Transcript of "Lesson 26: The Fundamental Theorem of Calculus (slides)"
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Sec on 5.4 The Fundamental Theorem of Calculus V63.0121.001: Calculus I Professor Ma hew Leingang New York University May 2, 2011.
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Announcements Today: 5.4 Wednesday 5/4: 5.5 Monday 5/9: Review and Movie Day! Thursday 5/12: Final Exam, 2:00–3:50pm
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Objectives State and explain the Fundamental Theorems of Calculus Use the ﬁrst fundamental theorem of calculus to ﬁnd deriva ves of func ons deﬁned as integrals. Compute the average value of an integrable func on over a closed interval.
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Outline Recall: The Evalua on Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Func on Statement and proof of 1FTC Biographies Diﬀeren a on of func ons deﬁned by integrals “Contrived” examples Erf Other applica ons
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The deﬁnite integral as a limit Deﬁni on If f is a func on 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 n→∞ i=1 b−a where ∆x = , and for each i, xi = a + i∆x, and ci is a point in n [xi−1 , xi ].
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The deﬁnite integral as a limit Theorem If f is con nuous on [a, b] or if f has only ﬁnitely many jump discon nui es, then f is integrable on [a, b]; that is, the deﬁnite ∫ b integral f(x) dx exists and is the same for any choice of ci . a
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Big time Theorem Theorem (The Second Fundamental Theorem of Calculus) Suppose f is integrable on [a, b] and f = F′ for another func on F, then ∫ b f(x) dx = F(b) − F(a). a
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The Integral as Net Change Corollary If v(t) represents the velocity of a par cle moving rec linearly, then ∫ t1 v(t) dt = s(t1 ) − s(t0 ). t0
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The Integral as Net Change Corollary If MC(x) represents the marginal cost of making x units of a product, then ∫ x C(x) = C(0) + MC(q) dq. 0
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The Integral as Net Change Corollary 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
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My ﬁrst table of integrals . ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx ∫ ∫ ∫ xn+1 xn dx = + C (n ̸= −1) cf(x) dx = c f(x) dx ∫ n+1 ∫ 1 ex dx = ex + C dx = ln |x| + C ∫ ∫ x ax sin x dx = − cos x + C ax 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
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Outline Recall: The Evalua on Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Func on Statement and proof of 1FTC Biographies Diﬀeren a on of func ons deﬁned by integrals “Contrived” examples Erf Other applica ons
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x Dividing the interval [0, x] into n pieces gives ∆t = and n ix ti = 0 + i∆t = . n . 0 x
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x x3 x (2x)3 x (nx)3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n . 0 x
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x x3 x (2x)3 x (nx)3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 4 ( ) x = 4 13 + 23 + 33 + · · · + n3 . n 0 x
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x x3 x (2x)3 x (nx)3 Rn = · 3 + · 3 + ··· + · 3 n n n n n n 4 ( ) x x4 [ 1 ]2 = 4 1 + 2 + 3 + · · · + n = 4 2 n(n + 1) 3 3 3 3 . n n 0 x
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x4 n2 (n + 1)2 Rn = 4n4 . 0 x
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x4 n2 (n + 1)2 Rn = 4n4 x4 . So g(x) = lim Rn = x x→∞ 4 0
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Area as a Function Example ∫ x 3 Let f(t) = t and deﬁne g(x) = f(t) dt. Find g(x) and g′ (x). 0 Solu on x4 n2 (n + 1)2 Rn = 4n4 x4 . So g(x) = lim Rn = and g′ (x) = x3 . x x→∞ 4 0
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The area function in general Let f be a func on which is integrable (i.e., con nuous or with ﬁnitely many jump discon nui es) 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. Ques on: What does f tell you about g?
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Envisioning the area function Example Suppose f(t) is the func on y graphed to the right. Let ∫ x g(x) = f(t) dt 0 g . x What can you say about g? 2 4 6 8 10f
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Features of g from f Interval sign monotonicity monotonicity concavity y of f of g of f of g [0, 2] + ↗ ↗ ⌣ g [2, 4.5] + ↗ ↘ ⌢ . fx 2 4 6 810 [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → none
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Features of g from f Interval sign monotonicity monotonicity concavity y of f of g of f of g [0, 2] + ↗ ↗ ⌣ g [2, 4.5] + ↗ ↘ ⌢ . fx 2 4 6 810 [4.5, 6] − ↘ ↘ ⌢ [6, 8] − ↘ ↗ ⌣ [8, 10] − ↘ → none We see that g is behaving a lot like an an deriva ve of f.
