Problem 39 - The Calculus with Analytic Geometry

Question

Suppose that the function \(f\) is continuous on \([a, b]\) and \(f^{\prime}(x)=1\) for all \(x\) in \((a, b)\). Prove that \(f(x)=x-a+f(a)\) for all \(x\) in \([a, b]\)

Step-by-Step Solution

Answer

Using the Fundamental Theorem of Calculus, integrate both sides; find \( f(x) = x - a + f(a) \).

Step 1 - Understand the Given

Given that the function is continuous on \( [a, b] \) and \( f^{\prime}(x) = 1 \) for all x in \( (a, b) \). We need to prove that \( f(x) = x - a + f(a) \).

Step 2 - Use the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( f^\backslashprime(x) = 1\) for all x in \( (a, b) \), then integrating both sides with respect to x over the interval \( [a, x] \) will give us \( \backslashint_a^x f^\backslashprime(t) dt = \backslashint_a^x 1 dt \).

Step 3 - Solve the Integral

Perform the integration on both sides. The left side gives us \( f(x) - f(a) \), and the right side is just the integral of a constant 1, which is \( x - a \). Therefore, \( f(x) - f(a) = x - a \).

Step 4 - Solve for f(x)

Rearrange the equation \( f(x) - f(a) = x - a \) to solve for \( f(x) \). This gives \( f(x) = x - a + f(a) \), as desired.

Key Concepts

Continuous Function

A continuous function is one that does not have any breaks, jumps, or holes in its graph. In simpler terms, you can draw its graph without lifting your pen from the paper. This property is crucial for understanding many concepts in calculus.

In the given exercise, the function \(f\) is continuous over the interval \([a, b]\). This means that for every point within this interval, the function smoothly transitions from one value to the next without any disruptions.

Why is continuity important? Continuity ensures that we can apply the Fundamental Theorem of Calculus, which relies on the idea that there are no sudden jumps in the function’s values.

Integration

Integration is a core concept in calculus that involves finding the accumulated area under a curve. It is the inverse process of differentiation. When you integrate a function, you are essentially summing up an infinite number of infinitesimally small areas to find the total area under the curve.

In our exercise, we used integration to prove a specific formula for the function \(f\). Given \(f^{\prime}(x) = 1\) for all \(x\) in \((a, b)\), we integrated both sides of this equation with respect to \(x\) over the interval \([a, x]\).

This is crucial because:

• On the left side, we integrated the derivative of \(f\), which gave us the original function back, minus a constant of integration, which in this context is \(f(a)\).
• On the right side, we integrated the constant 1, which simply gives us \(x - a\).

Therefore, the integration step helps us build the relationship between the values of the function.

Proof Technique

The proof technique used in this exercise is a direct application of the Fundamental Theorem of Calculus. Here are the critical steps involved:

• Understand the given conditions: Recognize that the function is continuous and that its derivative is a constant value across an interval.
• Apply the Fundamental Theorem of Calculus: This theorem links differentiation and integration, allowing us to solve the integral and find the relationship between the function and its derivative.
• Solve the integral: Perform the integration on both sides of the equation.
• Rearrange to prove the desired result: Finally, manipulate the resulting equation to solve for \(f(x)\).

By following these steps, we can rigorously justify that \(f(x) = x - a + f(a)\) based on the given conditions.
This proof technique highlights the interconnected nature of different calculus concepts and shows how we can use them together to solve problems.