Question

Use Theorem 3.5 .6 and Exercise 3.5 .3 to show that if \(f\) and \(g\) are integrable on \([a, b],\) then so are \(f+g\) and \(f g\).

Short answer

#Question# Prove that if two functions, f and g, are integrable on the interval [a, b], then their sum, f+g, and their product, fg, are also integrable on [a, b]. Use Theorem 3.5.6 and Exercise 3.5.3. #Answer# To prove that both the sum, f+g, and the product, fg, are integrable on the interval [a, b], we can use Theorem 3.5.6 and Exercise 3.5.3: 1. Theorem 3.5.6 states: Let f and g be integrable functions on [a, b]. Then, the following properties hold: a. The sum, f+g, is integrable on [a, b]. b. The product, fg, is integrable on [a, b]. 2. Exercise 3.5.3 states: A function h is Riemann integrable on [a, b] if and only if, for all ε > 0, there exists a partition P of [a, b] such that U(h, P) - L(h, P) < ε, where U(h, P) and L(h, P) are the upper and lower Riemann sums of the function h with respect to the partition P. Using these results, we can show that the sum, f+g, is integrable on [a, b] by showing that for all ε > 0, there exists a partition P such that U(f+g, P) - L(f+g, P) < ε. Similarly, we can show that the product, fg, is integrable on [a, b] by showing that for all ε > 0, there exists a partition P such that U(fg, P) - L(fg, P) < ε.

Step-by-step solution

Theorem 3.5.6 and Exercise 3.5.3

Before diving into the main proof, let's recall Theorem 3.5.6 and Exercise 3.5.3. Theorem 3.5.6 states: Let f and g be integrable functions on [a, b]. Then, the following properties hold: 1. The sum, f+g, is integrable on [a, b]. 2. The product, fg, is integrable on [a, b]. 3. A constant times a function, cf, is integrable on [a, b]. Exercise 3.5.3 states: A function h is Riemann integrable on [a, b] if and only if, for all ε > 0, there exists a partition P of [a, b] such that U(h, P) - L(h, P) < ε, where U(h, P) and L(h, P) are the upper and lower Riemann sums of the function h with respect to the partition P. Now let's prove that both the sum, f+g, and the product, fg, are integrable.

Prove f + g is integrable

To show that f+g is integrable, we have to show that for all ε > 0, there exists a partition P of [a, b] such that the upper and lower Riemann sums of f+g with respect to P satisfy U(f+g, P) - L(f+g, P) < ε. 1. Since f and g are integrable, by Exercise 3.5.3, for ε/2 > 0, there exist partitions P1 and P2 of [a, b] such that U(f, P1) - L(f, P1) < ε/2 and U(g, P2) - L(g, P2) < ε/2. 2. Let P be the common refinement of P1 and P2 (combine the points of both partitions into one partition without repetition). 3. Then, using the properties of upper and lower sums, we have: U(f+g, P) - L(f+g, P) = (U(f, P) + U(g, P)) - (L(f, P) + L(g, P)) ≤ (U(f, P1) - L(f, P1)) + (U(g, P2) - L(g, P2)) < ε/2 + ε/2 = ε. Since for all ε > 0, there exists a partition P such that U(f+g, P) - L(f+g, P) < ε, we can conclude that f+g is integrable on [a, b], which verifies the first part of Theorem 3.5.6.

Prove fg is integrable

To show that the product fg is integrable, we'll again use Exercise 3.5.3. We need to show that for all ε > 0, there exists a partition P of [a, b] such that U(fg, P) - L(fg, P) < ε. 1. Start by noting that the functions |f| and |g| are also integrable since their absolute values are continuous and the interval [a, b] is compact (closed and bounded). 2. By the properties of integrable functions, there exists a constant M such that |f(x)g(x)| ≤ M for all x in [a, b]. 3. Since f and g are integrable, by Exercise 3.5.3, for ε/M > 0, there exist partitions P1 and P2 of [a, b] such that U(f, P1) - L(f, P1) < ε/(2M) and U(g, P2) - L(g, P2) < ε/(2M). 4. Let P be the common refinement of P1 and P2 (combine the points of both partitions into one partition without repetition). 5. Then, applying the properties of upper and lower sums, we have U(fg, P) - L(fg, P) ≤ M(U(f, P1) - L(f, P1) + U(g, P2) - L(g, P2)) < ε/2 + ε/2 = ε. Since for all ε > 0, there exists a partition P such that U(fg, P) - L(fg, P) < ε, we can conclude that fg is integrable on [a, b], which verifies the second part of Theorem 3.5.6.

Key concepts

Integrable Functions

Integrable functions are those which can be measured using integration. This is a key aspect in calculus because it allows us to calculate areas under curves. Let's break this down a bit more:

• An "integrable function" over an interval \([a, b]\), means we can find its integral or total area from "a" to "b".
• Both continuous and some non-continuous functions can be integrable, as long as they have well-defined areas.
• The functions need to satisfy the condition that for any small number \(\varepsilon > 0\), we can find a partition P such that the difference between the upper and lower sums is less than \(\varepsilon\).

To put it simply, these functions behave well under integration, following the rules of calculus without producing infinite or undefined areas.

Riemann Sums

Riemann sums help us approximate the integral of a function. They are built from partitions of the main interval and are crucial for understanding integrable functions.

• Upper Riemann sum (\(U(f, P)\)) is the sum where each piece of the partition takes the maximum function value in the interval, creating a potential overestimate of the area.
• Lower Riemann sum (\(L(f, P)\)) makes use of the minimum function values, creating a potential underestimate.
• To have a function be Riemann integrable, these sums need to converge as the partitions get finer, such that the difference \(U(f, P) - L(f, P)\) can be made smaller than any specified tolerance \(\varepsilon\).

Riemann sums serve as the stepping stones towards achieving exact integration, starting with approximations.

Integration Theorems

Integration theorems provide rules that guide the integration process and are foundational for solving complex integrals.

Key Points of Theorem

• If \(f\) and \(g\) are integrable on \([a, b]\), then their sum \(f+g\) and product \(fg\) are also integrable. This is because when both functions behave nicely under integration, combining them does not disrupt this property.
• Moreover, multiplying an integrable function by a constant will also yield another integrable function, allowing us flexibility in scaling results.

By using these theorems, we can break down complex functions into simpler parts, integrate each, and then combine results to solve more involved problems.