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Another Big Time Theorem Theorem (The First Fundamental Theorem of Calculus) Let f be an integrable func on on [a, b] and deﬁne ∫ x g(x) = f(t) dt. a If f is con nuous at x in (a, b), then g is diﬀeren able at x and g′ (x) = f(x).
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Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h
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Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = f(t) dt. h h x
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Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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].
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Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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 f(t) dt x
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Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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 f(t) dt ≤ Mh · h x
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Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have ∫ g(x + h) − g(x) 1 x+h = 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
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Proving the Fundamental Theorem Proof. From §5.2 we have ∫ x+h mh · h ≤ f(t) dt ≤ Mh · h x g(x + h) − g(x) =⇒ mh ≤ ≤ Mh . h As h → 0, both mh and Mh tend to f(x).
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About Mh and mh Mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, Mh and x + h → x, we have lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, Mh and x + h → x, we have lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have Mh lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have Mh lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have Mh lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) Mh h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) Mh h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx + h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx + h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x +h x
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx+ h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx h +
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x+h x
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 This is not necessarily true when f is not con nuous. . x
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About Mh and mh Since f is con nuous at x, Mh and x + h → x, we have lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have Mh lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have Mh lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have Mh lim Mh = f(x) h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) Mh h→0 lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx + h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx + h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x +h x
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x+h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx+ h
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . xx h +
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About Mh and mh Since f is con nuous at x, and x + h → x, we have lim Mh = f(x) h→0 Mh lim mh = f(x) f(x) h→0 mh . x+h x
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Meet the MathematicianJames Gregory Sco sh, 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
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Meet the MathematicianIsaac Barrow English, 1630-1677 Professor of Greek, theology, and mathema cs at Cambridge Had a famous student
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Meet the MathematicianIsaac Newton English, 1643–1727 Professor at Cambridge (England) invented calculus 1665–66 Tractus de Quadratura Curvararum published 1704
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Meet the MathematicianGottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathema cian invented calculus 1672–1676 published in papers 1684 and 1686
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Diﬀerentiation and Integration asreverse processes Pu ng together 1FTC and 2FTC, we get a beau ful rela onship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) I. If f is a con nuous func on, then ∫ d x f(t) dt = f(x) dx a So the deriva ve of the integral is the original func on.
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Diﬀerentiation and Integration asreverse processes Pu ng together 1FTC and 2FTC, we get a beau ful rela onship between the two fundamental concepts in calculus. Theorem (The Fundamental Theorem(s) of Calculus) II. If f is a diﬀeren able func on, then ∫ b f′ (x) dx = f(b) − f(a). a So the integral of the deriva ve of is (an evalua on of) the original func on.
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Outline Recall: The Evalua on Theorem a/k/a 2nd FTC The First Fundamental Theorem of Calculus Area as a Func on Statement and proof of 1FTC Biographies Diﬀeren a on of func ons deﬁned by integrals “Contrived” examples Erf Other applica ons
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Diﬀerentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0
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Diﬀerentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solu on (Using 2FTC) 3x t4 1 h(x) = = (3x)4 = 1 4 · 81x4 , so h′ (x) = 81x3 . 4 0 4
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Diﬀerentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solu on (Using 1FTC) ∫ u We can think of h as the composi on g k, where g(u) = ◦ t3 dt 0 and k(x) = 3x.
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Diﬀerentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0 Solu on (Using 1FTC) ∫ u We can think of h as the composi on g k, where g(u) = ◦ t3 dt 0 and 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 = 81x3 .
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Diﬀerentiation of area functions, in general by 1FTC ∫ d k(x) f(t) dt = f(k(x))k′ (x) dx a by reversing the order of integra on: ∫ ∫ d b d h(x) f(t) dt = − f(t) dt = −f(h(x))h′ (x) dx h(x) dx b by combining the two above: ∫ (∫ ∫ 0 ) d k(x) d k(x) 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)
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Another Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0
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Another Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 Solu on (2FTC) sin2 x 17 3 17 h(x) = t + 2t2 − 4t = (sin2 x)3 + 2(sin2 x)2 − 4(sin2 x) 3 3 3 17 6 = sin x + 2 sin4 x − 4 sin2 x (3 ) ′ 17 · 6 5 h (x) = sin x + 2 · 4 sin x − 4 · 2 sin x cos x 3 3
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Another Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0 Solu on (1FTC) ∫ d sin2 x ( ) d (17t2 + 4t − 4) dt = 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x dx dx 0 ( ) = 17 sin x + 4 sin x − 4 · 2 sin x cos x 4 2
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A Similar Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3
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A Similar Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3 Solu on We have ∫ 2 d sin x ( ) d (17t2 + 4t − 4) dt = 17(sin2 x)2 + 4(sin2 x) − 4 · sin2 x dx 3 dx ( ) = 17 sin x + 4 sin x − 4 · 2 sin x cos x 4 2
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Compare Ques on ∫ 2 ∫ 2 d sin x d sin x Why is (17t + 4t − 4) dt = 2 (17t2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the deriva ve?
