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Lesson 26: The Fundamental Theorem of Calculus (Section 021 handout)
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Lesson 26: The Fundamental Theorem of Calculus (Section 021 handout)

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The fundamental theorem shows that differentiation and integration are inverse processes. …

The fundamental theorem shows that differentiation and integration are inverse processes.

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  • 1. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Section 5.4 Notes The Fundamental Theorem of Calculus V63.0121.021, Calculus I New York University December 9, 2010 Announcements Today: Section 5.4 ”Thursday,” December 14: Section 5.5 ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam (location still TBD) Announcements Notes Today: Section 5.4 ”Thursday,” December 14: Section 5.5 ”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and Movie Day! Monday, December 20, 12:00–1:50pm: Final Exam (location still TBD) V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 2 / 32 Objectives Notes State and explain the Fundemental Theorems of Calculus Use the first fundamental theorem of calculus to find derivatives of functions defined as integrals. Compute the average value of an integrable function over a closed interval. V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 3 / 32 1
  • 2. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Outline Notes Recall: The Evaluation Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Function Statement and proof of 1FTC Biographies Differentiation of functions defined by integrals “Contrived” examples Erf Other applications V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 4 / 32 The definite integral as a limit Notes Definition If f is a function defined on [a, b], the definite integral of f from a to b is the number b n f (x) dx = lim f (ci ) ∆x a ∆x→0 i=1 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 5 / 32 Big time Theorem Notes Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F for another function F , then b f (x) dx = F (b) − F (a). a V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 6 / 32 2
  • 3. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 The Integral as Total Change Notes Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If v (t) represents the velocity of a particle moving rectilinearly, then t1 v (t) dt = s(t1 ) − s(t0 ). t0 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32 The Integral as Total Change Notes Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If MC (x) represents the marginal cost of making x units of a product, then x C (x) = C (0) + MC (q) dq. 0 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32 The Integral as Total Change Notes Another way to state this theorem is: b F (x) dx = F (b) − F (a), a or the integral of a derivative along an interval is the total change between the sides of that interval. This has many ramifications: Theorem If ρ(x) represents the density of a thin rod at a distance of x from its end, then the mass of the rod up to x is x m(x) = ρ(s) ds. 0 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 7 / 32 3
  • 4. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 My first table of integrals Notes [f (x) + g (x)] dx = f (x) dx + g (x) dx x n+1 x n dx = + C (n = −1) cf (x) dx = c f (x) dx n+1 1 e x dx = e x + C dx = ln |x| + C x x ax sin x dx = − cos x + C a dx = +C ln a cos x dx = sin x + C csc2 x dx = − cot x + C sec2 x dx = tan x + C csc x cot x dx = − csc x + C 1 sec x tan x dx = sec x + C √ dx = arcsin x + C 1 − x2 1 dx = arctan x + C 1 + x2 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 8 / 32 Outline Notes Recall: The Evaluation Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Function Statement and proof of 1FTC Biographies Differentiation of functions defined by integrals “Contrived” examples Erf Other applications V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 9 / 32 Area as a Function Notes Example x Let f (t) = t 3 and define g (x) = f (t) dt. Find g (x) and g (x). 0 Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x x 3 x (2x)3 x (nx)3 Rn = · + · + ··· + · n n3 n n3 n n3 x4 3 3 3 3 = 4 1 + 2 + 3 + ··· + n n x x4 1 2 0 = 4 2 n(n + 1) n V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32 4
  • 5. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Area as a Function Notes Example x Let f (t) = t 3 and define g (x) = f (t) dt. Find g (x) and g (x). 0 Solution Dividing the interval [0, x] into n pieces x ix gives ∆t = and ti = 0 + i∆t = . So n n x 4 n2 (n + 1)2 Rn = 4n4 x4 So g (x) = lim Rn = and g (x) = x 3 . x→∞ 4 0 x V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 10 / 32 The area function in general Notes Let f be a function which is integrable (i.e., continuous or with finitely many jump discontinuities) on [a, b]. Define x g (x) = f (t) dt. a The variable is x; t is a “dummy” variable that’s integrated over. Picture changing x and taking more of less of the region under the curve. Question: What does f tell you about g ? V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 11 / 32 Envisioning the area function Notes Example Suppose f (t) is the function graphed below: y x 2 4 6 8 10f x Let g (x) = f (t) dt. What can you say about g ? 0 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 12 / 32 5
