Explanation Given: FTC II Key concepts applied: FTC II or The second Fundamental Theorem of Calculus deals with derivatives of definite integrals with respect to variable upper limits of integration . It states that if and f is continuous, then . Proof: Part (a) From FTC II, we have A ' ( x ) = f ( x ) . Therefore, A ( x ) and F ( x ) are both antiderivatives of f ( x ) . Therefore, it follows that F ( x ) = A ( x ) + C for some constant C . Part (b) Consider F ( b ) − F ( a ) ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 5.5, Problem 52E

To determine

To prove:

FTC I anew using FTC II,

a) Show that $F (x) = A (x) + C$ for some constant

b) Show that $F (b) - F (a) = A (b) - A (a) = \int_{a}^{b} f (t) d t$

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Chapter 5.5, Problem 51E

Chapter 5.5, Problem 53E

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Chapter 5 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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To prove: FTC I anew using FTC II, a) Show that F ( x ) = A ( x ) + C for some constant b) Show that F ( b ) − F ( a ) = A ( b ) − A ( a ) = ∫ a b f ( t ) d t

Solution Summary: The author illustrates how FTC II deals with derivatives of definite integrals with respect to variable upper limits of integration.

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To prove:

FTC I anew using FTC II,

a) Show that $F (x) = A (x) + C$ for some constant

b) Show that $F (b) - F (a) = A (b) - A (a) = \int_{a}^{b} f (t) d t$

Want to see the full answer?

Chapter 5 Solutions

To prove: FTC I anew using FTC II, a) Show that F ( x ) = A ( x ) + C for some constant b) Show that F ( b ) − F ( a ) = A ( b ) − A ( a ) = ∫ a b f ( t ) d t

Solution Summary: The author illustrates how FTC II deals with derivatives of definite integrals with respect to variable upper limits of integration.

Concept explainers

Videos

To prove: FTC I anew using FTC II, a) Show that F(x)=A(x)+C for some constant b) Show that F(b)−F(a)=A(b)−A(a)=∫abf(t)dt

Want to see the full answer?

Chapter 5 Solutions

To prove:

FTC I anew using FTC II,

a) Show that $F (x) = A (x) + C$ for some constant

b) Show that $F (b) - F (a) = A (b) - A (a) = \int_{a}^{b} f (t) d t$