π) + cos(0)) = 0. 2π 0 2π − 0 2π − Z h i We may wonder if there is a time of a day when the temperature is the same as the average temperature of the day, or, in general, if there is a number c such that f(c) = fave. The following theorem tells us that the answer is affirmative for continuous functions.

Theorem 13 (Mean Value Theorem). If f is continuous on [a, b], then there exists a num- ber c in [a, b] such that b f(x) dx = f(c)(b a). − Za x Proof. Let F (x) = a f(t) dt. By the Mean Value Theorem for derivatives, there exists a number c between aRand b such that

F (b) F (a) = F 0(c)(b a). − − By the Fundamental Theorem of Calculus, we get

b f(x) dx = f(c)(b a). − Za

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