  • 6. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 features of g from f Notes y Interval sign monotonicity monotonicity concavity of f of g of f of g g [0, 2] + x 2 4 6 8 10f [2, 4.5] + [4.5, 6] − [6, 8] − [8, 10] − → none We see that g is behaving a lot like an antiderivative of f . V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 13 / 32 Another Big Time Theorem Notes Theorem (The First Fundamental Theorem of Calculus) Let f be an integrable function on [a, b] and define x g (x) = f (t) dt. a If f is continuous at x in (a, b), then g is differentiable at x and g (x) = f (x). V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 14 / 32 Proving the Fundamental Theorem Notes Proof. Let h > 0 be given so that x + h < b. We have x+h g (x + h) − g (x) 1 = f (t) dt. h h x Let Mh be the maximum value of f on [x, x + h], and let mh the minimum value of f on [x, x + h]. From §5.2 we have x+h mh · h ≤ f (t) dt ≤ Mh · h x So g (x + h) − g (x) mh ≤ ≤ Mh . h As h → 0, both mh and Mh tend to f (x). V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 15 / 32 6
  • 7. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Meet the Mathematician: James Gregory Notes Scottish, 1638-1675 Astronomer and Geometer Conceived transcendental numbers and found evidence that π was transcendental Proved a geometric version of 1FTC as a lemma but didn’t take it further V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 16 / 32 Meet the Mathematician: Isaac Barrow Notes English, 1630-1677 Professor of Greek, theology, and mathematics at Cambridge Had a famous student V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 17 / 32 Meet the Mathematician: Isaac Newton Notes English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687 V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 18 / 32 7
  • 8. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Meet the Mathematician: Gottfried Leibniz Notes German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 19 / 32 Differentiation and Integration as reverse processes Notes Putting together 1FTC and 2FTC, we get a beautiful relationship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) I. If f is a continuous function, then x d f (t) dt = f (x) dx a So the derivative of the integral is the original function. II. If f is a differentiable function, then b f (x) dx = f (b) − f (a). a So the integral of the derivative of is (an evaluation of) the original function. V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 20 / 32 Outline Notes Recall: The Evaluation Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Function Statement and proof of 1FTC Biographies Differentiation of functions defined by integrals “Contrived” examples Erf Other applications V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 21 / 32 8
  • 9. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Differentiation of area functions Notes Example 3x Let h(x) = t 3 dt. What is h (x)? 0 Solution (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x 4 , so h (x) = 81x 3 . 4 0 4 Solution (Using 1FTC) u We can think of h as the composition g ◦ k, where g (u) = t 3 dt and 0 k(x) = 3x. Then h (x) = g (u) · k (x), or h (x) = g (k(x)) · k (x) = (k(x))3 · 3 = (3x)3 · 3 = 81x 3 . V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 22 / 32 Differentiation of area functions, in general Notes by 1FTC k(x) d f (t) dt = f (k(x))k (x) dx a by reversing the order of integration: b h(x) d d f (t) dt = − f (t) dt = −f (h(x))h (x) dx h(x) dx b by combining the two above: k(x) k(x) 0 d d f (t) dt = f (t) dt + f (t) dt dx h(x) dx 0 h(x) = f (k(x))k (x) − f (h(x))h (x) V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 23 / 32 Another Example Notes Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 0 Solution V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 24 / 32 9
  • 10. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 A Similar Example Notes Example sin2 x Let h(x) = (17t 2 + 4t − 4) dt. What is h (x)? 3 Solution V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 25 / 32 Compare Notes Question Why is sin2 x sin2 x d d (17t 2 + 4t − 4) dt = (17t 2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the derivative? Answer V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 26 / 32 The Full Nasty Notes Example ex Find the derivative of F (x) = sin4 t dt. x3 Solution Notice here it’s much easier than finding an antiderivative for sin4 . V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 27 / 32 10
  • 11. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Why use 1FTC? Notes Question Why would we use 1FTC to find the derivative of an integral? It seems like confusion for its own sake. Answer Some functions are difficult or impossible to integrate in elementary terms. Some functions are naturally defined in terms of other integrals. V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 28 / 32 Erf Notes Here’s a function with a funny name but an important role: x 2 2 erf(x) = √ e −t dt. π 0 It turns out erf is the shape of the bell curve. We can’t find erf(x), 2 2 explicitly, but we do know its derivative: erf (x) = √ e −x . π Example d Find erf(x 2 ). dx Solution V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 29 / 32 Other functions defined by integrals Notes The future value of an asset: ∞ FV (t) = π(s)e −rs ds t where π(s) is the profitability at time s and r is the discount rate. The consumer surplus of a good: q∗ CS(q ∗ ) = (f (q) − p ∗ ) dq 0 where f (q) is the demand function and p ∗ and q ∗ the equilibrium price and quantity. V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 30 / 32 11
  • 12. V63.0121.021, Calculus I Section 5.4 : The Fundamental Theorem December 9, 2010 Surplus by picture Notes consumer surplus price (p) producer surplus supply p∗ equilibrium market revenue demand f (q) q∗ quantity (q) V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 31 / 32 Summary Notes Functions defined as integrals can be differentiated using the first FTC: x d f (t) dt = f (x) dx a The two FTCs link the two major processes in calculus: differentiation and integration F (x) dx = F (x) + C Follow the calculus wars on twitter: #calcwars V63.0121.021, Calculus I (NYU) Section 5.4 The Fundamental Theorem December 9, 2010 32 / 32 Notes 12