Question

Continuity at the endpoints Assume \(f\) is continuous on \([a, b]\) and let \(A\) be the area function for \(f\) with left endpoint \(a\). Let \(m\) and \(M^{*}\) be the absolute minimum and maximum values of \(f\) on \([a, b],\) respectively. a. Prove that \(m^{*}(x-a) \leq A(x) \leq M^{*}(x-a),\) for all \(x\) in \([a, b] .\) Use this result and the Squeeze Theorem to show that \(A\) is continuous from the right at \(x=a\) b. Prove that \(m^{*}(b-x) \leq A(b)-A(x) \leq M^{*}(b-x),\) for all \(x\) in \([a, b] .\) Use this result to show that \(A\) is continuous from the left at \(x=b\)

Short answer

Question: Prove that the area function A(x) is continuous at the endpoints of the interval [a, b] for a continuous function f on the interval [a, b]. Answer: We showed that A(x) is continuous from the right at x=a and continuous from the left at x=b by using the inequalities m*(x-a) ≤ A(x) ≤ M*(x-a) for x in [a, b] and m*(b-x) ≤ A(b)-A(x) ≤ M*(b-x) for x in [a, b] along with the Squeeze Theorem. Therefore, the area function A(x) is continuous at the endpoints of the interval [a, b].

Step-by-step solution

a. Proving the inequality for \(A(x)\)

Given that \(f\) is continuous on \([a, b]\), we know that \(m \leq f(x) \leq M^{*}\) for all \(x\) in \([a, b]\). Multiply both sides of this inequality by \((x-a)\), noting that \((x-a)\geq0\) on \([a, b]\), and we get: \(m^{*}(x-a) \leq f(x)(x-a) \leq M^{*}(x-a)\) Now, since \(A(x)\) is the area function for \(f\) with left endpoint \(a\), by definition, we have the following integral: \(A(x) = \int_{a}^x f(t) dt\) By the properties of integrals, we have: \(m^{*}(x-a) \leq \int_{a}^x f(t) dt \leq M^{*}(x-a)\) So, we have proven the inequality: \(m^{*}(x-a) \leq A(x) \leq M^{*}(x-a)\), for all \(x\) in \([a, b]\) Now, we will use this result to prove the continuity of \(A\) from the right at \(x=a\).

a. Using Squeeze Theorem to show A is continuous from the right at \(x=a\)

We know that \(\lim_{x\to a} m^{*}(x-a) = m^{*}(a-a) = 0\). Also, since \(M^{*}\) is a constant, we have \(\lim_{x\to a} M^{*}(x-a) = M^{*}(a-a) = 0\). Since \(m^{*}(x-a) \leq A(x) \leq M^{*}(x-a)\), for all \(x\) in \([a, b]\) we can apply the Squeeze Theorem to \(A(x)\), so: \(\lim_{x\to a^{+}} A(x) = 0\) Therefore, \(A\) is continuous from the right at \(x=a\). Now, let's move to part b.

b. Proving the inequality for \(A(b)-A(x)\)

Just like part a, since \(f\) is continuous on \([a, b]\), we know that \(m \leq f(x) \leq M^{*}\) for all \(x\) in \([a, b]\). Multiply both sides of this inequality by \((b-x)\), noting that \((b-x)\geq0\) on \([a, b]\), and we get: \(m^{*}(b-x) \leq f(x)(b-x) \leq M^{*}(b-x)\) Now, since \(A(x)\) is the area function for \(f\) with left endpoint \(a\), by definition, we have the following integral: \(A(b) - A(x) = \int_{x}^b f(t) dt\) By the properties of integrals, we have: \(m^{*}(b-x) \leq \int_{x}^b f(t) dt \leq M^{*}(b-x)\) So, we have proven the inequality: \(m^{*}(b-x) \leq A(b)-A(x) \leq M^{*}(b-x)\), for all \(x\) in \([a, b]\) Now, we will use this result to prove the continuity of \(A\) from the left at \(x=b\).

b. Using Squeeze Theorem to show A is continuous from the left at \(x=b\)

We know that \(\lim_{x\to b} m^{*}(b-x) = m^{*}(b-b) = 0\). Also, since \(M^{*}\) is a constant, we have \(\lim_{x\to b} M^{*}(b-x) = M^{*}(b-b) = 0\). Since \(m^{*}(b-x) \leq A(b)-A(x) \leq M^{*}(b-x)\), for all \(x\) in \([a, b]\) we can apply the Squeeze Theorem to \((A(b)-A(x))\), so: \(\lim_{x\to b^{-}} (A(b) - A(x)) = 0\) This implies that \(\lim_{x\to b^{-}} A(x) = A(b)\), which means that \(A\) is continuous from the left at \(x=b\). To summarize, we have proven that the area function \(A\) is continuous from the right at \(x=a\) and continuous from the left at \(x=b\), proving the continuity of \(A\) at the endpoints of the interval \([a, b]\).

Key concepts

Area function

An area function, denoted as \(A(x)\), is a special type of function particularly used in calculus, where it represents the accumulation of areas under a curve of another function \(f(t)\). Think of it like tracking how much area has accumulated as you move from a starting point \(a\) to another point \(x\) along the curve of the function \(f(t)\).

The integral of \(f(t)\) from \(a\) to \(x\), \(\int_{a}^{x} f(t) \, dt\), gives us the value of the area function at \(x\). Important features of an area function include:

• The area function \(A(x)\) provides a continuous picture of the cumulative area beneath the graph of \(f(x)\) from \(a\) to \(x\).
• It’s fundamentally linked to the concept of definite integrals.
• In a broader sense, the area function demonstrates how integrals serve as an accumulation function, adding up tiny slices of area to form a total.

Understanding this idea is crucial when discussing continuity and applying the Squeeze Theorem in mathematics.

Squeeze Theorem

The Squeeze Theorem is an essential tool in calculus used to determine the limit of a function. The key idea is simple: if a function is squeezed between two other functions that converge to the same limit at a particular point, then the squeezed function must also converge to that limit.

Applying the Squeeze Theorem requires:

• Establishing two bounding functions \(g(x)\) and \(h(x)\) where \(g(x) \leq f(x) \leq h(x)\) for values of \(x\) near the point of interest.
• Proving that \(\lim_{x \to c} g(x) = L\) and \(\lim_{x \to c} h(x) = L\).
• Concluding that \(\lim_{x \to c} f(x) = L\).

This theorem is particularly valuable for proving the continuity of functions and ensures that functions behaving irregularly at certain points can be effectively analyzed. In our example, the Squeeze Theorem was used to validate the continuity of the area function at the endpoints of the interval.

Continuous functions

Continuous functions are fundamental in mathematics due to their unbroken nature. A function \(f(x)\) is termed continuous at a point \(c\) if the following are true:

• \(f(c)\) is defined, meaning \(c\) is within the domain of \(f\).
• The limit of \(f(x)\) as \(x\) approaches \(c\) exists, stated formally as \(\lim_{x \to c} f(x)\).
• The limit of \(f(x)\) as \(x\) approaches \(c\) equals the value of the function at \(c\), or \(\lim_{x \to c} f(x) = f(c)\).

These properties must be maintained across every point in the interval for a function to be continuous over that range.

Understanding the continuity of functions helps predict their behavior over intervals and allows utilization of important calculus tools, like the Squeeze Theorem, which rely on continuity. Whether a function is continuous across an entire range or purely at endpoints can affect computation and interpretation within calculus-related problems, such as those involving area functions and limits. Knowing these principles will help in properly establishing limits and finding solutions to calculus problems involving area computations.