Lecture 15 Notes:The Fundamental Theorem of Calculus

18.01 Calculus Jason Starr Fall 2005 Lecture 15. October 18, 2005 Homework. Problem Set 4 Part I: (d) and (e); Part II: Problem 2. Practice Problems. Course Reader: 3B­6, 3C­2, 3C­3, 3C­4, 3C­6. 1. The Riemann sum for the exponential function. The problem is to compute the Riemann integral, � b ex dx, 0 using Riemann sums. Choose the partition of [0, b] into a sequence of n equally­spaced subintervals of length b/n. So the partition numbers are xk = kb/n. Also the length of each partition is Δxk = b/n. Because ex is increasing, the minimum value of ex on the interval [xk−1 , xk ] occurs at the left endpoint, yk,min = exk−1 = e(k−1)b/n . Similarly, the maximum value occurs at the right endpoint, yk,max = exk = ekb/n . Thus the lower sum is, Amin = n � yk,min Δxk = k=1 n � b e(k−1)b/n . n k=1 And the upper sum is, Amax = n � yk,max Δxk = k=1 n � b ekb/n . n k=1 To evaluate each of the sums, make the substitution c = eb/n . Then the lower sum is, n Amin n−1 b � k−1 b� l = c = c. n k=1 n l=0 The sum is a geometric sum, (1 + c + c2 + · · · + cn−2 + cn−1 ) = cn − 1 . c−1 Plugging this in gives, Amin = b cn − 1 b ebn/n − 1 = . n c−1 n eb/n − 1 18.01 Calculus Jason Starr Fall 2005 This simplifies to, Amin = (eb − 1) b/n . −1 eb/n A similar computation gives, Amax = (eb − 1)eb/n b/n . −1 eb/n Now make the substitution, h = b/n. This gives, Amin = (eb − 1) eh h , −1 h . −1 Taking the limit of Amin , respectively Amax , as n tends to infinity is the same as taking the limit as h tends to 0. Amax = (eb − 1)eh eh Now observe that, eh − 1 , h→0 h is the difference quotient limit giving the derivative of ex at x = 0. Since dex /dx equals ex , and since e0 equals 1, this gives, eh − 1 lim = 1. h→0 h Inverting gives, � �−1 h eh − 1 lim = lim = (1)−1 = 1. h→0 eh − 1 h→0 h lim Also, because ex is continuous, lim eh = e0 = 1. h→0 Putting this together gives, lim Amin = (eb − 1) lim n→∞ h→0 eh h = (eb − 1)(1) = eb − 1. −1 Similarly, h ) = (eb − 1)(1)(1) = eb − 1. −1 exist and are equal, the Riemann integral exists and lim Amax = (eb − 1)(lim eh )(lim n→∞ h→0 eh h→0 Since the limit of Amin and the limit of Amax equals, � b ex dx = eb − 1. 0 18.01 Calculus Jason Starr Fall 2005 2. The Riemann sum for xr . Let r > 0 be a positive real number. The problem is to compute the Riemann integral, � b xr dx, 1 using Riemann sums. For this particular integral, a different partition than usual is more efficient. Let n be a positive integer, and let q be the real number, q = b1/n . Choose the partition of [1, b] into n subintervals with partition numbers, xk = q k . Observe that, 1 = x0 < x1 < · · · < xn−1 < xn = (b1/n )n = b. The length of the k th subinterval is, Δxk = xk − xk−1 = q k − q k−1 = q k−1 (q − 1). Observe this increases as k increases. So this is not the partition of [1, b] into n equal subintervals. The mesh size is, mesh = max(Δx1 , . . . , Δxn ) = Δxn = (q − 1)b(n−1)/n ≤ q − 1. As n tends to infinity, the mesh size tends to, lim mesh = lim q − 1 = lim b1/n − 1 = 0. n→0 n→0 n→0 Thus, even though this isn’t the most obvious choice of partition, it can be used to compute the Riemann integral. Because xr is increasing, the minimum value of xr on the interval [xk−1 , xk ] occurs at the left endpoint, yk,min = xrk−1 = q (k−1)r . Similarly, the maximum value occurs at the right endpoint, yk,max = xrk = q kr . Thus the lower sum is, Amin = n � k=1 yk,min Δxk = n � k=1 q (k−1)r · q (k−1) (q − 1). 