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exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative. 149.              x (d/dx) ∫ e^(cos t) dt             1

exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative. 149.              x (d/dx) ∫ e^(cos t) dt             1

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 80E
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exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

149. 

            x

(d/dx) ∫ e^(cos t) dt

            1

O The Most Critical Unpopular Op x b Answered: now | bartleby = 5.3 The Fundamental Theorem O Course Modules: ITAL V01 - Ele O Course Modules: MATH R120 + A https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus#fs-id1170571704350 < Calculus Volume 1 5.3 The Fundamental Theorem of Calculus E Table of contents Search this book 9 My highlights B Print Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, F (x), as the definite integral of another function, f (t), from the point a to the point x. At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it's a function. The key here is to notice that for any particular value of x, the definite integral is a number. So the function F (x) returns a number (the value of the definite integral) for each value of x. Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative. Proof Applying the definition of the derivative, we have F'(x) = lim F(z+h)–F(z) h h0 ra+h = lim 1 Af (t) dt – • x+h = lim 1 h+0 h f (t) dt + M I 11:04
Transcribed Image Text:O The Most Critical Unpopular Op x b Answered: now | bartleby = 5.3 The Fundamental Theorem O Course Modules: ITAL V01 - Ele O Course Modules: MATH R120 + A https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus#fs-id1170571704350 < Calculus Volume 1 5.3 The Fundamental Theorem of Calculus E Table of contents Search this book 9 My highlights B Print Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, F (x), as the definite integral of another function, f (t), from the point a to the point x. At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it's a function. The key here is to notice that for any particular value of x, the definite integral is a number. So the function F (x) returns a number (the value of the definite integral) for each value of x. Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative. Proof Applying the definition of the derivative, we have F'(x) = lim F(z+h)–F(z) h h0 ra+h = lim 1 Af (t) dt – • x+h = lim 1 h+0 h f (t) dt + M I 11:04
O The Most Critical Unpopular Op x b Answered: now | bartleby = 5.3 The Fundamental Theorem O Course Modules: ITAL V01 - Ele O Course Modules: MATH R120 + A https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus#fs-id1170571704350 < Calculus Volume 1 5.3 The Fundamental Theorem of Calculus E Table of contents Search this book 9 My highlights B Print nU LJa Ja z+h = lim ! h0 h f (t) dt + = lim 1 h+0 I+h Looking carefully at this last expression, we see f (t) dt is just the average value of the function f (x) over the interval [x, x + h]. Therefore, by The Mean Value Theorem for Integrals, there is some number c in [x, x + h] such that 1 f (2) da = f (e). In addition, since c is between x and x + h, c approaches x as h approaches zero. Also, since f (x) is continuous, we have limf (c) = lim f (c) = f (x). Putting all these pieces together, we have h-0" I+h F'(x) = lim 1 h0 h limf (c) h-0 = f (x), and the proof is complete. M I 11:03
Transcribed Image Text:O The Most Critical Unpopular Op x b Answered: now | bartleby = 5.3 The Fundamental Theorem O Course Modules: ITAL V01 - Ele O Course Modules: MATH R120 + A https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus#fs-id1170571704350 < Calculus Volume 1 5.3 The Fundamental Theorem of Calculus E Table of contents Search this book 9 My highlights B Print nU LJa Ja z+h = lim ! h0 h f (t) dt + = lim 1 h+0 I+h Looking carefully at this last expression, we see f (t) dt is just the average value of the function f (x) over the interval [x, x + h]. Therefore, by The Mean Value Theorem for Integrals, there is some number c in [x, x + h] such that 1 f (2) da = f (e). In addition, since c is between x and x + h, c approaches x as h approaches zero. Also, since f (x) is continuous, we have limf (c) = lim f (c) = f (x). Putting all these pieces together, we have h-0" I+h F'(x) = lim 1 h0 h limf (c) h-0 = f (x), and the proof is complete. M I 11:03
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