Fundamental Theorem Of Calculus (Second Form) - B&S Theorem 7.3.5 .... ....

In summary: Theorem 7.3.5 states that the integral of a function f on an interval is bounded by the values of the function at the endpoints of the interval multiplied by the length of the interval. In this case, since f(c) - ε is less than f(x) for all x in c ≤ x < c + h, the integral must also be less than or equal to (f(c) - ε) * h. Therefore, (f(c) - ε) * h is a lower bound for the integral, which is why it appears on the left side of the inequality. The same reasoning applies for the upper bound, which is why (f(c) + ε) * h appears on the right side.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 7: The Riemann Integral ...

I need help in fully understanding an aspect of the proof of Theorem 7.3.5 ...Theorem 7.3.5 and its proof ... ... read as follows:
View attachment 7325In the above proof from Bartle and Sherbert we read the following:

" ... ... Now on the interval \(\displaystyle [c, c + h]\) the function \(\displaystyle f\) satisfies inequality (4), so that we have

\(\displaystyle ( f(c) - \epsilon ) \cdot h \le F( c + h ) - F(c) = \int^{ c + h }_c f \le ( f(c) + \epsilon ) \cdot h\)

... ... "Can someone please demonstrate rigorously and in detail how Bartle and Sherbert arrived at

\(\displaystyle ( f(c) - \epsilon ) \cdot h \le F( c + h ) - F(c) = \int^{ c + h }_c f \le ( f(c) + \epsilon ) \cdot h\) ... ... ?Peter================================================================================

It may help readers of the above post to have access to B&S's definition of the indefinite integral of \(\displaystyle f\) ... ... so I am providing the same ... ... as follows:View attachment 7326
 
Physics news on Phys.org
  • #2
Peter said:
I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 7: The Riemann Integral ...

I need help in fully understanding an aspect of the proof of Theorem 7.3.5 ...Theorem 7.3.5 and its proof ... ... read as follows:
In the above proof from Bartle and Sherbert we read the following:

" ... ... Now on the interval \(\displaystyle [c, c + h]\) the function \(\displaystyle f\) satisfies inequality (4), so that we have

\(\displaystyle ( f(c) - \epsilon ) \cdot h \le F( c + h ) - F(c) = \int^{ c + h }_c f \le ( f(c) + \epsilon ) \cdot h\)

... ... "Can someone please demonstrate rigorously and in detail how Bartle and Sherbert arrived at

\(\displaystyle ( f(c) - \epsilon ) \cdot h \le F( c + h ) - F(c) = \int^{ c + h }_c f \le ( f(c) + \epsilon ) \cdot h\) ... ... ?Peter================================================================================

It may help readers of the above post to have access to B&S's definition of the indefinite integral of \(\displaystyle f\) ... ... so I am providing the same ... ... as follows:
On reflection I think that the explanation for my question is as follows:

Since \(\displaystyle f(c) - \epsilon\) is less than \(\displaystyle f(x)\) for all \(\displaystyle x\) in \(\displaystyle c \le x \lt c + h\) ... ... ... we have that \(\displaystyle ( f(c) - \epsilon ) \cdot h \le \int^{ c + h }_c f\) ... ...Is that basically the correct explanation ... ... ?Peter
 
  • #3
Peter said:
On reflection I think that the explanation for my question is as follows:

Since \(\displaystyle f(c) - \epsilon\) is less than \(\displaystyle f(x)\) for all \(\displaystyle x\) in \(\displaystyle c \le x \lt c + h\) ... ... ... we have that \(\displaystyle ( f(c) - \epsilon ) \cdot h \le \int^{ c + h }_c f\) ... ...Is that basically the correct explanation ... ... ?
Yes. :)
 

Related to Fundamental Theorem Of Calculus (Second Form) - B&S Theorem 7.3.5 .... ....

What is the Fundamental Theorem of Calculus (Second Form)?

The Fundamental Theorem of Calculus (Second Form) is a mathematical theorem that establishes the relationship between differentiation and integration. It states that if a function f is continuous on an interval [a,b] and F is any function that satisfies F'(x) = f(x), then the definite integral of f from a to b is equal to F(b) - F(a).

What is the significance of the Fundamental Theorem of Calculus (Second Form)?

The significance of the Fundamental Theorem of Calculus (Second Form) lies in its application to finding the area under a curve. It allows us to evaluate definite integrals without having to use the limit definition of integration, making calculations much easier and more efficient.

How is the Fundamental Theorem of Calculus (Second Form) related to the First Form?

The Fundamental Theorem of Calculus (Second Form) is essentially an extension of the First Form. The First Form states that integration can be used to find the area under a curve, while the Second Form adds the notion of a function's anti-derivative to the equation. This allows us to solve definite integrals in a more general sense.

What are the conditions for the Fundamental Theorem of Calculus (Second Form) to hold?

Theorem 7.3.5, or the Fundamental Theorem of Calculus (Second Form), holds if the function f is continuous on the interval [a,b] and F is an anti-derivative of f on that same interval. Additionally, both a and b must be within the domain of F.

How can the Fundamental Theorem of Calculus (Second Form) be applied to real-life situations?

The Fundamental Theorem of Calculus (Second Form) has many real-world applications, particularly in the fields of physics and engineering. It allows us to calculate quantities such as displacement, velocity, and acceleration from position-time graphs, and is also used in economic and financial analysis to calculate total revenue and profit.

Similar threads

Replies
2
Views
1K
  • Topology and Analysis
Replies
3
Views
1K
  • Topology and Analysis
Replies
2
Views
1K
  • Topology and Analysis
Replies
5
Views
2K
Replies
2
Views
1K
  • Topology and Analysis
Replies
8
Views
4K
  • Topology and Analysis
Replies
3
Views
3K
Replies
3
Views
2K
Replies
2
Views
2K
  • Topology and Analysis
Replies
2
Views
1K
Back
Top