18.01 Calculus Jason Starr Fall 2005 This simplifies to, Amin = (q − 1) n � q (k−1)(r+1) . k=1 And the upper sum is, Amax = n � yk,max Δxk = k=1 n � q kr q (k−1) (q − 1). k=1 This simplifies to, Amax = (q − 1)q r n � q (k−1)(r+1) . k=1 To evaluate the sum, make the substitution c = q r+1 . Then the sum is, n � ck−1 = 1 + c + c2 + · · · + cn−2 + cn−1 . k=1 This geometric sum equals, q n(r+1) − 1 cn − 1 = r+1 . c−1 q −1 Thus the upper and lower sums simplify to, Amin = (q − 1)(q n(r+1) − 1)/(q r+1 − 1), Amax = q r (q − 1)(q n(r+1) − 1)/(q r+1 − 1). Now back­substitute q = b1/n to get that q n(r+1) = br+1 . Simplifying gives, Amin = (br+1 − 1) 1 , (q r+1 − 1)/(q − 1) Amax = (br+1 − 1)q r (q r+1 1 . − 1)/(q − 1) As n tends to infinity, the quantity q = b1/n tends to 1. The fraction, q r+1 − 1 , q−1 is the difference quotient for y = xr+1 for x = 1. As q tends to 1, the limit of the difference quotient is the derivative of y = xr+1 at x = 1, d(xr+1 ) q r+1 − 1 |x=1 = ((r + 1)xr |x=1 = (r + 1). = q→1 q − 1 dx lim 18.01 Calculus Jason Starr Fall 2005 Also, since xr is continuous, lim q r = 1r = 1. q→1 Substituting this in gives, lim Amin = (b n→∞ lim Amax = (b r+1 n→∞ r+1 q r+1 − 1 − 1) lim q→1 q − 1 � � − 1) lim q r �−1 = q r+1 − 1 lim q→1 q − 1 �� q→1 br+1 − 1 , r+1 �−1 = br+1 − 1 , r+1 Since the limit of Amin and the limit of Amax exist and are equal, the Riemann integral exists and equals, � b xr dx = (br+1 − 1)/(r + 1). 1 3. The Fundamental Theorem of Calculus. There is a single theorem that it is at the heart of almost all applications involving Riemann integrals. The theorem answers two question simul­ taneously: Which functions are Riemann integrable? What is the Riemann integral of a function? The answer to the first question is: Every function you are likely to encounter is Riemann inte­ grable. Precisely, every continuous function, and every piecewise continuous function is Riemann integrable. The answer to the second question is more interesting. Assume f (x) is a continuous function. Let x = a be a fixed point where f (x) is defined. Form the function, � x F (x) = f (t)dt. a The function F (x) is defined whenever f (t) is defined on all of [a, x]. If f (x) is continuous, the Fundamental Theorem of Calculus asserts F (x) is differentiable and, � x dF d (x) = f (t)dt = f (x). dx dx a The proof of the second part is very easy. Consider the increment in F from x to x + Δx, � x � x+Δx � x+Δx F (x + Δx) − F (x) = f (t)dt − f (t)dt = f (t)dt. a a x Let ymin be the minimum value of f (t) on the interval [x, x + Δx]. Let ymax be the maximum value of f (t) on the interval [x, x + Δx]. Then for every choice of partition t0 < t1 < · · · < tn of [x, x + Δx], and every choice of values yk∗ on the subintervals, ymin ≤ yk∗ ≤ ymax , 18.01 Calculus Jason Starr Fall 2005 for every k. Thus the Riemann sum is squeezed between, n � ymin Δtk ≤ k=1 n � yk∗ Δtk k=1 ≤ n � ymax Δtk . k=1 Of course the lower bound is, n � ymin Δtk = ymin k=1 n � Δtk = ymin Δx, k=1 because the total length of the interval [x, x + Δx] is Δx. Similarly, the upper bound is, n � ymax Δtk = ymax Δx. k=1 Thus the Riemann sum is squeezed between, ymin Δx ≤ n � yk∗ Δxk ≤ ymax Δx. k=1 Because the Riemann integral is a limit of Riemann sums, it is also squeezed, � x+Δx ymin Δx ≤ f (t)dt ≤ ymax Δx. x Substituting in F (x + Δx) − F (x) and dividing each term by Δx gives, ymin ≤ F (x + Δx) − F (x) ≤ ymax . Δx The middle term is the difference quotient. Consider what happens as Δx tends to 0. Because f (t) is continuous, both the maximum and minimum values of f (t) on [x, x + Δx] simply limit to the value f (x). Thus, lim ymin = lim ymax = f (x). Δx Δx By the Squeezing Lemma for limits, since these two limits exist and are equals, the middle limit also exists and equals f (x), F (x + Δx) − F (x) lim = f (x). Δx→0 Δx This is precisely what the Fundamental Theorem of Calculus asserts, � x d f (t)dt = f (x) . dx a 18.01 Calculus Jason Starr Fall 2005 4. Algorithm for computing Riemann integrals. The Fundamental Theorem of Calculus has many important applications. The most obvious is to give us a simpler method for computing Riemann integrals, under the hypothesis that we can compute the antiderivative. If f (x) is a continuous function and G(x) is a known antiderivative of f (x), then, � b f (t)dt = G(b) − G(a). a To see this, observe that, � F (x) = x f (t)dt, a is also an antiderivative of f (t) by the Fundamental Theorem of Calculus. Thus, since the general antiderivative is G(x) + C, there is a constant C such that F (x) = G(x) + C. But also, � a F (a) = f (t)dt = 0. a Thus, F (x) = G(x) − G(a). Now plug in x = b to get, � b f (t)dt = F (b) = G(b) − G(a). a