Lesson 26: The Fundamental Theorem of Calculus (slides)
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The document discusses the Fundamental Theorem of Calculus, which has two parts. The first part states that if a function f is continuous, then the derivative of the integral of f is equal to f. This is proven using Riemann sums. The second part relates the integral of a function f to the anti-derivative F of f. Examples are provided to illustrate how to use the Fundamental Theorem to find derivatives and integrals.
![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 ].](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-5-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-6-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-7-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-12-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-15-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-18-320.jpg)
![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?](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-22-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-35-320.jpg)
![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.](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-36-320.jpg)
![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).](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-37-320.jpg)
![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].](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-40-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-41-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-42-320.jpg)
![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](https://image.slidesharecdn.com/lesson26-thefundamentaltheoremofcalculus001slides-110503222918-phpapp01-121002031400-phpapp01/85/Lesson-26-The-Fundamental-Theorem-of-Calculus-slides-43-320.jpg)
Lesson 26: The Fundamental Theorem of Calculus (slides)
- 1. Sec on 5.4 The Fundamental Theorem of Calculus V63.0121.001: Calculus I Professor Ma hew Leingang New York University May 2, 2011 .
- 2. 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
- 3. 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.
- 4. 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
- 5. 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 ].
- 6. 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
- 7. 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
- 8. The Integral as Net Change
- 9. 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
- 10. 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
- 11. 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
- 12. 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
- 13. 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
- 14. 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
- 15. 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
- 16. 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
- 17. 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
- 18. 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
- 19. 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
- 20. 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
- 21. 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
- 22. 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?
- 23. 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
- 24. 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
- 25. 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
- 26. 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
- 27. 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
- 28. 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
- 29. 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
- 30. 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
- 31. 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
- 32. 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
- 33. 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
- 34. 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
- 35. 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
- 36. 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.
- 37. 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).
- 38. Proving the Fundamental Theorem Proof. Let h > 0 be given so that x + h < b. We have g(x + h) − g(x) = h
- 39. 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
- 40. 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].
- 41. 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
- 42. 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
- 43. 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
- 44. 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).
- 45. 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
- 46. 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
- 47. 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
- 48. 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
- 49. 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
- 50. 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
- 51. 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
- 52. 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
- 53. 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
- 54. 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
- 55. 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
- 56. 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
- 57. 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
- 58. 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
- 59. 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
- 60. 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
- 61. 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
- 62. 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
- 63. 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
- 64. 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
- 65. 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
- 66. 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
- 67. 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
- 68. 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
- 69. 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
- 70. 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
- 71. 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
- 72. 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
- 73. 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 +
- 74. 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
- 75. 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
- 76. 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
- 77. 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
- 78. 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
- 79. 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
- 80. 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
- 81. 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
- 82. 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
- 83. 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
- 84. 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
- 85. 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
- 86. 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
- 87. 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
- 88. 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
- 89. 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
- 90. 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
- 91. 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
- 92. 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
- 93. 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
- 94. 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
- 95. 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
- 96. 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
- 97. 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
- 98. 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
- 99. 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
- 100. 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
- 101. 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
- 102. 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
- 103. 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
- 104. 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 +
- 105. 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
- 106. Meet the Mathematician James 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
- 107. Meet the Mathematician Isaac Barrow English, 1630-1677 Professor of Greek, theology, and mathema cs at Cambridge Had a famous student
- 108. Meet the Mathematician Isaac Newton English, 1643–1727 Professor at Cambridge (England) invented calculus 1665–66 Tractus de Quadratura Curvararum published 1704
- 109. Meet the Mathematician Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathema cian invented calculus 1672–1676 published in papers 1684 and 1686
- 110. Differentiation and Integration as reverse 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.
- 111. Differentiation and Integration as reverse 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.
- 112. 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
- 113. Differentiation of area functions Example ∫ 3x Let h(x) = t3 dt. What is h′ (x)? 0
- 114. 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
- 115. 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.
- 116. 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 .
- 117. 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)
- 118. Another Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 0
- 119. 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
- 120. 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
- 121. A Similar Example Example ∫ sin2 x Let h(x) = (17t2 + 4t − 4) dt. What is h′ (x)? 3
- 122. 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
- 123. 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?
- 124. 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.
- 125. The Full Nasty Example ∫ ex Find the deriva ve of F(x) = sin4 t dt. x3
- 126. 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
- 127. 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 .
- 128. 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.
- 129. 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.
- 130. 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.
- 131. Erf ∫ x 2 e−t dt 2 erf(x) = √ π 0
- 132. Erf ∫ x 2 e−t dt 2 erf(x) = √ π 0 erf measures area the bell curve. erf(x) . x
- 133. 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) =
- 134. 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
- 135. Example of erf Example d Find erf(x2 ). dx
- 136. 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
- 137. 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.
- 138. Surplus by picture price (p) . quan ty (q)
- 139. Surplus by picture price (p) demand f(q) . quan ty (q)
- 140. Surplus by picture price (p) supply demand f(q) . quan ty (q)
- 141. Surplus by picture price (p) supply p∗ equilibrium demand f(q) . q∗ quan ty (q)
- 142. Surplus by picture price (p) supply p∗ equilibrium market revenue demand f(q) . q∗ quan ty (q)
- 143. Surplus by picture consumer surplus price (p) supply p∗ equilibrium market revenue demand f(q) . q∗ quan ty (q)
- 144. Surplus by picture consumer surplus price (p) producer surplus supply p∗ equilibrium demand f(q) . q∗ quan ty (q)
- 145. 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