Problem 67 - Calculus: Early Transcendental Functions

Question

Let \(f(x)=\left\\{\begin{array}{cl}x & \text { if } x<2 \\ x+1 & \text { if } x \geq 2\end{array} \text { and define } F(x)=\int_{0}^{x} f(t) d t\right.\) Show that \(F(x)\) is continuous but that it is not true that \(F^{\prime}(x)=f(x)\) for all \(x .\) Explain why this does not contradict the Fundamental Theorem of Calculus.

Step-by-Step Solution

Answer

The function \(F(x)\) is continuous for all x because the limit of \(F(x)\) exists and equals to \(F(2)\) from both sides of 2. However, \(F'(x) ≠ f(x)\) for all \(x\) because the function \(f(x)\) is discontinuous at \(x=2\), which doesn't satisfy the conditions of the Fundamental Theorem of Calculus, meaning there is no contradiction.

Step 1: Evaluating the Integral

Start by evaluating the indefinite integral of the function \( f(t) \). This involves two substeps: \n 1. When \(x < 2\), \( f(t) = t \) and the integral would be \( F(x) = \int_{0}^{x} t \, dt = \frac{1}{2}t^2 \Big|_0^x = \frac{1}{2}x^2 \) . \n 2. When \(x \geq 2\), \( f(t) = t+1 \) and the integral would be \( F(x) = \int_{0}^{x} (t+1) \, dt = \frac{1}{2}t^2 + t \Big|_0^x = \frac{1}{2}x^2 + x - 2 .

Step 2: Evaluating Continuity

We define a function to be continuous at a point if the limit of the function as it approaches that point exists and is equal to the value of the function at that point.\n Evaluate \( F(x) \) at \( x=2 \) from both sides of 2 :\n \(\lim_{{x \to 2^-}} F(x) = \lim_{{x \to 2^-}} \frac{1}{2}x^2 = 2\) and \(\lim_{{x \to 2^+}} F(x) = \lim_{{x \to 2^+}} (\frac{1}{2}x^2 + x - 2) = 2\) . Since these two values match and also equal to \(F(2) = \frac{1}{2} \cdot 2^2 = 2\), it can be concluded that \( F(x) \) is continuous for all \( x \).

Step 3: Evaluating the Derivative

The Fundamental Theorem of Calculus states that if a function \( F(x) \) is an antiderivative of \( f(x) \), then the derivative of \( F(x) \) should give \( f(x) \) back.\n Take the derivative of the function \( F(x) \) as follows: For \( x<2 \), \( F'(x) = x = f(x) \) is valid. However, for \( x \geq 2 \), \( F'(x) = x + 1 \neq f(x) \). The function \( F'(x) \) is discontinuous at \( x = 2 \) as its left-hand limit value does not match its right-hand limit value. Thus, the Fundamental Theorem of Calculus is not applicable at this point.

Step 4: Interpreting the Outcome

The fact that \( F'(x) \neq f(x) \) for all \( x \) does not contradict the Fundamental Theorem of Calculus because the function \( f(x) \) is not continuous at \( x = 2 \). The Fundamental Theorem of Calculus only applies to continuous functions on a given interval.

Key Concepts

Piecewise Functions

A piecewise function is a function that is defined by different expressions based on the input value. In simpler terms, it's like having different rules for different situations. For example, when you look at the function \( f(x) \) from the exercise:

• For \( x < 2 \), the rule is \( f(x) = x \).
• For \( x \geq 2 \), the rule changes to \( f(x) = x + 1 \).

Each segment or piece of the function operates under its own rule, depending on the value of \( x \). This kind of setup is useful because real-world scenarios often have different behaviors at different times or conditions.
The challenge with piecewise functions lies in analyzing points where the function might change its rule, like \( x = 2 \) in this exercise. It requires careful consideration to evaluate the continuity or find the derivative.

Continuity

Continuity in a function means that there are no sudden jumps or breaks at any point. For the function \( F(x) \), you need to check whether \( F(x) \) is smooth at \( x = 2 \), where the rule of the piecewise function changes.
To establish continuity at a given point, say \( x = 2 \), you must verify if:

• The left-hand limit \( \lim_{{x \to 2^-}} F(x) \) matches the right-hand limit \( \lim_{{x \to 2^+}} F(x) \).
• These limiting values are the same as the actual function value \( F(2) \).

For this exercise, both the left and right-hand limits at \( x = 2 \) computed to 2, and \( F(2) = 2 \), demonstrating that \( F(x) \) is continuous even though the derivative might not exist at this junction. This outcome helps us understand why some functions can be continuous even if their derivatives aren't continuous at specific points.

Derivatives

Derivatives provide the rate of change of a function. In the given exercise, we're interested in finding \( F'(x) \), the derivative of \( F(x) \), to check if it matches the original piecewise function \( f(x) \) according to the Fundamental Theorem of Calculus.
The theorem generally holds when the function \( f(x) \) is continuous over the interval used in the integral. However, for this exercise:

• For \( x < 2 \), you find \( F'(x) = x \), which does match \( f(x) \).
• But at \( x \geq 2 \), \( F'(x) = x + 1 \) does not match \( f(x) = x \).

The discrepancy arises because the function \( f(x) \) doesn't meet the continuity requirement at \( x = 2 \). Therefore, the Fundamental Theorem of Calculus cannot be applied right at the point where the piecewise function changes its rule, leading us to break down and analyze each segment separately.