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Compare Ques on ∫ 2 ∫ 2 d sin x d sin x Why is (17t + 4t − 4) dt = 2 (17t2 + 4t − 4) dt? dx 0 dx 3 Or, why doesn’t the lower limit appear in the deriva ve? Answer ∫ sin2 x ∫ 3 ∫ sin2 x 2 2 (17t +4t−4) dt = (17t +4t−4) dt+ (17t2 +4t−4) dt 0 0 3 So the two func ons diﬀer by a constant.
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The Full Nasty Example ∫ ex Find the deriva ve of F(x) = sin4 t dt. x3
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The Full Nasty Example ∫ ex Find the deriva ve of F(x) = sin4 t dt. x3 Solu on ∫ ex d sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2 dx x3
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The Full Nasty Example ∫ ex Find the deriva ve of F(x) = sin4 t dt. x3 Solu on ∫ ex d sin4 t dt = sin4 (ex ) · ex − sin4 (x3 ) · 3x2 dx x3 No ce here it’s much easier than ﬁnding an an deriva ve for sin4 .
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Why use 1FTC? Ques on Why would we use 1FTC to ﬁnd the deriva ve of an integral? It seems like confusion for its own sake.
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Why use 1FTC? Ques on Why would we use 1FTC to ﬁnd the deriva ve of an integral? It seems like confusion for its own sake. Answer Some func ons are diﬃcult or impossible to integrate in elementary terms.
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Why use 1FTC? Ques on Why would we use 1FTC to ﬁnd the deriva ve of an integral? It seems like confusion for its own sake. Answer Some func ons are diﬃcult or impossible to integrate in elementary terms. Some func ons are naturally deﬁned in terms of other integrals.
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Erf ∫ x 2 e−t dt 2 erf(x) = √ π 0 erf measures area the bell curve. erf(x) . x
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Erf ∫ x 2 e−t dt 2 erf(x) = √ π 0 erf measures area the bell curve. erf(x) We can’t ﬁnd erf(x), explicitly, but we do know its deriva ve: . x erf′ (x) =
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Erf ∫ x 2 e−t dt 2 erf(x) = √ π 0 erf measures area the bell curve. erf(x) We can’t ﬁnd erf(x), explicitly, but we do know its deriva ve: . 2 x erf′ (x) = √ e−x . 2 π
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Example of erf Example d Find erf(x2 ). dx Solu on By the chain rule we have d d 2 4 erf(x2 ) = erf′ (x2 ) x2 = √ e−(x ) 2x = √ xe−x . 2 2 4 dx dx π π
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Other functions deﬁned by integrals The future value of an asset: ∫ ∞ FV(t) = π(s)e−rs ds t where π(s) is the proﬁtability at me 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 func on and p∗ and q∗ the equilibrium price and quan ty.
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Surplus by picture price (p) demand f(q) . quan ty (q)
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Surplus by picture price (p) supply demand f(q) . quan ty (q)
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Surplus by picture price (p) supply p∗ equilibrium demand f(q) . q∗ quan ty (q)
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Surplus by picture price (p) supply p∗ equilibrium market revenue demand f(q) . q∗ quan ty (q)
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Surplus by picture consumer surplus price (p) supply p∗ equilibrium market revenue demand f(q) . q∗ quan ty (q)
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Surplus by picture consumer surplus price (p) producer surplus supply p∗ equilibrium demand f(q) . q∗ quan ty (q)
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Summary Func ons deﬁned as integrals can be diﬀeren ated using the ﬁrst FTC: ∫ d x f(t) dt = f(x) dx a The two FTCs link the two major processes in calculus: diﬀeren a on and integra on ∫ F′ (x) dx = F(x) + C
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